From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Newsgroups: sci.math
Subject: Re: bounded operator
Date: 4 Mar 1998 01:32:37 -0500

In article <34FC6D6E.6655@student.utwente.nl>,
Wilbert Dijkhof  <w.j.dijkhof@student.utwente.nl> wrote:
>Hi all,
>
>Suppose we got the integraloperator A, 
>(Af)(x) = int(f(y),y=0..x) in L_2]0,1[ with the norm f,
>||f||^2 = int(|f(x)|^2,x=0..1). What is ||A||?
>If it isn't possible to determine ||A||, an upper and lower bound
>will be good.
[...]

The answer is well-known: ||A|| = 2/pi.

 A general recipe for norms of operators in a Hilbert space (such as
L_2(0,1)) is   

     ||A|| = sqrt(||A' * A|| 

     = max{sqrt(lambda): lambda is in the spectrum of A'*A}

where A' is the (Hermitian) adjoint of A.

By interchange of order of integration, we find that

 (A'f)(x) =  int(f(y),y=x..1) (an easy exercise)

A' * A is a Hermitian compact positive definite operator, so its norm
equals its maximal eigenvalue.  The equation

                A' * Af = lambda * f

can be re-written as a second order differential equation with boundary
conditions, and the eigenfunctions will be certain trigonometric
functions.

(Remark: Observe that A itself is the inverse of the differential operator
D = d/dx with boundary condition f(0)=0). A' can also be identified with
the inverse of another differential operator, and with some patience, we
manage to obtain the composition.) 

Cheers, ZVK (Slavek).