MATHEMATICS-PAPER-II-Fourier-Analysis-munotes

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FOURIER SERIES
Unit Structure
1.1 Periodic function
1.2 Dirichlet’s conditions
1.3 Fourier Series of periodic continuous functions
1.4 Fourier Series of even and odd functions
1.5 Fourier series of periodic function s having arbitrary period
1.1 DEFI NITIO N:P E R I O D I CF U NCTIO N:
A real or complex value dfunction fis said to be periodic with
period0T,if, x and f x nt f x n.
Example : 1)sin 2 sinxn x
2)cos 2 cosxn x
hencesinxandcosxare periodic function with period2.
TheOrthog onality Relations of Trigonometric functions :
1)0, 0 , 1 , 2 , . . . .cos cos 1,2,...
20mn m n
mx nx dx m n
mn

  
 
 
2)0, 1 , 2 , . . .sin sin 1, 2,...
00mn m n
mx nx dx m n
mn

  
 
 
3)cos sin 0, , 0,1, 2,...mx nxdx m n
 
4)0
2imx inxmnee d xmn
 

Definition : Trigonometric Series : A series of the form
0
112 2cos sin cos 2 sin 2 ......2aax b x a x b x  ................ cos sin ...........nnan x b n x 
where,01 122,,,,, . . . . . . . . . ,,, . . . . . . . . . .nn aa b ab abare constants is called as
trigonometric series.munotes.in

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1.2 DIRICHLET’S CONDITIO NS:
Iffxisaperiodic function of period2defined in the interval2CxCwhere Cis any constant then following c ondition are
known to be Dirichlet ’sconditions
i)Funct ionfxand its integrals are finite and single valued in the
interval.
ii)Functionfxhas at most finite num ber of finite discontinuities
inthe interval.
iii)Functionfxhas at most finite number of maxima and minima
in the interval.
1.3FOURIER SERIES OF PERIODIC CO NTINUOUS
FUNCTIO NS:
Definition : Iffxis a periodic function of period2define din
the interval2Cx Cand satisfies the Dirichlet ’scondition s
then, functionfxcan be represented by the trigonometric series
as  0
1cos sin2nn
naan x b n x
   .This representation of a functionfxas a trig onometric series is known as Fourier series
expansion of functionfxanditsco-efficients0,,nnaabare called
Fourier coefficients .
Example :
1)tanfx xcannot be expanded as a Fourier se ries in the
interval0, 2since tan2.
2)axfx ewhere ais consta ntcan be expressed in terms of
Fourier series in any interval.
Note : The Fourier series expansion offxconverges to
12fx fx   ,i.e.Right hand limit +left hand limit2at the point
of discontinuity.munotes.in

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Calculation of Fourier coefficient s:
Letfxbe a periodic function of period2defined in the interval2CxCsatisfy ingDirichlet’s condition st h e n itsFourier
series expansion is given by
 0
1cos sin2nn
nafx a n x b n x
  (1)
1)To calculate Fourier coefficient0,aintegrate equation (1)fromCto2C.
22 2 2
0
1cos sin2CC C C
nn
n CC C Cafx d x d x a n x d x b n x d x        
         
2
0020 02C
Cafx d x a 
 
2
01C
Caf x d x


2)To determine the Fourier coefficientnamultiply equation (1) bycosnxand the integrate fromCto2C.
  2 0
1cos cos cos sin cos2nn
nafx n x n x a n xb n x n x
 
22 2 2
2 0
1cos cos cos sin cos2CC C C
nn
n CC C Cafx n x d x n x d x a n x d x b n x n x d x        
          
21cosC
n
Caf x n x d x
 
3)To determine the Fourier coefficientnbmultiply equation (1) bysinnxand integrate fromCto2C.
22 2 2
2 0
1sin sin cos sin sin2CC C C
nn
n CC C Cafx n x d x n x d x a n x n x d x b n x        
         2 22
0
1sin 2 cos 1 cos
22C CC
nn
n C CCnx a nx nxad x b d xnn      
              
21sinC
n
Cbf x n x d x
 munotes.in

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Thus we have complete set of formulation for Fourier series
expansion of periodic functionfxof period2satisfying
Dirichlet ’scondition sas
 0
1cos sin2nn
nafx a n x b n x
 
where


2
0
2
21
1cos
1sin 2C
C
C
n
C
C
n
Caf x d x
af x n x d x
b f x nx dx for C x C









Note:
(1) If0Cthen02xand


2
0
0
2
0
2
01
1cos
1sin 2n
naf x d x
af x n x d x
b f x nx dx for x







2)IfCthenxthen


01
1cos
1sinn
naf x d x
af x n x d x
bf x n x d x f o r x











1.4FOURIER SERIES EXPA NSIONOF EVE NANDO D D
FUNCTIO NS:
Definition :
The function fi ss a i dt ob ee v e n , if,fx f x x c x c.
The function fi ss a i dt ob eo d d , if,fx f x x c x c.munotes.in

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Example :cosis even function sincecos cos.sinis odd function s incesin sin.
Property : 
02i f i s e v e n0i f i s o d da
a
afx d x ffx d x
f 
 
Hence Fourier series e xpansion of even function defined in the
intervalxisgiven by


0
0
0
0
12
2cos
0
cos2n
n
n
naf x d x
af x n x d x
b
afx a n x x







This series is also called as Fourier C osine series .
Fourier Series expansion of odd function defined in the intervalxis given by
0
020, 0, sinnnaab f x n x d x
 

1sinn
nfx b n x x

This series is also known as Fourier S ineseries .
1.5FOURIER SERIES EXPA NSIONOF A PERIODIC
FUNCTIO NHAVI NG ARBITRARY PERIOD:
Letfxbe a periodic function of period2Ldefined in the interval2CxC Lthen substitutexzLorzLx
whenz( )CxC d s a yL
when 2z 2 2xC L C L dLThusfzis a periodic function of period2defined in the
interval2dzd.munotes.in

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Hence Fourier series expansion of a periodic functionfxof a
period 2L defined in the interval2Cx C Lis given by
0
1cos sin2nn
nanx nxfx a bLL 
    
where the Fourier coefficients are given by
2
01CL
Caf x d xL
21cosCL
n
Cnxaf x d xLL
21sinCL
n
Cnxbf x d xLL
Note :
IfCLthenLxL.In this case we can verify whether the
given periodic function is given either even or odd.
Hence Fourier series e xpansion of even function defined in the
intervalLxLis given by


0
0
0
0
12
2cos
0
cos2L
L
n
n
n
naf x d xL
nxaf x d xLL
b
a nxfx a L x LL


 

 


This series is also called as Fourier Cosine series .
Fourier series expansion of odd function defined in the intervalxis given by
0
020, 0, sinL
nnnxaab f x d xL
  

1sinn
nnxfx b L x LL
 
This series is also known as Fourier Sine series .munotes.in

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Example s
Ex.1.Find Fourier s eries expansion offx x xand
show that2222111.....135 8.
Solution :fx x x f xfis even function.
0
0
2
0 02
22
2af x d x
xxd x
 



0
02cos
2cosnaf x n x d xxn x d x



0 0
0
222 sin sin
21 c o s2112111n
nxn x n xdxnn
nx
nn
nn


  
   
   
           

02sin0nbf x n x d x
 

0
1
2
1cos2
211cos2n
n
n
nafx a n xxn xn




 

Note that 0112o d dnif n is evenif n is
hence r eplace n by 2n –1,w eh a v emunotes.in

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8

2
1
2
122cos 2 12 21
2cos 2 1 2
2 210n
nxn xn
nx
nPut x





   
 




2
1
2
1
22222 c o s 002 21
22
2 21
41 1 1.......21 3 5n
nn
n










   

222
2
22241 1 1.......135 2111......135 8
    

Ex 3.Find Fourier series expansion of2fx x x.
Evaluate series atxandfind211nn
Solution :2fx x2 2fx x x f xfxis eve n function
0332
0
22
22 23323o
oaf x d x
xxd x

  
 
  


2
22cos
2 sin sin2
220 sinno
o o
oax n x d x
nx nxxx d xnn
xn x d xn



         

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22 c o s c o s
o
onx nxxd xnn n

          




21 22 1 s i n
1 22 10
1 4
41
0n
o
n
n
n
nnx
nn n n
nn n
nn
n
b



                         


TheFourier Cosine series is given by
 
1cos sin2o
nn
nafx a n xb n x
 2
2
2
141cos3n
nxn xnat x

222
2
22
1141 41133nnn
nn nn  
     222
2
1
2
2
142331
6n
nn
n






Ex.4.Compute Fourier seri eso faxfx ewhere a is +ve and
hence prove that
2221sinh11 2 1n
naaan a

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Solution :Letaxfx e
1
1o
axaf x d xed x




 

12
2
2sinh
1cosax a a
nee e
aa
aa
af x n x d x 


 


    
 
Let1cosax
naI e n x d x
 
1cos sin
1cos cos sinax ax
aa
axeenx nx n dxaa
ee nnn n x e d xaa a

 

 

         
(by LI ATE )
111 s i n c o saa a x a x
nnee n eenxa n nxaa a a a 



         




2
2
2
2
2
2111 s i n c o s
1 1cos
1 1cos
1
1aa a x a x
nn
n
aa a x
n
aa a x
n
aa
nee n eenx n nxaa a a a
nnee e n x d xaa a
nee e n x d xaa
nIe e Iaa
nIIaa 


 

 

 

 








                       
 





22
21aa
n
aaee
anIe eaa 
 


   
2
naa221n
a an 


22 2212 1sinhaa
nn
aaee
a aee aan an 
  

  
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1sin
1sinn
ax
nbf x n x d xbe n x d x




 
(by LI ATE)
1sin cos .
10c o s .
1cosax ax
n
ax
axeebn x n x n d xaa
nen x n d xa
nen x d xa
n
a










     
   
 



a



22
11
22 221
12 1sinhn
aa
nn
aaee
an
nnee aan an 
  
 


  
 
Thus the F ourier series expansion offisgiven by
 






1
1
22 22
1
22
1
2
22
1cos sin2
2 1 sinh 2 1 sinh sinhcos sin0
21 sinh1s i n h
1 sinh11 2
sin1o
nn
n
nn
ax
n
n
n
n
nafx a n xb n x
aa n a aen x n xa an an
at x
a aaa an
a a
a an  
  
 








 
               



    
   




2
22
1h12 1n
naa
a an
Hence proved

 
  
   

Ex. 6. Show that  22220
11
2nn
nafx d x a b

  
where&nnabare Fourier coefficients o f Fourier series expansion of
periodic function f defined in,
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Solution: TheFourier series expansion of a periodic functionfxof period2define din the intervalxsatisfying Dirichlet’s
conditions is given by
 0
1cos sin2nn
nafx a n xb n x
 
On squaring both sides we get
2
22 2 2 2 0
11
00
11 1cos sin4
cos sin sinnn
nn
nn n n
nn nafx a n x b n xaa n x ab n x ab n x c o n x 
 
  
   
   
  
Assuming term by term integration on R.H.S. of above equation is
permissible.
Integrating both side of above equation with the limitto.
2222 0
1
22
0
11
0
11cos4
sin cos
sin 2 sin cosn
n
nn
nn
nn n
nnafx d x d x a n x d x
bn x d x a a n x d xab n x d x ab n x n xd x  
  
 
 
 
 
  
 
  
 
    
       
   
  
Using orthogo nality relations we get
 22220
1 2nnnafx d x a b

    
 22220
11
2nn
nafx d x a b

  
This relation is known as Parseval’s Identity .
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2
BASIC PROPERTIES OF FOURIER SERIES
Unit Structure
2.1 Complex form of Fourier series
2.2 Properties of Fourier Coefficient
2.3 Riemann Lebes gue Lemma
2.4 Good kernel s
2.1COMPLEX FORM OF FOURIER SERIES :
Letfxbe a periodic function of period2define din the interval2CxCthen its Fourier series expansion is given by
 0
1cos sin2nn
nafx a n x b n x


We have cos2iiee
sin2iieei
0
1 22 2inx inx inx inxnn
na ee eefx a bi  
        
0
1 22 2inx inxnn nn
na a ib a ibfx e e

      
Setting002aC;22nn nnnnai b ai bCC 
  0
1inx inxnn
nfx C C e Ce


 
inxn
nfx C e
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This is Complex form of Fourier series where,nCisFourier
coefficient which is given by,2nnnai bC.
Using value of Fourier coefficient&nnabwe can simplify fornCas.
 2211cos sin2CC
n
CCiCf x n x d x f x n x d x 
       
 21cos sin2C
n
CCf x n x i n x d x
    
21
2C
inx
n
CCf x e d x

 
This is general formula for Fourier coefficient in the complex
form .
Note :
1)The Fourier series coefficientsnCin complex form is also
denoted byˆfn.
i.e.  21ˆ
2C
inx
n
Cfn C fx e d x

 
2)Iffxis a periodic function of period2define din the
intervalxthen  1ˆ
2inx
nfn C fx e d x

 .
3)We have2nnnai bCand2nnnai bC
2nn nnn nnn no
oCC aCC i b
bi C C
aC




4)Similarly, we can find the Fourier series expansion of aperiodic
functionfxof arbitrar yperiod 2L define din the interval2CxC Lin complex form of as
in xLn
nfx C e
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where,
 21ˆ
2CL in x
L
n
Cfn C fx e d xL  
Ex.1.Find complex form of Fourier series ofgSolution :We have Fourier coeffi cient in Complex Fourier series
expansion as

221
2
1
2
1
2
11
2
11 1cos2in
n
in
in in
in in
in
in in
in inCg e d
ed
eedin in
eeein in in
eeee n iin in n n



 

 

 


 

 


 

 


  

 
     
      
 



1
1sin cos sin
1 112
1 122
1n
n
n
nnn i n
in in
inin

           
  
   


11in
n
n
ninngC eein





0At nTo find the value0Cconsider Fourier coefficient in comple xform
12in
nCg e d

 
Put n012Cg d
…….. { since gi so d df u n c t i o n } .munotes.in

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16
Thus complex form of Fourier series of a given function is given by

11n in
n
noein
ga t n



We have

1100ninn
noegin
ga t n



Since 11nnas n vari es from -veto+ve integer.
Hence we can combine nthterm & ( -nth)t e r ma s .
1
1 11n in in in
n e eein in in  
      

1
1
11
1
1cosn
in in
n
in in
nie ien
ie en
inn 
 




   
 sin cosin n

1
1
1
1sin
12 sin
21sin
21sin nn
n
n
nin
iinn
nn
gn

 


  


Note : The functiongis odd function. Hence we
can expand this function in terms of Fourier S ine series.
Ex.2.Show that 2 21
2n
nCf x d x


wherenCis complex Fourier coefficient of Fourier series expansion of
periodic function f defined in,
(This relation is known as Bessel’s Inequality .)munotes.in

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Solution :The complex form of Fourier series expansion of periodic
functionfxis given by inxn
nfx C e
.
We have property of the complex number2,zz z zConsider,
  2NN Ninx inx inxnnn
NN Nfx C e fx C e fx Ce
            
  NNim nxinx inx
nnnm
Nm n Nfxfx C fx e Cfx e C Ce 
          
Divide both side of above equation by2and integrate within limittoalso using
 11&22inx inxnnfx e d x C fx e d x C 
   
   and
0 1
1 2im nxmned xmn


We obtains
 
 

2
2
2222 211
22
122
1
2N
inx
n
N
NN
nn nnn n
NN
NNnnNN
N
n
Nfx C e d x fx d x
CC CC CCfx d x C Cfx d x C 
 



 

 
 
 
 
 
 
  
 
 

2 2
2 21
2
1
2N
n
N
n
nCf x d xLetting N we getCf x d x





 



wherenCis complex Fourier coefficient.
This relation is known as Bessel’s Inequality .munotes.in

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Note :
1)22 20
111
42nn
naa b
 
2 21
2n
nCf x d x

.
2)From above Bessel’s Inequality the series222,nnnab Care convergent.
2.2PROPERTIES OF FOURIER COEFFICIE NT
Thefollowing statements are equivalent
1)2Perio dicfunction on Rlike exponential function.
2)Function defined on the interval of length2.
3)Function defined on the unit circle.
Since a point on the unit circle takes the formie,is real and
unique upto integer multiple of2.IfFis a function on the circle
then we may define for each real number()ifF e
Observe that2( )fffora l l.
Thus fis periodic of period2.The integrability, continuity and
other smoothness properties of Fare determined by those of f.
Definition : The Fourier coefficient of an integrable period ic
functionfare the complex numberˆfndefine dby the integral.
 1,2inxfn fx e d x n z

 
The1Lnorm of anintegrable periodic functionfis given by
112ff x d x
 .
The2Lnorm of square integrable periodic functionfis given by
122
21
2ff x d x
    .
Properties of Fourier Coefficient :
Theor em1:Suppose thatfis an integrable periodic function then
1ˆ ,fn f n Z.munotes.in

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Proof:
We have,
 1ˆ2inxfn fx e d x

 
Taking mod on both sides
 
  
1ˆ
2
1ˆ
2
12inx
inx
inxfn fx e d xfn fx e d x f ffxe d x










 
 
  
since
 
22
1
1cos sin cos sin 11ˆ
2
ˆinxen x i n x n x n x
fn fxd x f
fn f n Z

 

Theorem 2:Translation Property :S u p p o s et h a tfis an
integrable periodic function. Given a in R.Letaftranslate functionfasafx f xathenˆ ˆ.ina
afn e f n nZProof : We have,
 1ˆ2inxfn fx e d x


 
    1ˆ
2
1ˆ
2inx
aa
inx
aafn fx e d xfn f xa e d x fx f xa






 
 

Putxa y xaydx dywhen,xy awhen,xy a 
1ˆ
2
2ain a y
aa
inaainy
afn fy e d yefy e d y






  
munotes.in

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Sincefis periodic function of period2.
 
1ˆ
2
ˆ ˆina iny
a
ina
aafn e f y e d yfn e fn
 

    
Theorem 3:Suppose thatfis continuous function with continuous
derivative'fthen'ˆ ˆ()fn i n fn n.
Proof : We have,
 1ˆ2inxfn fx e d x

 
On integrating by part s
' 1ˆ
2inx inxfx e efn f x d xin in


 

          
Sincefis periodic function of period2,w eh a v e2 ff f
The 1stterm inabove equation vanish es
 

  
'
'
'' '
'11ˆ
2
11
2
11ˆ ˆ}2
ˆ ˆinx
inx
inxfn f x e d xin
fx e d xinfn fn fx e d xin
fn i n f n n











    
 
 



Notation :
21ˆfn Onasnmeans L.H.S. is bounded by constant
multiple of R.H.S. i.e. there exist constant C>0such that
2ˆCfn
nlargen.
In general,fx O g xasxameans for some +ve constant C,fx C g xasxa.
Note :1fx O meansfis bounded function.munotes.in

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Theorem 4:Suppose that functionfis twicecontinu ously
differentiable function defined on the circle then 21ˆfn Onas
n.So that Fourier series offconverges absolutely &
uniformly tof.
Proof : We have
 1ˆ2inxfn fx e d x

 
Integrating R.H.S. by part
 
 '
'1ˆ
2
ˆ2inx inx
inx inxeefn fx f x d xin in
eefn fx f x d xin in







 


 

                  

Sincefis periodic function with period2.1stterm of R.H.S.
Vanishes
 ' 1ˆ2inxfn e f x d xin

  
Once again integrating by parts,
 
 '' '
'' 'ˆ2
ˆ2inx inx
inx inxeein f n f x f x dxin in
eefn f x f x d xin in







 


 

       
        

Since'fis periodic and  cos sin cos sin 0inx in inee e n i n n i n
  

2' '
2' 'ˆ2
ˆ2inx
inxnfn f x e d xnfn f x e d x






  

2' 'ˆ21inx inxnf n fx e d x e
 
 
2' 'ˆ2.nf n fx d xC

 .
where Ci sac o n s tantandindependent of n. and s incefis twice
continuously differentiable,''fis bounded function .
Setting2CBwhere, B is bound of''fmunotes.in

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
2
2ˆ22ˆnf n BBfn
n 

21ˆfn onasn2.3THE RIEMA NN-LEBESGUE LEMMA :
Statement : Iffis integrable function defined on a circle thenˆ0fnasn.
OR
Iffis integrable periodic function of period2thenˆlim 0nfn
.
Proof : Since for any0, we can choose a continuous periodic
function g with fg.
Since ˆ ,fn f n Zˆˆfn g n f g(1)
i.e.theFourier coefficient of fu nctionfand g differ by less than.
So thatˆfnare eventually less thanin modulus if0gnas
n.
If gis continuous periodic function and athen we haveagx g x aˆ ˆina
agn e g n nZ(2)
Choose an ˆ ˆinn
agn e g nˆ ˆ1agn g n (3)
Weh a v e ,
 11ˆ2gn g gx d x
 (4)munotes.in

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Now consider,ˆˆˆ 2gn gn gnˆˆagn g n (byequation (3))
12agx g x d x
 (byequation (4))
12gx gx a d x
 
Putan 1ˆ22gn gx g x d xn

 
As0nanhence, 0gx gxnˆ20gn asnˆ0gnasnBy (1), ˆ0fnasnˆlim 0nfn
 hence proof.
2.4GOOD KER NELS :
Definition : A family of Kernels 1nnKxdefined on the circle is
said to be fa mily of good Kernel if it sa tisfies the following property
1)for all 11, 12nnK X d x
 
2)There exist M > 0 Such that for1nnKx d xM

3)for every0,n
xKx d x oasnConvolution : Letfandgbe2periodic integrable functions then
the convolution of functionfand g on interval,is denoted
anddefined asmunotes.in

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 12fgx f y g xy d y
 
OR
 12fgx f xy g y d y
 
Note :fg gfTheorem : Let1nnKbe a family of Good Kernels andfis an
integrable periodic function defined on the circle thenlimnnfK x f xwhenever,fis continuous atx.
Iffis continuous everywhere then the above limit is uniform.
Proof : If0andfis continuous atxthen we can choose,So
that y.fx y fx(1)
Consider,
 1*2nnfK x f x Ky f xy d y f x
  
(Definition of convolution )
AsnKis a good Kernel 112nKy d y
 
  

 
11*22
1
2
1*2
1
2nn n
n
nn
nfK x f x Ky f xy d y f x Ky d yKy f xy f xd y
fK x f x Ky f xy f x d y
Kyf xy f x d y 
 





 


 


       
     
 
 




1
2
1
2n
y
n
yKyf xy f x d y
Kyf xy f x d y


 
 
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Note that
&yy
yy y  2
22nn n
yyBfK x f x Ky d y Ky d y       (2)
Clearly, 1stterm is bounded by2M(by 2ndproperty of good
Kernel) and b y3rdproperty of Good Kernel for large value of n,2nd
term will be less than.
Hence for some constant C we have,nfK x f x CnfK x f x asn.
Iffis continuous everywhere then is it uniform lycontinuous.
Hence,can be chosen independent of x which proves desired
conclusion.
nfK f
i.e.limnnfK x f x

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3
DIRICHLET KER NEL
Unit Structure
3.1 Dirich let’s Kernel
3.2 Properties of Dirich let’s Kernel
3.3 Diric hlet Theorem on point wise convergence of Fourier
series
3.1DIRICHLET’S KER NEL :
We have complex form of a Fourier series expansion of a periodic
functionfof a period2defined on,.
ˆinnff n e
  (1)
TheNthpartial sum of Fourier series expansion of a series (1) is
denoted and defined as,
ˆNinN
nNSf fn e
  (2)
We have Fourier series coefficient.
 1ˆ2infn f e d

 (3)
Using equation (3) in equation (2) we have,
 
 
 ()1
2
1
2
1
2N
in in
N
nN
Nin inN
N
in
NNSf f e d efe defe d










    


 1
2N
in
N
NSf f e d

   
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Put,ddWhen , When ,  1
2N
in
N
NSf f e d
 

 
Sincefis periodic function of period2 1
2N
in
N
NSf f e d
 
  
 12NNSf f D d
 
 (4)
where NinN
NDe(5)
and it is known as NthDirichlet Kernel .
Equation (4) represents Nthpartial sum of Fourier series in terms of
Dirichlet Kernel.
3.2PROPE RTIES OF DIRICHLET’S KER NEL :
Theorem 1:The NthDirichlets’s kernel is given by
1sin2
1sin2N
in
N
NN
De


Proof : We have
NinN
NDe 
(1 ) (2 ) 0 2
2( ) ( 2 ) 2
2
0
2
0.... .....
1. . . . . . . . . . . .iN i N i N i i iN
N
iN i i iN nH i i N iN
N
iN in
n
NniN i
nDe ee e e e e
ee e e e e e
ee
ee          

 
 




The above series is a geometric series with first term a=1 and
common ratio,r 1ire.
we have
1
011K K
n
nrrr
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
21
111
1Ni
iN
N i
iN iN
N ie
Dee
eeDe



        Multiply Numerator as well as Denominator by/2ie
(1 ) / 2/211
22
/2 /21iN i N i
N ii
iN iN
iiee eDee
eeee 
 
    
      

11
22
/2 /22
2
1sin2... sin1 2sin2iN iN
N ii
ii
Nee
iDee
i
NeeDi    
 


      
Theorem 2:Suppose th atfis periodic and integrable then nth
partial sum of Fourier series expansion offis given by
  1122NN NSf D x y f y d y D y f x y d y
   
i.e.NN NSf x D f x f D x
Proof : TheNthpartial sum of Fourie r series is given by
ˆNinxN
nNSf x f n e (1)
whereˆfnis aFourier coefficient given by
 12inyfn fy e d y

 (2)
Put (2) in (1) we get
 1
2Niny inxN
NSf x f y ed y e

    
 1
2N
in x y
N
NSf x f y e d y

 (3)munotes.in

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Putxyzdy dzWhen,yz x When,yz x   1
2
1
2Nxinz
NxN
Nxinz
xNSf x f x z e d zfx z e d z






  
 

Sincefis periodic function of period2defined on the interval,
 1
2N
inz
N
NSfx f x z e d z

  
 1
2N
iny
N
NSf x f x y e d y
  (4)
Put (3) and(4) we get,
 11
22NN
in x y iny
N
NNSf x f y e d y f x y e d y 
   
       
Since Nin xN
NDx e
 11
22
1122NN
in x y in y
N
NN N
NNSfx fy e d y fx y e d yfy D x y d y fx y D y d y 
 
 
  
  
          
 By definition of convolution,NN NSf x f D x D f x
Theorem 3: 112NDd
where,NDNthDirich let Kernel.
Proof : We have NthDirichilet Kernel

  cos sinN
in
N
nN
N
N
nNDeDn i n 


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  cos 0 sin 0 cos si nNDi i  cos si n i  cos 2 sin 2 i
  
  cos 2 sin i    2. . . . . . . . .
cos sinNi N
  
 cos si n Ni   

112 c o s 2 c o s 2 . . . . .2 c o s
12 c o sN
N
N
nN
DN
Dn
  
 
  

On Integrating both side fromto
  112 c o s22 0 . . . . . c o s 0N
N
n
NDd d n d
Dd n d  
  
 
  
   
     
 
112NDd
 
Theorem 4: logNDx d xc N
 asNwhere, C is any
constant andNDxis NthDirichlet Kerne l
Proof : Step (1)
We have 112ff x d x
 
Similarly  112NNDx Dx d x
 
SinceNDxis even ,
 122NNoDx Dx d x  
We have, 1sin21sin2NNxDx
x
11sin1 2
1sin2NoNxDx d xx
 
Put2xy2dx dyWhen0, 0xymunotes.in

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When ,2xy

2
11sin 21 22sinNoNyDx d yy
  2sin 2 1 2
sinoNydyy
 (1)
Sin y can be approximated as y
i.e.sinyy2
1sin 2 1 21NoNyDx d y oy
   
Step (2) :
Put21Ny t21Nd y d tWhen0, 0ytWhen21,2 2Nyt 

21
2
1
0
21
2
1
0
112 2
1
01
2sin
2(1)21 21
2s i n(1)
2 sin(1)N
N
N
N
K
N
N
KKt
dt tDx oNN
tDx d t ot
tDx d t ot








  
  
 


Step (3) :
Put2KtSdt dsWhen1,02tK S When 11,22tK Smunotes.in

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2 2
1
00sin2 2(1)2N
N
KKSDx d s oKS
  
(2)
We have,sin sin sinsin2cos sin cos2si f K i s e v e n S n SKS ns if K is odd S

      
sin
cosKs if K is evenuSsi fK i s o d d
2 2
1
002(1)2N
K
N
KuSDx d s oKS
  
(3)
The value2KScan be approximated to2K.
Since11022KKS 

2222022 220
24KKSS
KK KKSS
S
KS K
  

The maximum value of224
2SK
K     is2 2
2SK K

22
2042SK SK K

Also211KK
is convergent and Hence bounded.munotes.in

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2 2
1
002(1)2N
K
N
KuSDx d s oKS
   
This equation can be written as
 2 2
1
0 021(1)2N
NK
KDx uS d s oK
  (4)
3)Step (4) :
Consider, 22sin 1KoouS d s s d s if K is even and
22cos 1KoouS d s s d s if K is odd use this value in (4).

2
1
0
2
2 1
021(1) (1)2
41(1)N
N
K
N
N
KDx O
K
Dx OK

  

Now we have,2
01logN
KNK
2 14log (1)NDx N O By using definition of1Lnom
214log (1)2NDx d x N O
   

28log (1)logN
NDx d x N ODx d x C N



 
 

Theorem 5:Dirichlet Kernel is not good Kernel.
Proof: By above property of Dirichlet Kernel, the 2ndproperty of
good Kernel fails and hence Dirichlet Kernel is not good Kern el.
3.3DIRICHLET’S THEOREM :
Statement : TheFourier series of real continuous periodic functionfwhich has only finite number of relative maxima and minima
converges everywhere tof(and hence conver ges uniformly)munotes.in

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OR
Suppose thatfis an integrable periodic function that is
differentiable at0xxthen00limNNSfx fx.
Proof : We have Nthpartial sum of integrable periodic functionfas
 12NNSfx fx yD y d y
 
at0xx 0012NNSfx fx yD y d y
 
Consider,
 00 0 012NNSfx fx fx yD y d y fx
  .
By property of Dirichlet Kernel,
  00 0 011()22NN NSfx fx fx yD y d y fxD y d y 
          
112NDy d y
    
 
  
  00
00
0011
22
1
2
12NN
NN
Nfx y D y d y D yfxd yfx y D y D yfx d y
fx y fx D y d y 
 



 
 

  
 
      


as again by property of Dirichlet Kernel .
 
00 0 01sin1 2
1 2sin2
11sin22NNySfx fx fx yfx d yy
Ny g y d y




                

where, 00sin2fx y fxgyymunotes.in

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i.e. 00 22sin2yfx y fxgyy y clearly,1sin2Ny   isbounded near zero and hence i ntegrabl eo n,.Also 2ndfactorgyis bounded and hence integrable on, {sincefisdiff at0x&
0sin2lim 12yy
y }
Hence it f ollows that000NSfx fxasN00lim ( )NNSfx fx 
Ex:Iffis2periodic and piecewise smooth onthen show that
1lim2NNSf f f   
   and hence show thatlimNNSf ffor everywherefis continuous.
Solution :We have,
Step (1) :
 ...... 2o
NNDd Dd
      
01122NDd
 022Nff
Dd
   (1)
Also N
oDd22N
off
Dd   (2)
Step (2) :
We have Nthpartial sum of Fourier series
1
2NNSf f D d
  
 0
011
22NNfD d fD d
     munotes.in

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Consider,

 0
01
2
11
22N
NNSf f ffD d fD d
 
   
       0
01
2NNffDd Dd
 
  
       
 1122o
NNff D d ff D d     
           
(4)
Step 3 :
We have,
11iN iN
N ieeDeConsider,
1 iN iNge e d
 
    
where 01
1i
iff
eg
ff
e 

 



 
  
g is well define d function defined on,andalso g is smooth
except at0Also,0ff at0.
Hence,gis in00form at0.By applying1LHospital rule,
00
'
0lim lim1
limi
iff
ge
f fie i  

 

 
Similarly,
00
' 1
0lim lim1
limi
iff
ge
f foie i  

 

 
Thus R.H.S. & L.H.S. limit exist.
Hence g is piecewise continuous on,.munotes.in

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Step (4) : Using equation (4) we have,

 111
2
11
22N
oiN iN iN iN
oSf f f
ge e d ge e
 
  
   
              
1 12iN iNge e d

    (5)
We have, Fourier coefficient  1ˆ2infn g e d

 
ByRiemann Lebesque lemma ,ˆfn oasn.
Consider,
 

1
11ˆ ˆ12
1
2
1lim2iN iN
iN iN
NNge e d f N f Nge e d o a s N
Sf f f




  

 

 

   
     

whenever iffis continuous atthenlimNNSf=f

munotes.in

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4
FEJER KER NEL
Unit Structure
4.1 Cesaro mean and Cesaro summation
4.2 Fejer’s kernel
4.3 Properties of Fejer’s kernel
4.4 Fejer’s theorem
4.5 Uniqueness theorem
4.6 Weirstrass approximation Theorems
4.1CESARO MEA NAND CESARO SUMMATIO N:
Let012
0...... .....KKCCC C
be a series of complex numbers.
Define nthpartial sum by
0nnkKSC.
This series converges to S if limNNSS.
The average of 1stN partial sum is denoted and defined by012 1.....N
NSSS SN
i.e.1
01NNnnSN
is called NthCesaro Mean of the series
0kKC
.
IfNconverges toasNthen we say thatnNC
isCesaro
summable to.
Example : Consider 
0111111. . . . . 1KK

Partial sum of the sequence11111 . . . . . . . .is1, 0,1, 0, .......which has no limit since partial sum fluctuate between 0 and 1.
So average value10 122N.
Therefore, above series is Cesaro summable to12.munotes.in

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4.2FEJER’S KER NEL
TheNthCesaro mean of Fourier series is given by
01 1 ....N
NSf x Sf x S f xfxN 
Weh a v e ,Nthpartial sum of Fourier series given byNNSf f D.
   
01 1
01 1......
........N
N
N
N
NNfDx fD x fD xfxN
fD xD x D x
fxN
fx f F x


      


 
where 01 1 .........N
NDx Dx D xFxN 
i.e.  1
01N
Nn
nFx D xN
is called theNthFejer’s kernel .
4.3PROPERTIES OF FEJER’S KER NEL
Theorem 1:The Nthis Fejer’s kernel is given by
2
2sin12sin2NNxFxxNProof : We have,
1
01NNnnFx D xN
1
01sin1 2
1sin2N
nnxN x
      
11
22 1
0
1 22
0
11
22
001
1 2sin2
1.
1 2sin2
1
12s i n2in x in x
N
n
ix ix
inx inx N
N
n
ix ixNNinx inxnnee
iNx
ee e eFxiNx
eee e
iN x     


 

  

                         

munotes.in

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Both of above series are in geometric progressive
for 1stseries, Common ratioixre,
for 2ndseries, Common ratioixre
Note that,1rUsing1
011K K
n
nrrr

 22111
1 112s i n2NNix ixix ix
N ix ixee
Fx e eeeiN x
               22
22 22
22 2211 1
2 sin 112
11 1
2 sin2
11 1
2 sin2iNx iNx
ix ix
ix ix
iNx iNx
ix ix ix ix
iNx iNx
ix ix ix ixee
xiN ee e e
ee
xiN ee e e
ee
xiN ee ee



 

                             
    
     
    

 2
2 2122 sin sin22
122 sin2iNx iNx
iNx iNxee
xxiN
eex iN
      
222
2 21
2 sin2iNx iNx
NFx e ex iN    222
21
2 sin2iNx iNx
ee
x i N     2
2sin12sin2NNxFxxNmunotes.in

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Theorem 2:The NthCesaro sum of Fourier series of continuous
periodic functionfis given by
  1122NN Nfx F x yfy d y F yfx y d y 
      
where,NFis NthFejer’s kernel.
i.e.NN Nfx F f x f F x
Proof : We have Nthpartial sum of Fourier series is given by
  1122NN NSfx D x yfy d y D yfx y d y 
       
where,NDis NthDirichlet Kernel.
Taking summation on both side.
 
11
00
1
01
2
1
2NN
nn
nn
N
n
nSf x D x yf yd yDy f xy d y



 
 

   

  11 1
00 011
22NN N
nn n
nn nSf x D x y f yd y D y f x yd y 
     
         
We have NthCesaro sum of Fourier seriesf1
01NNnnfSN
and also
we have, Fejer’s Kernel  1
01N
Nn
nFx D xN
.
 
1
2
1
2NN
NNf x N F x y f y d yNF y f x y d y




   

  1122NN Nfx F x yfy d y F yfx y d y 
       
ThusNN Nfx F fx f F x
Theorem 3: 112NFx d x
whereNFxisNthis Fejer’s kernel
Proof : NthFejer’s Kernel is given by,  1
01N
Nn
nFx D xN
.
Now integrating using limittomunotes.in

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 

 1
0
1
0
1
01
1
1122 2N
Nn
n
N
n
n
N
N
nFx d x D x d xN
Dx d xN
Fx d x NNN 
 




 



 
 
    

 
12NFx d x
 
Hence proved,
Theorem 4:lim 0NN
xFx d x

if0.
Proof : We have, NthFejer Kernel
2
2sin2sin2NNxFxxN

The maximum value of2sin2xis one.
Also,2sin2xincreases as x goes away from the origin in,.
Hence, 
21
sin2NFx
N
where x0N
xFx d x
 asN.
Theorem 5:Fejer KernelNFxis good kernel
Proof : Since we have
1)0NFx x2) 112NFx d x

3) Msuch that NFx d xM

4)for every0, 0N
xFx d x

 asNThus Fejer ’sKernel is good kernel .munotes.in

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4.4FEJER’S THEOREM :
Theorem: Iffis integrable on the circle then Fourier Series offisCesaro summable tofat every point of continuity off.
Moreover, if is continuous on the circle then Fourier series offis
uniformly Cesaro summable tofProof :
Step (1) :Iffis integrable function defined on the circle then it can
be approximated as a Fourier seriesinxn
nf(x) a e

The NthCesar omean of Fourier Series is given by
N1
Nn
n01f(x) S f(x)N

Where,NSf ( x )isthNPartial sum of Fourier series .thNCesaro mean of Fourier series offcan be written as
convolution
NNf(x) ( f*F )(x)where,NFisthNFejer kernel
Step(2) :
We have property of good kernel i.e. letnn1Kbe a family of good
kernel andfis integrable function defined on the circle then
nnlim ( f * )( x ) f ( x )
Whenever,fis continuous atx.
Moreover, iffis continuous everywhere then above limit is
uniform.
Step(3): We know thatthNFejer kern elNFis good kernelBy propert ym e n t i o ni n step ( 2) we can write
NN
NNlim ( f * F )( x ) f ( x )lim f ( x ) f ( x )

Hence, Fourier series of an integrable function defined on the circle
isCesaro summable tofat every point of continuity Also, by
step(2), iffis continuous on the circle then the Fourier series offis uniformly Cesaro summable tof.munotes.in

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Fejer’s Theorem: Alter native Form
Alternatively the statement of Fejers theorem may be written as
Statement : Iffis continuous and periodic then averagesNfof
partial sum of Fourier series offconve rges uniformly tofas
N.
i.e.lim ( ) ( )NNfx fx

Proof : Claim :NffasNi.e.lim ( ) ( )NNfx fx

We have NthCesaro mean of Fourier se ries offis given by,
 12NNfx F yfx y d y
 
Consider,
 
 
1
22
11
22
12NN N
NN
Nfxfx fx F yfx y d y F y d yFy f xy d y Fy f x d yFyf xy f xd y 
 
 
 

 
  
 
         
 

 1
2NN
yfx fx F y fx y fx d y   
1
2N
yFyf xy f xd y   (1)
For any choice ofsuch t hato.By the properties of Fejer
Kernel, the 1stintegral,
1
2N
yFy f xy f xd y  has modulus bounded by12sup
/ fx y fx y (2)
A continuous periodic function is uniformly con tinuous so giveno, we fixso small so that the bound of equation (2) is2N.
The modulus of 2ndintegral 1
2N
yFy f xy f xd y  is
bounded by 12s u p2N
yfy F y d y  (3)munotes.in

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For large N, the bound of equation (3) is2.
Sincelim 0NN
yFy d y


Now using equation (1), (2) and(3),
22Nfx fxasNlimNNfx fx

Alternative Pr oof of Fejer’s Theorem
Step 1: We have theorem
Let1nnKbe a family of Good Kernels andfis an integrable
periodic function defined on the circle thenlimnnfK x f xwhenever,fis continuous atx.
Iffis continuous everywhere then the above limit is uniform.
Step 2: We know that Fejer Kernel is a good kernel and hence by
above theorem, we havelimnnfF x f xwhenever,fis continuous atx.
Iffis continuous everywhere then the above limit is uniform.
Step 3: We also know that,NN Nfx F fx f F x
Hence by abov es t e p2 ,w eh a v elimNnfx fx
whenever,fis continuous atx.
Iffis continuous everywhere then the above limit is uniform.
4.5UNIQUE NESS OF FOURIER SERIES
Theorem : Iffis integrable periodic function defined on the circle
andˆ0fn nthen0fat all points of continuity of a
functionf.munotes.in

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Proof : Wehave Nthpartial sum of Fourier series offˆNinxN
nNSfx fn e 
Sinceˆ0fn n0NSfx n (1)
i.e. all partial sum of Fourier series of functionfare zero
Also, we have NthCesaro mean of Fourier series of functionf.
 1
01N
Nn
nfx S fxN

By equation (1)0Nfx n(2)
i.e. NthCesaro mean of Fourier series offare zero we have,
property of Fejer Kernel.*NNfx f F x
By equation (2) *000N
NfFx
fx F
Uniqueness ofFourier Series :
Since Fourier series of a continuous periodic functionfconverges
tof, the functionfis uniquely determined by its Fourier
coefficient s.
Iffandgaretwo functions having same Fourier coefficient sthen
functionsfand g a re necessarily equal i.e. iff(n) g(n)thenfg 0fg 0{By above then i.e. iff(n) 0 f 0}fg4.6THE WEIERSTRASS APPROXIMATIO N
THEOREM :
Statem ent :
Any continuous periodic functionfcan be approximated by
trigonometric polynomial.
ORmunotes.in

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Iffis continuous function defined on the inter val,withf( ) f( )and0then there exist trigonometric polynomial P
such that f(x) p (x) , xProof:
By Fejer’s Theorem, iffis continuous and periodic then averages
Nfof partial sum of Fourier series of functionfconverges
uniformly tof.
i.e.Nf(x) f(x)for 0xHere,Nf(x)itselfproves existence of trigonometric polynomialP( x ).




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5
POISSO NKER NEL
Unit Structure
5.1 Abel mean and Abel summation
5.2 Poisson Kernel
5.3 Properties of Poisson Kernel
5.4 Abel sum mability of Fourier series
5.1 ABEL MEA NAND SUMMATIO N:
Definition : As e r i e so fc omplex numberkk0C
is said to be Abel
Summable to S if for every0r1the seriesA( r )kk
k0Cr
is
convergent and if
r1lim A( r ) S.T h eq u a n t i t yA( r )iscalled Abel
mean of the series .
Example :consider the Series
1-2+3-4+5----------k
k0(1 ) ( k 1 )
kkk0
2A( x ) ( 1 ) ( k 1 ) r1
(1 r)

r11lim A( r )4Hence Series 1 -2+3-4+5-6+........ is Abel summable to14.
5.2 POISSO NKER NEL
ThePoisson kernel is denoted and defined as ninr
nPr e

Definition :Let us define Abel Mean of the Fourier seriesinn
nf( ) a e
munotes.in

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where,nais Complex Four ier coefficient, is given by
n in
rn
nAf ( ) r ae

Since ntakes positive andnegative integer value, we considernhere. Herefis integrable andnaac o m p l e x Fourier c oefficient
which is uniformly bounded. Hence SeriesrAf ( )converges
absolutely and uniformly for eachr,0r1.
Theorem: The Abel Mean can be written as convolution of periodic
integrab le function f and the Poisson kernelrPasrrAf f PProof : We have,
n in
rn
nAf ( ) r a e 
where, complex Fourier coefficient
 in
n1af ( n ) f ( ) ed2


nin inr
n1Af ( ) r f ( ) e d e2       
n in
n
n in ( )
n
n in ( )
n1rf ( ) e d2
1rf ( ) e d2
1f( ) r e d2 

 
 


 

           


since weh a v e , Poisson Kernelninr
nP( ) r e
rr
rr r1Af ( ) f ( ) P ( ) d2
Af ( ) f P P f

5.3PROPERTIES OF POISSO NKER NEL
Theorem 1: If0r1then Poisson kernel2
r21rP( )12 r c o s rmunotes.in

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Proof: We have byd e f i n i t ion of poisson kernel
ninr
nP( ) r eni n n i nr
nn 1P( ) r e r e     (1)
Both of above Series are geometric Series .
For 1stSeries,
First term =a= 1 and Common Ratio = R =ireiiR re r e 1 Sincei0r1 r e 1For 2ndSeries,
First team =iar eandCommon ratio =iRr e
iiRr e r e We have sum of infinite term of geometric Series whose 1stteami sa
and common ratio is Ris given byaS1R,providedR1.
Use this in equation (1)
i
rii1r eP( )1r e 1r e ii 2
ii 2
2
ii21r e r e r
1r e r e r
1r
ee12 r r2

 
    
221r12 r c o s riieecos2   Theorem 2:The Poisson kernelrP0Proof:
2
r 21rP( ) 0 r 112 r c o s r Since20r1 1 r 0Also1 cos 1.Hence in any case
212 r c o s r 0Hence0rP.munotes.in

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Theorem 3:r1P( )d 12
whererPis the Poisson kernel
Proof:
2
r21rP( )12 r c o s r2
r 21rP( ) d d12 r c o s r 
   
SincerP( )is even function
2
r 2
01rP( )d 2 d12 r c o s r 
  
Also we can wri te
2 2
r 2
011 rP( ) d 2 d21 2 r c o s r 
     
2 2
2
01rd1 )12 r c o s r 
By applying contour integration Method
PutiZe z1idz ie d i z ddzdiz
ii1zeezcos22 
Put in (1)
2
r
c
21r d zIP ( ) d1izzz12 r r2
  


2
2c1r 1.dz1 iz1rz rz 
2
22
c(1 r )z 1.dziz zr z 1 r z munotes.in

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2
22
c11 rdzizr z rr z 
2
22
c11 rId z ( 2 )ir z ( 1 r ) z r  
To Find poles and residues :
Let221rz r z r o
22
1
1rz z r z r o
rz r z r z o
rz r z ozr a n d z a r e p o l e sr



Since 1,zr z rlies inside circle1Cz.
11,zrso1zrlies outside circle C.
By Cauchy Residue theorem,
2112 lim1zrrIi z riz r r z
    2211212riir

From (1), rP2d

112rPd
 
Theorem 4:For0, 0 1rPd a s r
 
 
Proof :
2
21,112 c o srrPrrr 2 212 c o s 1 21c o srrr r 
As1r,212 c o s 2 1c o srr munotes.in

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which is bounded ascosisbounded.
Hence2r1rP( )C(asapproaches towards,cosdecreases)
2
r1rP( )d 0 a s r 1C 
Theorem 5:ThePoisson Kernel is agood ke rnel.
Proof: Since we have proved
1)rP( ) 02)r1P( ) d 12

3)M0Such that0r1rP( ) d M

4) for every0,rP( ) d 0 a sr 1
Henc ePoisson Ke rnel is agood ke rnel.
5.4ABEL SUMMABILITY OF FOURIER SERIES:
Theorem: TheFourier Series of an integrable function on circle is
Abel summable tofat every point of continuity, Moreover, iffis
continuous on the circle then the Fourier series offis uniformly
Abel summable tof.
Proof: Step 1:We have, Abel mean of the functionf( )which is
approximated by the Fourier series wherefis integrable function
defined on the circle.inn
nf( ) a e
ninrn
nAf ( ) r a e
Abel Mean of Fourier Series offcan be written as convolution
rrAf ( ) (f P)( )Where,rP( )is the Poisson kernelmunotes.in

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Step 2:We have property of a good kernel,
Letnn1Kbe a family of good kernel andfis integrable function
defined on the circle then
whenever,fis continuous atx. If f is continuous everywhere then
above limit is uniform.
Step 3: We know that Poisson kernelrP( )isagood ke rnel
Therefore by above property men tion in step (2)
rr1
rr1lim ( f P )( ) f ( ) 0 r 1lim A f ( ) f ( )

Hence, Fourier series of an integrable function defined on the circle
isAbel summable tofat every point of continuity.
Also, by step (2)
Iffis continuou s on the circle then the Fourier series offis
uniformly Abel summable tof.
Ex: IfrPdenotes the Poisson kernel, show that the function
rPur , ,0 r 1 , R satisfies
(i)u0in the disc where2222 211rr r r(ii)
1lim ( , ) 0rur
for eachHowever uis not identically zero.
Solution: (i)We have ninr
nPr e

On differentiating w.r.t,w eh a v er ninnPinr e


r ninnPu r, inr e

  (1)
Consider2222 2u1 u 1 uurr r rnnlim ( f )( x ) f ( x )munotes.in

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On differentiating (1)term by term , we obtain


 3
n2 n1 nin in in2
3
n2 2i n
2
n2 3i nin n inu in n n 1 r e r e r err
in n inui n n n 1 r r rerr
u in n n 1 in n in r e
u0    
  
    
   


(ii)We have 2
21,112 c o srrPrrr 
 2
r22
22P 1rur ,12 r c o s r1r 2 r s i n
ur ,
12 r c o s r     


Consider
 2211211 2 sin
lim ( , ) lim
12 c o s
lim ( , ) 0rr
rrr
ur
rr
ur

 




Since1ruis not identically zero.


munotes.in

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6
DIRICHLET PROBLEM
Unit Structure
6.1 Laplacian operator andHarmonic function s
6.2 Dirich let problem for the unit disc
6.3 TheSolution for Dirichlet problem
6.4 The Poisson integral
6.1LAPLACIA NOPERATOR A NDH A R M O NIC
FUNCTIO NS:
Two d imensional transient (time dependent) heat equation is given
by
22
22uu uxy k t whereu(x,y, t)is the temperature at point(x ,y)at timet.Transient mean stemperature depends o nt i m e .The&kare
physical quantities namely specific heat and thermal conductivity of
the material respectively .
If temperature is independent of time thenu0tand such a
physical situation is known as ste ady state. Hence a bove Heat
Equation can be written as
22
22uu0xy
This equation is known as Laplace equation .
Laplace equation can be written as :
22
22uuu0xy
The Operator2222xyis known as Laplacia n operator.
The Solution of Laplac eequationu0is known as Harmonic
function.munotes.in

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6.2DIRICHLET’S PROBLEM FOR U NIT DISC :
Consider unit disc in the plane22 2Dx , y / x y 1 whose
boundary is unit circle22 2,/ 1Cx y x y .
In polar co -ordinate(r, )with0r1 & 0 2,wehave unit
discD( r , ) / 0 r 1 , 0 2whose boundary is a unit circleC( r , ) / r 1 , 0 2.
The boundary value problemu0withuf ( )atr10 2is known as Dirichlet problem in the unit disc.
Note:The Laplace equationu0where2222xywhich is in
Cartesian form can be convert in term s of polar form(r, )as
22
22 2u1 u 1 u0rr r ri.e.2222 211rr r r6.3SOLUTIO NOF DIRICHLET PROBLEM FOR U NIT
DISC :
Problem Statement:
Consider unit discD( r , ) / 0 r 1 , 0 2whose boundary is unit ci rcleC( r , ) / r 1 , 0 2The governing steady -state heat equation given by the Laplace
equationu0i.e.22
22 2u1 u 1 u0rr r r(1)
subject edto boundary condition.,uf ( )atr1 , 0 2(2)
Solution: Let us apply separation of variable smethod to solve
Dirichlet problem.
Letu( r, ) F( r )G( )(3)
where ,F(r )is some function ofrandG( )is some function ofmunotes.in

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Using equation (3) in equation (1)
22
22 211(F G) (F G) (F G) 0rr r r     211F"G F' G FG" 0rr211F"G F'G FG"rrDivide both sides by FG
211F"G F'G FG"rrFG FG

211F" F' G"rrFG
 
2rF" F' G"rF r GrF" F ' G"Fr G
2rF " r F ' G "FG 
which is separation form of given D.E.
Sincerandare independent variables we can write
2rF "r F ' G "FG(4)
Whereis constant
Consider,G"( )
G( )G"( ) G( ) 02d(D ) G ( ) h e r e , Dd(5)
Consider Auxiliary equation2D02DSinceGis a function ofand02i.e.Gis defined ona
circle i.e .Gis periodic of paired22
220let m , m zDmDm imunotes.in

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Hence solution of equation (5)can be written asG( ) Acos m B sin mOrim imG( ) Ae Be  (6)
whereA&Bare constant s.
Now consider,
2rF " ( r ) r F ' ( r )
F(r )2rF " ( r ) r F ' ( r ) F ( r ) 0(7)
zput r e i.e. z log r2r.F'(r ) DF zr. F " ( r ) D ( D 1 ) F zwheredDdr
Put th esevalue sin equation (7)
2
2D( D 1 ) F( z ) DF( z ) F( z ) 0(D D D ) F (z) 0
(D )F (z) 0

Auxiliary equation
222D0Dm
Dmmz mzF( x) Ce Dewhere C andD arbitrary constant s.
PutZl o g rmlogr mlogrmm
m
mF(r ) Ce De
F(r ) Cr Dr
DF(r ) Crr


(8)
Using equation (6)and(8)in(3)i . e .u( r, ) F( r )G( )we have
 mi m i m
mDu( r, ) cr Ae Ber  (9)
Since0r1asr0thenmDrandFwill be unbounded at center and hence
arbitrary constantD0.munotes.in

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60Solutio n (9) can be written as
mi m i mu( r, ) Cr ( Ae Be )  
m imu( r, ) Er e , m z(10)
whereEis new constant combining all the solutions
m im
m
mu( r, ) a r e  (11)
wheremais arbitrary con stant.
Equation (11) g ives general solution of Dirich let problem to find
particular solution we need to find cons tantsmawhich can be
determined by boundary condition given by equ ation (2),uf ( )atr1.imm
mu(1, ) a e (12)
The above equation is complex form of Fourier series of periodic
functionf( )of period2.
Hence,mais aFourier coeffi cient which is given by,
2
im
m
01af ( ) e d2
(13)
6.4 THE POISSO NINTEGRAL:
Theorem: Letfbe integrable function define on the unit circle
then the function u defined in the unit disc given by the Poisson
integra la sru( r, ) ( f P )( )has the following property
1)uhas two continuous derivatives in the unit disc and satisfiesu0(i.e.usatisfies Laplace equation)
2) Ifis any point of continuity of functionfthen
r1lim u( r, ) f ( )Iffis continuous everywhere then this limit is uniform .
3) Iffis continuou st h e nu( r, )is the unique solution to the steady
state heat equation equation in the disc which satisfies above
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Proof :
Step (1):
Claim :u( r, )has two continuous derivatives in unit dis ka n di t
satisfies Laplace equation
we have,ru( r, ) ( f P )( )
Fix1inside each discr1centered at origin .
The SeriesuCan be differentiated term by term and the
differentiated series is uniformly and absolutely convergent .Thus,ucan be differentiated twice. (Infact,ucan be differentiated
infinitely ma ny times) and since this holds for for all1,wec a n
conclude that u is twice differentiable inside the unit disc.
In polar co -ordinates we have2222 2u1 u 1 uurr r rPutru( fP ) ( )Term by te rm different iation gives usu0Step (2) :
Claim :
a)1lim ,
rur f,w h e n e v e rfi sc o n t i n u e sa t.
b) If f is continuous everywhere then above limit is uniform.
We have, property of a good kernel,
Letnn1Kbe a family of good kernel andfis integrable function
defined on the circle then
whenever,fis continuous at. If f is continuous everywhere then
above limit is uniform.
We know that Poisson kernelrP( )isagood ke rnel Therefore by
above property mention in step (2)nnlim ( f )( x ) f ( x )munotes.in

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rr1
r1lim ( f P )( ) f ( ) 0 r 1lim u r, f ( )

whenever,fis continuous atx. If f is continuous everywhere then
above limit is uniform.
Hence claim.
Step (3) :
Supposev( r, )is another solution of steady state heat equation0in the unit disc and converges tofasr1i.e.
r1lim V( r, ) f ( )Sub claim :V(r, ) u(r, )For each fixrwith0r1the functionV(r, )has a Fourier series
expansioninn
nV(r, ) a (r) e
in
n1a( x ) V ( r , ) e d2

 
SinceV(r, )satisfies Laplace equation
i.e.22
22 2v1 v 1 v0rr r r(1)
Putin
n va ( r ) e n2
in in in
nn n 2
2
in in in
nn n 21na "( r ) e a '( r ) e a ( r ) e 0rr
1na "( r ) e a '( r ) e a ( r ) e 0rr  
  
2
nn n 21na" ( r ) a' ( r ) a( r ) 0rr(2)
The solution of above equation (2) is given by,
nn
nn na( r ) Ar Br n 0{seesolution of Dirchlet problem }
wherennA& Bare arbitrary constants.
To evaluate constantnnAa n d Bwe observe thatnA( r )is bounded
becausevis bounded
Since,
nnnnnBa( r ) A rrSincena( r )boundednB0Hence,nB0munotes.in

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Also to findnAif we take limitr1Sincevconverges uniformly
tof,we can writenAas a Fourier coefficient
in
n1Af ( ) e d2

 
By similar arguments abo ve formula holds forn0Hence, for each0r1,the Fourier Series ofvis given byu( r, ).So by the
uniqueness of Fourier series of continuous function, we must have,v( r, ) u( r, )Note: IfuSatisfies Laplace equ ationu0in the unit disc and
converges to zero uniformly asr1thenumust be identi cally
zero.However if uniform convergence is replaced by pointwise
convergence then this conclusion may fail.
Ex1:In a semicircular plate of radius 1 cm ,the bounding diameter
is kept at00Cand the c ircumference is at fixed t emperature0
0uCuntil steady state condition revels . Find the temperature distribution
in the semi -circular plate.
Solution :The steady state temperature with the semi -circular plate
is given by Laplace equation (Polar form)
22
22 2u1 u 1 u0rr r r(1)
where,u( r, )represent temperature within semi -circular plate with
boundary conditionu( r,0 ) u( r, ) 0(2)0u(1, ) u(3)
We have general solution of diri chelet problem as
mimm
mu(r, ) a r e
This solution may be written asmmm
m0u( r, ) ( A cos m B sin m ) r
 (4)
wheremmA& Bare arbitrary constants.u( r,0 ) 0i.e.u0 a t 0m
m0A c o s 0A0munotes.in

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put in (4)
m
m
m0u( r, ) B r sin m
 (5)
Nowu( r, ) 0i.e. atu0atm
m 0 B Sin ( m ) r
Sin( m )mn n 0 , 1 , 2 , . . . . . . . . .i.e.m n

Also from (3)
0 u(1 , ) uWhere0r1 , uuput in (5)
0m
m0u B sin m

Which represents the sine series andmBrepresent the Fourier
coefficient of sine series.
m0
02B u sin m d
0
m
0
m 02u cos mBm
2u[( 1) 1]m 
Put this value ofmBin equation (5)mm0
m02uu( r, ) [1 ( 1) ] sin m rm

The solution is not defined atm0m0
02B u sin m d
Putm000
02B u sin m d 0mm0
m12uu( r, ) [1 ( 1) ] sin m rm
 
m1(1 ) 0ifmis even2ifmis odd2m 10
m14u sin [( 2m 1) ]ru( r, )2m 1
 
Which gives temperature distributionu( r, )within the semicircular
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Ex. 2 :Solve Dirichlet Problem on unit disc defined byD{ ( r , ) / 0r1 , 0 2 }Whose boundary is unit circleC{ ( r , ) / r1 , 0 2 }Subject to boundary conditionu sinonC.
Solution :Consider Dirichlet Prob lem on unit disc D whose
boundary is unit circle C given byu0subject tou sin on C..
We have general solution of Di richlet problem,mmm
m0u( r, ) ( A cos m B sin m ) r
  (1)
On the boundaryCwe haveu sinatr1mm
m0sin ( A cos m B sin m )
 
Which is a Fourier series expansion where,mmA& Brepresents
fourier coefficient s.
2
m
01Af ( ) c o s m d
2
01sin cos m d
mA0...... { By Orthogonality property of circular function}
2
m
0
2
01B f ( )sin m d
1sin sin m d 0



02sin sin m d0m 1
1m 1

1m
mB1 & 0 m 1&A 0 m1u( r, ) B r sin =r sinEx. 3 :Find the solution of Dirichelet problem on unit disc D whose
boundary is unit circle C as defined before subjected to boundary
conditions.
0
0u0f( )u2munotes.in

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Solution :We have dirichlet problemu0Where2222 211rr r ron unit discD{ ( r , ) / 0r1 , 0 2 }Whose boundary is unit circleC{ ( r , ) / r1 , 0 2 }subject
to boundary condition
0
0u0f( )u2
We have general solution of Dirichlet problem
m
mm
m0
mm
m0u( r, ) ( A cos B sin m )r
at r 1
u(1, ) f ( )
f( ) (A c o sm B s i nm )






 
(1)
which is Fourier Series expansion off( )wheremmA& Bare
Fourier coefficients
we have , 2
m
01Af c o s m d
2
m0 0
011Au c o s m c o s m d 
   200
0uusin m Sin m
mm0    2
m
01Bf ( ) s i n m d
2
00
011u sin m u sin m d 
   200
0uu cos m cos m
m          mm00
m0uu(1 ) 1 1(1 )mm
2u11m   

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2
0
0
21Bf ( ) 0 . d 0
Also,1 ( 1) 2 if m is odd0 if m is even




0
m2m 10
m1
2m 1 0
m14uB(2 m 1)
4uf ( ) sin ( 2m 1) r(2 m 1)
sin ( 2m 1) 4ur2m 1



 
  





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7
HILBERT SPACES
Unit Structure
7.1 Hilbert Spaces -Definiti on and its properties
7.2 Standard examples of Hilbert spaces
7.3 Properties of Hilbert Space
7.4 Cauchy -Schwar zi n e q u a l i t y
7.5 Orthonormal bas is
7.6 Equivalent character ization :Bessel’s inequality and
Parse val’s i dentity
7.1DEFI NITIO N: HILBERT SPACE
Definition 1 :
Let H be a complex Banach space then H is called Hilbert space if,xyassociated to each of two vectors x&yHin such a way that
i),,xy yx
ii),,,xy z x z y z 
iii)2,, ,xx x xyz Hfor all scalars,Definition 2 :
Thevector space with their inner product and norm satisfying :
i)The inner product is strictly positive definite.
i.e.00xxii)The vector space is complete.
i.e. Every Cauchy sequence in the norm converges to a limit in the
vector space, is called Hilbert Space.munotes.in

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Definition 3 :
As e tH is called Hilbert Space if it satisfied the foll owing properties
i)Hi sav e c t o rs p a c eo v e r2orii)H is an inner product space satisfying.
a) ,fg g f(conjugate symmetry)
b),, ,fg h f h g h (linearity property)
c) ,0 , , , , ,ff f Hf g h H iii)Let12,ff f0fif and only if0fi.e. Inner product is strictly positive
definite.
iv)The Cauchy -Schwarz inequality and Triangle inequality
Cauchy -Schwarz inequality
,fg f g
Triangle inequality
,fg f g f gH.
V)H is complete in the metric,df g f gNote : In the above definition of Hilbert space, the Cauchy -Schwarz
inequality and triangle inequality are direct consequence of property
(I) & (II).
7.2 EXAMPLES OF HILBERT SPACE :
1)The spacedRLet12,, . . . . . . . ,dXx x x12,, . . . . . . . ,dYy y y
Then inner product of X & Y
11 2 2 ,. . . . .ddXY x y xy xyand
1222 212,
....dXX Xxx xWhich is usual Euclidean distance .munotes.in

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2)The spacedCLet12,, . . . . . . . ,dZz z z12,, . . . . . . . ,ddWw w w C 
Then11 2 2 ,. . . . .ddzw z w zw zwand
 
 11221112221,. . . . . .
......dd
dZZ Z z z z z
zz 
2zz Z
3)The sequence space2The sequence space2overis set of all infinite sequences of
complex number as1012...... ,...... , ,...... ,.....nnaa a a a a  such that
2
n
na
Let101.... ......Aa a a101.... ......Bb b bbe the elements in2Then ,nnnAB ab1112222,nn n
nnAA A a a a
   4)The sequence space2NThe sequence space2Noveris set of all infinite sequence of
complex number as12...... ,.....n aa aone sided such that2
n
nNa
Let12......Aa a12......Bb b
1,nnnAB ab
1112222
11,nn n
nnAA A a a a 
  
5)Square Integrable function2LE.
Let E be measurable subset ofdwith0mE.L e t2LEdenote
the space of square integrable function that are supported on E.
i.e.2LE= {f supported on E such that 2
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The norm & Inner product is defined as 11222,
E
EEfg f xgxd xff x f x d x f x d x       
 
7.3PROPERTIES OF HILBERT SPACE :
Theorem 1: Let ,,XYZ H,,are scalars then
i),, ,XY Z X Z Y Z   
ii),, ,XY Z X Y X Z   
iii),, ,XY Z X Y X Z   
iv) ,0 0 0, ,XX X HProof:
i)Consider
,,,,,,,,XY Z X Y ZXZ YZXZ YZXZ YZ
 
 
 
 
 
 
ii),,XY Z Y Z X ,,,,,,,,YX ZXYX ZXYX ZXXY XZ 
 
 
  
 
 
 
iii),,XY Y Z XY Z  ,,,1 ,,,
,,XY XZXY XZXY XZ
XY XZ 
 
 
  
 
 
 
iv)Consider0, 0. 0, 0 0, 0XX X,0 0, 0 0XXmunotes.in

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Definition : Orthogonality : Let V be vector space overwith
inner product and associated norm.T h et w oe l e m e n tX andY are
said to be orthogonal if,0XYand we writeXY.
Theorem 2: The Pythagorean Theorem :
If X & YHare orthogonal then22 2 2XY X Y XY
Proof :
2
22,, , , ,,,XY XY XY X X X Y Y X Y YXX Y Y X Y

Since,, 0XY X Y Y X22 2XY X Y
2
22,, , , ,00XY XY XY X X X Y Y X Y YXY

SinceXY,, 0XY YX22 2XY X Y
7.4THE CAUCHY -SCHWARZ I NEQUALITY :
Theorem 3: For any ,XY H,XY X Y
Proof : Case (i) if00YYand,, 0 0XY X.
and obviously Cauchy -Schwarz inequality holds.
Case (ii) If0YFor any scalarwe have,0XY XY……. {+ve definite prop.},, 0XX Y YX Y ……{Linearity prop.},, , , 0XX X Y YX Y Y      22 2,, 0XX Y Y XY     Since0Yput2,XYymunotes.in

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2
22
22 22, ,, ,,0XY XY XY XYXY X Y
YY Y   222
2
222,,,0XY XY XYX
YYY   2
2
2222,
,
,XYX
YXY X YXY X Y
 

Theorem 4: Triangle Inequality :
For any ,XY H,XY X Y
Proof :2,XY XY XY,, , ,XX X Y YX Y Y22,, ,XX X Y Y Y 
ByCauchy Schwarz inequality us have,,XY X Y
22 2,,2XY YX X Y X YXY X X Y Y 
 22XY X YXY X Y

Theorem 5: Parallelogram Law
If ,XY Hthen22 2 222XY XY X Y
Proof :
Consider,
22
22,,
,, , ,,, , ,22XY XY XY XY XY XYXX X Y YX Y YXX X Y YX Y YXY


munotes.in

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7.5ORTHO NORMAL BASIS :
Definition : A finite or countably infinite subset12,. . . . . . . .ee of
Hilbert Space H is said to be orthonormal if
1,0kwhen keewhen kand 1keki.e. Eachkehas unit norm and is orthogonal toewhenever k.
Property :Let H be a non -zero Hilbert space so that the class of all
its orthonormal set is non -empty. This class is a partially ordered set
w.r.t. set inclusion relation.
Definition :
An orthonormal setiein Hilbert space H is said to be complete if
it is maximal in partial order set i.e. if it is impossible to adjoin the
vector e to collectioniein such a way that,ieeis an orthonormal
set which properly containsie.
Theorem : Every non -zero Hilbert space contains a complete
orthonormal set.
Proof :
We know that
i)An orthonormal setiein Hilbert space H is said to be complete
if it is maximal in partial order set w.r.t. set inclusion relation.
ii)Zorn’s Lemma states that if P is partially ordered set in which
every chain has an upper bound then P posses a maximal
element.
iii)Since the union of any chain of orthonormal set is clearly an
upper bound for the chain in the partially ordered set of all
orthonormal set.
The above three statements shows that every non -zero Hilbert space
contains complete orthonormal se t.munotes.in

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Theorem : If1kkeis orthonormal andkkfa e Hwhere sum
is finite then22kfa.
Proof :
2,,kk llff f a e a e2,
..... , 1
0kk
kkaa e eae e k lkl  

Orthonor mal Basis:
Given an orthonormal subset 121,. . .kkee eof Hilbert Space H
Spans H i.e. Linear Combination of elements in12,. . . . . .ee are dense
in H and12,. . . . . .ee are linearly independent then we say that12,. . . . . .ee is an orthonormal basis for H.
Note : For anyfHand1kkeis orthonormal basis for H then
1kk k
kfa e a C

i.e.fcan be written as linear combination of elements in12,. . . . . .ee .
Consider,
1
1,,
,jk k jk
kkj
kfe a e eaee


  

When,, 1 & , , 0kk k j for k j e e for k j e e  i.e. ,jjfe aHence, whenever
1kkkfa e
then,kkaf e.
7.6EQUIVALE NT CHARACTERIZATIO N:
Theorem : The following property of an orthonormal set 1kkeare
equivalent.
1)Finite linear combination of elements in 1kkeare dense in H.
2)IffHand ,0jfe jthen0f.munotes.in

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3)IffHand
1NNk kkSf a ethenNSf fasNin norm of
Hilbert space H.
4)If,kkaf ethen22
1kkfa
.
Proof:S t e p( 1 ):12
Given : finite linear combination of elements in1ekkare dense
in H.
LetfHand ,0jfe jClaim :0fProof : Since fini te linear combination of elements in 1kkeare
dense in H, there exist a sequencengof elements in H which is
finite linear combination of elements in 1kkesuch that
nfg oasn.
Since ,0jfe j,0 . . . .nfg n {ngis finite linear combination of elements in
1kke}
By Cauchy -Schwarz inequality.
Consider,
2,,nnff f f f g f f g,, ,
,,,. . . , 0nn
n
nff g ff f g
ff f g
ff f g
 Lettingn200n ff g a s n00f
fStep 2 :23
Given ,, 0 ,jfHf e jthen0fAlso we have
1NNk kkSf a ewhere,,kkaf e.munotes.in

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Claim :0NSf fASNConsider,11 1
1, 1
1,, ,
,,
,,
,NN N NN
NN N
kk kk
Kk
NN
kk k k
Kk
kKfS fS f f S f S fS f
fa e a e a e
af e a a e e
aa e e  
 

  
 
 



 


1, 1
11
22
11,
,1.....NN
kk
Kk
NN
k
kk kk
KK
NN
kk
KKaa e eee kaa aaokaa 
 
      

 


0,0NN
NNfS fS ffS f S f 
By Pythagorean theorem,

222
2 2
1NN
N
Nk
Kff S fS ffS f a
 
 221N
k
Kfa
Letting N221k
Kaf
 {This is known as Bessel’s Inequality }
Bessel’s inequality implies that series21k
Ka
is convergent.
Therefore, pa rtial sum 1NNSfforms Cauchy seq. in H.
Since 
11NMNM k k k kKKSf Sf a e a e 
1
2
1N
kk
KM
N
k
KMae N Ma whenever N M
  
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Since H is completegHsuch thatNSf gasN.
Fix j and note that for all sufficiently larger N,,, ,Nj j NjfS fe f e S fe 
,()0jk k j
jjaa e ea a orthonormality

SinceNSf gwe can write
,0jfg e jfgo…………. {By given hypothesis (2)}fg,0 , 0jfe j fHenceNSf fasNi.e.0NSf fasNStep 3 : (3)(4)
Given 
1NNk kkfH Sf a eNSf f oasNClaim :221k
kfa

We have 2221NNkKff S f a 
Letting Nand using0NSf fasN221k
Kfa

This is known as Parseval’s Identity .
Step 4 : (4)(1)221k
kfa

Claim : finite..cofelements in 1kkeare dense in H.
We have from equation
2221NNkKff S f a munotes.in

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asN, we have Parseval’s identity.221k
Kfa
0N fS fasNSince eachNSfis finite linear combination of elements in 1kke.
Hence finite linear combination of elements in 1kkeare dense
in H.
Ex1:Let H be Hilbert Space. Sh ow that for any ,xy H22 2 24,x y x y x y i x iy i x iy
Solution:
Consider22 2 2x y x y i x iy i x iy
 
 22
22,,,,,
,xyx y y xxyx y y xi x iy x iy
i x iy x iy    
  
2
2
22
222, 2, , , ,
,, ,
2, 2, , ,
,,
2, 2, , , , ,
4,x y y x i x x iy iy x iy iyi x x iy iy x iy iy
xy yx i x ixy i yx y
ix i x y i y x y
xy yx xy yx xy yx
xy          


Ex2:Let12,, . . . . ,nee ebe a finite orthonormal set in a Hilbe rt space
H. If xis any vector in H. Then show that
221,n
i
ixe x
Also show
1,n
erii jixx e e e  for each j.munotes.in

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Solution :Consider
2
1
11
2
11 1 1
2
11 1 1
2
110,
,, ,
,, , , , , ,
,, ,, , ,,
,, , , , ,n
ii
i
nn
ii j j
ij
nn n njj i i i i jjji i j
nn n n
ji i i ij i j
ji i j
nn
jj i i i i
jixx e e
xx e e xx e exx x e e x e e x x e ex e exx e x e x e e x x e x e e e
xx e x ex e x ex e x e
 
   
  
 
 
  
  
  

  
  
 
1
2
111
2 22 2
11,, ,, ,,
,,n
i
nnn
ii ii ii
iii
nn
ii
iixx e x ex e x ex e x e
xx e x e x
  
   
  
  

Consider1
1,,niijixx e e e
1,, ,nji i jixe xe e e
1,, ,nji i jixe xe e e
1,, ,
,,
0,jj j j
jj
n
ii j
ixe e e xe
xe xexx e e e j

 
Ex3:Let H be Hilbert space. Letiebe an orthonormal set in H.
Then show that the following conditions are equivalent.
1)ieis complete
2)ixethen0x3)IfxHthen ,iiixx e e 4)IfxHthen22,i
ixx emunotes.in

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Solution :
Step -I:12
Letiebe complete.
Supposeixe iSub claim -0xSuppose that0xDefinexex
Clearly,0iiee x e i.T h u s,ieeis ort honormal set which
properly containsieWhich is contradiction toiebe complete.
Hence our assumption is wrong.0xStep -II :23
Supposeixe ithen0xSub claim : ,iiixx e e We know that ,iixx e e is orthogonal toieBy hypothesis, ,ii xx e e ,iixx e e Step III :34
Suppose for ,,iixH x x e e Sub claim :22,ixx e
Consider 2,xx x
 
2,, , ,
,, ,
,, 1
,ii j j
ij i j
ij
ii
i
i
ixe e xe e
xe xe e e
xe xee orthonormal setxe

 

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Step IV :41
Suppose22,,i xHx x e 
Sub claim :ieis complete.
Supposeieis not complete then it is proper subset of an
orthonormal set,iee.S i n c eer
iee iPutxein above identity.22
2,
00iee e


This is contradiction to e is a unit vector
Hence our assumption is wrong.
Thusieis complete.
Note : Letiebe complete orthonormal set in Hilbert space H. Let
x be an arbitrary vector in H. Then,ixeare Fourier coefficients of
x and the expression ,iiixx e e is called Fourier series expansion
of x and the equation,22,ixx e is called Parseval’s iden tity.
(all w.r.t. complete orthonormal setieunder consideration.)
Ex4:If1niieis an orthonormal set in Hilbert space H and if x is
any vector in H then,0ii Se x e is either empty or c ountable.
Solution :
For each +ve integer n, consider2
2,ni ixSe x en      .W eh a v e
Bessel’s inequality.
221,n
i
ixe x
Bessel’s inequality gives us,nScontains at most1nvectors since
1.nnSS
Si se i t h e re m p t yo rc o u n t a b l e .
Ex5:Show that a closed convex subset C of a Hilbert space H
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Solution : We recall from the definition in Problem 32 -5t h a ts i n c e
Ci sc o n v e x ,i ti sn o n -empty and contains/2xywhenever it
contains x and y. Let d = inf:xxC. There clearly exists a
sequencenxof vectors in C such thatnxd.B yt h ec o n v exity
of C,/2mnxxis in C and/2mnxx d,s o2mnxx d.
Using the parallelogram law, we obtain22 2 222 222
22 4 ;mn m n mn
mnxx x x xxxx d

and since22 22 2 222 4 2 2 4 0mnxx d d d d, it follows thatnxis a Cauchy sequence in C. Since H is complete and C is closed
Ci sc o m p l e t e ,a n dt h e r ee x i s t sav e c t o rxi nCs u c ht h a tnxx.I ti s
clear by the fact that lim limnnxx x dthat x is a vector in C
with smallest n orm. To see that x is unique, suppose thatxis a
vector in C other than x which also has norm d. Then/2xxis
also in C, and another application of the parallelogram law yields.
222222
222 22
,22xxxx xxxxd  

which contradicts the definition of d.













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8
HILBERT SPACE2,LUnit Structure
8.1 Hilbert Spaces20, 2Lor2,L
8.2 Existence of orthonormal basis
8.3 Orthonormal basis for20,2Lor2,L
8.4 Mean Square Convergence
8.5 Best Approximation Lemma
8.1 HILBERT SPACE2,LConsider the Hilbert space2L, associated with measure space0, 2where measure is Lebesgue measure and integrals are Lebesgue
integrals. This space essentially consist of all complex functions f
defined on0, 2which are Lebesgue measurable and square
integrable.
i.e. 2
2
0fx d x
Its norm and inner product is defined as 1222
2 0ff x d x 
2
0,fg f x gxd x
The function2inxewhere0, 1, 2,....nforms an orthonormal
basis for H since2
02.0imx inxmnee d xmn
This gives us ,
2inx
neex n
For any2fL,t h en u m b e r , 2
01,
2inx
nnCf e f x e d x
 gives
Fourier coefficient of the Fourier series expansion of f given by,
1
2inxn
nfx C e
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Definition:
The Hilbert Space20, 2Lor2,L.
Let R denote set of complex valued Riemann integrable function s
defined on a circle then the inner product and norm is defined as
2
01,2fg f g d and 1222
2
01
2ff d    .
Similarly, for interval,.
1,2fg f g d
 and 122
21
2ff d
    .
8.2 EXISTE NCE OF ORTH ONORMAL BASIS OF
HILBERT SPACE
Theorem : Any Hilbert Space has on orthonormal basis.
Proof : The proof of this theorem is follows from gram Schmidt
process.
Given finite family of elements12,. . . . . . ,kff f,t h es p a no ft h i s
family is set of all elements which are finite linear combination of
elements12,. . . . . . ,kff fWe denote it by span12,. . . . . . ,kff f.N o ww e
construct a sequence of orthonormal vectors say12,. . . . . .ee such that
span12,. . . . . . ,nee e=s p a n12,. . . . . . , 1n ff f n.
Let us prove this by induction on n.
Step 1 : By Linear independent hypothesis,10fthen we can take111fef.
Step 2 : Assume that orthonormal vectors12,. . . . . . ,kee ehas been
found such that span12,. . . . . . ,k ee espan12,. . . . . . ,kff f.
Claim : span12 1,. . . . . . ,k ee espan12 1,. . . . . . ,kff f
i.e.'
11
1kkk j jjef a e 
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'
11
1
1
1
1
1
'
11,,
,,,,,,k
kj k i i j
i
kkj i i jj
kkj i i ji
kj kj jee f a e efe a e efe a e eee fe a 





     


To have :'
1,kjee0jWe must have1,kj jfe aThis choice of for 1jaj kassure that'1keis orthogonal to1,....kee.
Moreover, our linear independent hypothesis assure that'
10keHence, the choice of1keis'11 '1k
k
keee

.
Hence span12,. . . . . . ,n ee espan12,. . . . . . ,nff f.
Thus, Every Hilbert space has an orthonormal Basis.
Example: Consider, Hilbert space H. Transform Basis12,. . . . . . ,nff finto orthonormal basis where,12 31, 1,1 , 2,1, 0 , 1, 1,1ff f .
(Take Euclidean inner product)
Solution:
1)1
1
11, 1,111 1,,33 3 3fef     2)Using11
1kkk j jjef a e 
222 1 1 , eff e e 21
2
2
254 1,, ,33 3
54 1,,
42 42 42jK jea f eee
e 
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3)Using11
1kkk j jjef a e 
333 1 13 2 2eff e ef e e
3
3
3
3123,,77 712 3,,14 14 14e
ee
e
   8.3 ORTHO NORMAL BASIS OF2,L:
Theorem 1:The setsinx
ne& cos sinnnnx nx   are
complete orthonormal basis for2,L.A l s ot h es e t s cosnnx&sinnnxare complete orthogonal basis for20,L.
Proof : Consider,inxnxe
Let2,fL
Let0(small)
Claim : Nthpartial sum of Fourier series of f approximate f in norm
withinwhen N is sufficiently large.
i.e.NSf fasN.
We can find2periodic functionfpossessing deriv atives of all
order such that3ff.
Let 12,nnCf 1 12,2inx
nnCf x e d xf e
 
    
and 12,nnCfbe Fourier coefficients of&ffrespectively.
We know that Fourier seriesnnCfuniformly.
Hence it converges tofin norm.
If we take N sufficiently large then3N
nn
NfC

By Bessel’s inequality
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2
nnCC N
223ff
Consider,
NN N Nnn nn nn nnNN N NfC f ffC C C
           
Taking norm on both side
Now using triangle inequality.333NN N Nnn nn nn nnNN N NfC f f fC C C
     
This proves completeness of setinxnein2,L.
Completeness of  cos sinnnnx nx   in2,Lcan be derived
by completeness ofinxe.
Similarly, completeness ofcosnx&sinnxin2,Lcan be
prove by c onsidering even & odd extension of2[0,2 ]fL to[, ].
Theorem 2:Let2,HLandintnft efor0, 1, 2,....nand,tthen0, 1, 2,......nft n is an orthonormal basis
for2,L.
Proof :
Step 1 : Lets verify0, 1, 2,......nft n is orthonormal
int i t 11,.22m
nm n mff ft ft d t ee d t 
       
int i t 1.2mee d t

 
in 1
2mted t

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






1
2
1
2
1cos sin cos sin2
2 sin
2
sin0in mt
in m in me
in m
eein m
nm i nm nm i nmin m
in m
in m
nm
nm

 
   





 
     
   

{Since nmand sin 0kk}
,0 ,nmff nm Now consider,

2
2
int int1,2
1
2
112nn n n nff f f t f t d tee d t






  


,1nnff nHence,
,0 ,nmff nmand21nfn.
Thus, set0, 1, 2,......nft n is orthonormal.
Step 2 : Claim :0, 1, 2,....nft n is basis for2,HL.
Since0, 1, 2,....nft n is linearly independent and it spans2,HL,h e n c e0, 1, 2,....nft n is basis for2,HL.munotes.in

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Theorem 3:The set 1c o s s i n,, 1 , 2 , . . . . . , ,2nt ntnt   isan
orthonormal basis for2,L.
Prove of this theorem is similar to above theorem so left as an
exercise
8.4 MEA NSQUARE CO NVERGE NCE:
Consider space R of integrable functions define on the circle.
Letinnee,n is an integer then clearly, Setnnzeis
orthonormal.
Consider, 2
01,2in
nfe f e d
 
2
01
2infe d
 ˆnfn a{Fourier coefficient}
where,ˆfnornaisFourier coefficient of complex Fourier series of
function f.
Consider the Nthpartial Sum ,Nn nnNSf a eThen orthonormal property of familyneand the fact that
,nnfe agives that t he differenceNfS fis ort hogonal tonei.e.NnfS f e nN.
Since,, ,Nn n NnfS fe f e S fe 
,
,
,1
00nm m n
mN
nm m n
mn
nnaa e e
aa e eee m naamn 
    

Hence,NfS fis orthogonal tonen NnnnNfa e    is orth ogonal tonnnNbewhere,nbis complex.munotes.in

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We have,nn nnnN nNff a e b e By Pythagorean theorem,222
222nn nn
nN nNff a e b e  
whennnabthe ort hogonal proper ty of familynnegives us
222nn nnN nNae a .
222
2 2NnnNff S f a .
This is called mean square approximation .
8.5BEST APPROXIMATIO NLEMMA :
Statement: If f is integrable function defined on a circle with
Fourie rc o -efficientnathen Nn nnNfS f f c efor any
complex numbernc. Moreover, equality holds when
nnca n N.
Proof :
Considernn n n nnN nNfc e f a b e wherenn nabcnn nn nnnN nN nNfc e fa e b e  Taking norm on both sides.nn nn nnnN nN nNfc e fa e b e     
Since a nis Fourier coefficientnn N
nNae S fnn N nnnN nNfc e f S f b e    .
Also we haveNfS fis orthogon al tonnnNbe.munotes.in

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By Pythagorean theorem.
222
nn N nn
nN nNfc ef S f b e     
This statement gives us, nn N
nNfc e f S f 
whennncawhere,nais Fourier coefficient given.0nn n nca b bNn nnNfS f f c e

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9
RIESZ FISHER THEOREM
Unit Structure :
9.1 Completeness of2dL9.2 Bessel’s inequality for2,Lfunction
9.3 TheRiesz Fisher Theorem
9.4 Unitary Isomorphism
9.5 Separability of2,L9.1COMPLETE NESSOF2dL:
Theorem : The space2dLis complete in its metric.
Proof : Let1nnfbe a Cauchy seque nce in2L.
Consider1knkfbe a subsequence of 1nnfwith the property
121kkk
nnff k(1)
Let   11
1kk nn n
kfx f x f x f x
  (2)
and 11
1kk nn n
kgx f x f x f x
  (3)
Consider partial sum
  11
1kkK
kn n n
kSf x f x f x f x
 
and 
11
1kkK
kn n n
kSgx f x f x f x
 
The triangle inequality implies that

11
1kkK
kn n n
kSg f f f

1
12Kkn
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Letting Kand applying mon otone convergence theorem we
have,
2gSince20kk1 knSg f x thkpartial sum of g is finitegis square summable & hence square integr able
2gfg
{by (2) & (3)
2f2dfLIn particular, the series defining f converges almost everywhere and
since 1thkpartial sum of this series is preciselyknf,weh a v e ,knff xalmost everywhere for all x.
To show
knffin2dLWe have, 2 22kfSf g kApplying dominated convergence theorem, we obtain,0knffask.
Since 1nnfis Cauchy sequence for given 0,Nsuch that
,nm N2nmff.
Ifknis chosen, so tha tknN2knffBy triangle inequality
22 kk nn n nff ff f f nffwhenevernNHence sequencenffin2dL2dLis complete.munotes.in

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9.2BESSEL’S I NEQUALITY FOR2,L:
Iffis2L-periodic function then 222ˆfn f .
Proof : LetNfS f gwhereNSfis Nthpartial sum offi.e.
ˆNinxN
nNSf f n e .
Consider,,,nN nge f S f e,,nN nfe S f e


ˆ ˆ,
ˆ ˆ ,
ˆ ˆ ,
ˆˆ
0
,0
1}N
imx
n
N
N
imx
n
N
Nimx inxN
inx
n
nmfn fm e efn fm e efn fm e efn fn e e
ee m m
mn










Consider,


 
2 2
2 2
,,, , ,,,N
NN
NN N N
NNfS f g
Sf g Sf gSf Sf Sf g g Sf g gSf Sf g g
 
   
 
,0
ˆ,0,0n
n
Nge
gf n e
gS f  
222222NfS f g 
22
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Consider,


2
2
,,
ˆ ˆ,
ˆˆ ,NN N
NNinx imxNN
N
inx imx
nm NSf Sf Sf
fn e fm e
fnfm e e 

  
 

2ˆˆ ˆNN
NNfnfn fn   (2)
Substituting (2) in (1) we get
22
2ˆN
Nff n
Writing N222ˆ
nfn f

Thus we proved,
222ˆ
nfn f

9.3THE RIESZ FISHER THEOREM:
Statement : Suppose that fis2L-periodic function then the Nth
partial sum of its Fourier SeriesNSfconverges to f in2L(I) where,I.
i.e.2lim 0NNSf f
Moreover, 222ˆ
nfn f
 {Parseval’s identity }
Conversely, suppose thatnnais two sided complex sequence
which issquare summable i.e.2
nathen there is unique
function f in2L(I) that hasnaas itsFourier coefficient.
Proof : Step (1) : Let2fL(I)
Givenochoose a continuous periodic function g such
fg…… (1)
Then22NN NSf f Sf gS g f gg222 2NN NSf f Sf g S gg g f munotes.in

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We haveNSf fNSf g f g 
 2222
2
222. . . . . 1
2NN
N
NNSf f f g S gg g fS g g from
Sf f S gg 

 
Since g is continuous periodic function ,2NSg gfor large N
2
2
23
0
lim 0N
N
NNSf f
Sf f a s N
Sf f
  
Step (2) We haveNNfS f S fi.e.NfS fis orthogonal toNSf.By Pythagorean theorem,
222
222NNff S fS f 
Also we have 22
2ˆN
N
NSf f n
Weg e t 22 2
2 2ˆN
N
Nff S f f n .
Letting Nand using 2
2lim 0NNfS f
 22
2ˆff n

(This is known as Parseval’s Identity )
Step (3) Converse part :
Suppose that nnais square summable two sided sequence of
complex numbers .
Let NinxNn
nNfx a e.
The orthonormality of exponential fun ctionneimplies that for M 22'NM n n
MnNf f a f a parseval s identity     andˆnnaf.
By the assumption of square summability i.e.2
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The right side of above equation converges to zero as ,MN.i . e .
20NMffas ,NM.nfis Cauchy sequence in2LI.
Let f be the limit
By orthonormality,,&Nn nfe a N nLetting Nlim ,
,
1
2Nn nN
nn
inx
nfe a n
fe a nfx e d x a n



nais Fourier coefficient of Fourier series of function f.A l s o by
uniqueness of Fourier series, we can conclude that there exists
unique f whose Fourier coefficient isna.
9.4UNITARY ISOMORPHISM :
Unitary Mappings : Suppose H & H’ be two given Hilbert spaces
with respect to inner product,H&1,Hand corresponding normH&1H.
Am a p p i n g1:UH His called unitary mapping if
1)Ui sl i n e a r
i.e.Uf g U f U g  where,are scalars &
,fg H.
2)Ui sb i j e c t i o n
3)1HHUf f f HNote:
1)Since unitary mapping U is bijective ,its inverse11:UH His
also unitary mapping .(prove it)
2)Property (3) of unitary mapping implies that1 ,, ,HHUf Ug f g f g HUnitary Isomorphism : Two Hilbert spaces1&HHare said to be
unitarily equivalent or unitary isomorphic ifau n i t a r ym a p p i n g1:UH H.
Note : Unitary isomorphism of Hilbert spaces is an equivalence
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Theorem : Any two infinite dime nsional Hilbert spaces are unitary
equivalent or unitary isomorphic.
Proof : If &HHare two infinite dimensional Hilbert spaces.
We may select orthonormal basis i.e.12,, . . . . . . .ee of H &''
12,, . . . . . . .ee ofH.
Consider the mapping :UH Hdefine das if
1kkkfa e
thenUf gwhere,
1,kk
kga e g H f H
 .
Claim :1:UH His unitary
1),, ,Uf h U f U h f h H are scalar s.
Let
11,kk kkkkfa e h b e 
  Consider



11
1
1kk kkkk
kk k
k
kk k
k
kk kk
kk kkUf h U a e b eUa b e
ab e
ae beae be 


 
  
 



         
     





 Uf U h  
2)Claim U is bijective
Clearly,Uf U hkk kkkk kk
kkUa e Ub eae be
ab k
fhU is one one


 
For anykkga e H,we havekkfa e Hsuch thatUf g Uis onto .
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3)Claim1HHUf f
Consider, 
11kk kkkkfa e U f a e 
   
 
 1
1
12
2
1
1..... '
.... 'kk H
k H
k
k
kk
k H
HUf a e
a By parseval s identitya e again by parseval s identityf
 




  
  



Hence by (1), (2) & (3) :UH His unitary and hence &HHare
unitary isomorphic.
Theorem : Suppose2,fLthen the mappingnfais
unitary correspondence between2,L&s q u a r e summable
sequence2Z.
Proof :
Step (1) : Let2,HLwith inner product
1,2fg f x gxd x
 
Let2,fL
Let1kkeis an orthonormal basis for H.
1kk k
kfa e a C

Step (2) : Let12HZ(sequence space) defined as
 2 2
101
1..... , , .... &jn
nZa a a a C a

      with inner
product.
,kkkab ab

Step (3) Consider a mapping :UH Hsuch that&nnfa f H a Hkk kUa e aClaim : :UH His unitarymunotes.in

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1)Sub-claim : U is linear
i.e.Uf g U f U g ,scalar ,fg HLet
11,gkk kkkkfa e b e 
  Uf gUf g  


kk kkkk k
kk
kkUa e b eUa b e
ab
ab
Uf U g 



 
 



2)Sub-claim : U is bijective
i.e. U is one -one and onto.
Clearly, U is one -one
SinceUf U gkk kkkkUa e Ub eab
kk
kk kkab Kae be f g Ui so n e o n e  
To Prove Ui so n t o
Consider,22
11N
Nk k k k
kkfS f a e a e
 21
2
1kk
kN
n
nNaea





If2
nazthen
2
2
11
2
1
2
1NM
NM k k k k
kk
N
kk
kM
N
k
kMSf Sf a e a e N Mae
a 

  
 

0NMSf Sf as ,NM.munotes.in

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Hence completeness of2Lguarantee that, there is2fLsuch that0N fS fasN.
As f hasnaas its Fourier coefficient we can conclude thatnfais onto (By the uniqueness of Fourier coefficient)
Hence U is bijective
3)Claim :1HHUf f
Consider,1122nHHUf a 22,nnnnn
Haaaaa
f


Hence by (1) ,(2) & (3) ,1:UH His unitary mapping.
9.5SEPARABLE HILBERT SPACE :
Definition : The space H is said to be separable if their exist
countable collectionkfof elements in the space H such that there
linear combination are dense in space H.
Theorem : A Hilbert Space H is separable if and only if it has
countable ortho normal basis.
Proof : Step 1 :Suppose that Hilbert space H is separable.
Claims : Hilbert space H has countable orthornormal basis.
Suppose Hilbert space H has uncountable orthornormal basis sayeThen 1, , & ee11,,22Se Se   ,&.
Hence there exist an uncountable family of disjoint open sphere with
radius ½.Hi snot separable which is a contradiction to our assumption.
Hence Hilbert space H has countable orthonormal basis.
Step (2) Converse part
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Claim : Hilbert space H is separable.
Let H has a countable ortho rnormal basis sayne.
LetfH1
1,
,nn
nnnnnnff e e f Hfe afa e


 


fHis a cluster point (i.e. limit point) of set of linear
combination of elements ofne.
Sinceneis complete orthonormal basis, set of linear combination
ofelements ofnecontains countable dense set of linear
combination ofnewith rational coeffi cients.
Hence H is separable Hilbert space.
Theorem : Hilbert Space2,Lis separable.
Proof : Step (1) : Let2,HL
We know that Hilbert space2,Lhas an orthonormal basis0, 1 , 2,........nfn.
Where,  int,,2neft n t .
Since set of integer is countable, hence set of orthonormal basis0, 1 , 2,........nfnis countable.
Step (2) : If Hilbert Space H has a countable orthonormal basis then
Hi ss e p a r a b l e .
Step (3) : Hilbert Space2,Lhas a countable orthonormal basis.
Hence H is separable.

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