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 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).