Define the operator K f ( x ) = ∫ 0 1 k ( x , y ) f ( y ) d y . by K f ( x ) = ∫ 0 1 k ( x , y ) f ( y ) d y .where k ( x , y ) is a continuous complex valued function on the unit square. [ 0 , 1 ] × [ 0 , 1 ] Also assume that s u p x ∈ [ 0 , 1 ] ∫ 0 1 | k ( x , y ) | d y &lt; 1How to show that given g ∈ ( C [ 0 , 1 ] , C ), the integral equation f ( x ) = g ( x ) + ∫ 0 1 k ( x , y ) f ( y ) d yhas exactly one continuous solution f ∈ ( C [ 0 , 1 ] , C )?Note: I only know introductory measure theory and functional analysis, so I am suppose to do this problem without heavy integral function theory.One way to get started is to show that K is compact and its operator norm | | K | | &lt; 1 because we are given that s u p x ∈ [ 0 , 1 ] ∫ 0 1 | k ( x , y ) | d y &lt; 1 Then we know that ( I − K ) is invertible and its inverse is given by the Neumann series. At this point, I'm not sure how to proceed.

Question

Define the operator   K  f  (  x  )  =      ∫    0    1    k  (  x  ,  y  )  f  (  y  )    d  y  . by  K  f  (  x  )  =      ∫    0    1    k  (  x  ,  y  )  f  (  y  )    d  y  .where   k  (  x  ,  y  ) is a continuous complex valued function on the unit square.   [  0  ,  1  ]  ×  [  0  ,  1  ] Also assume that  s  u      p          x      ∈      [      0      ,      1      ]            ∫          0              1            |    k  (  x  ,  y  )      |    d  y  &amp;lt;  1How to show that given   g  ∈  (  C  [  0  ,  1  ]  ,      C    ), the integral equation  f  (  x  )  =  g  (  x  )  +      ∫    0    1    k  (  x  ,  y  )  f  (  y  )    d  yhas exactly one continuous solution   f  ∈  (  C  [  0  ,  1  ]  ,      C    )?Note: I only know introductory measure theory and functional analysis, so I am suppose to do this problem without heavy integral function theory.One way to get started is to show that K is compact and its operator norm       |        |    K      |        |    &amp;lt;  1 because we are given that   s  u      p          x      ∈      [      0      ,      1      ]            ∫          0              1            |    k  (  x  ,  y  )      |    d  y  &amp;lt;  1 Then we know that   (  I  −  K  ) is invertible and its inverse is given by the Neumann series. At this point, I&#039;m not sure how to proceed.

EreneDreaceaw · Accepted Answer

The equation  f  (  x  )  =  g  (  x  )  +      ∫    0    1    k  (  x  ,  y  )  f  (  y  )    d  yis equivalent to  f  =  g  +  K  f  .Hence   (  I  −  K  )  f  =  g. Since   I  −  K is invertble, we get  f  =  (  I  −  K      )          −      1        g  .

Define the operator K f ( x ) = <msubsup> ∫ 0

Answered question

Answer & Explanation

New Questions in Pre-Algebra