0
$\begingroup$

Given $l$ is an even, $2\pi-$periodic, $L^2$ function. An operator $L: H \to H$, where $H$ is a Hilbert space, is defined as $$Lf:=\int_{-\pi}^\pi l(x-y)f(y)dy.$$ To show that $L$ does not have bounded inverse.

I don't know where to begin with to solve this problem. Any steps of hints towards showing this? Thanks in advance!

Update: Does it suffice to say that, $L$ is compact, so $0\in\sigma(L)$, and thus the inverse of $L$ isn't bounded?

$\endgroup$
  • 1
    $\begingroup$ Your argument is fine. Compact operators on an infinite dimensional HS cannot have abounded inverse. $\endgroup$ – Kabo Murphy Nov 17 at 0:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.