11
$\begingroup$

Let $T$ be the operator from $L^2(\mathbb R^n)$ to $L^2(\mathbb R^n)$
defined as composition of convolution and multiplication, $Tf := (af) * g$ where $g$ is in $L^2$ and $a$ is a bounded function.

Can we find the spectrum of $T$? For $a$ identically equal 1, the spectrum is the essential range of the Fourier transform of $g$. I am interested in the more general case. If both $a$ and the Fourier transform of $g$ are positive functions, I assume that the spectrum of $T$ will also be positive but don't have a proof.

  • 0
    In the general case, is your operator self-adjoint? Even when both $a$ and $\hat{g}$ are positive functions, what rules out complex eigenvalues? (If $a = \hat{g}$ then the operator is self-adjoint, but I don't see it in the general case.)2011-02-03
  • 0
    Also, in general, convolution is not involutive on $L^2$, so if you just assume $g$ is in $L^2$, $T$ is not guaranteed to be an operator from $L^2$ to itself. See http://math.stackexchange.com/questions/11104/is-l2-mathbbr-with-convolution-a-banach-algebra for example. Do you perhaps want $g\in L^1$?2011-02-03
  • 0
    Yes, sorry I forgot to write that $g$ is also $L^1$ function,2011-02-03
  • 1
    Hum, let $n = 1$, and consider $a$ to be the characteristic function of $(-1/2,1/2)$, $\hat{g}$ be the characteristic function $(1,3/2)$. Test against $f$ the characteristic function of $(-1,1)$. You have $\langle Tf,f\rangle < 0$ and so the spectrum must contain something in the left half plane.2011-02-03

1 Answers 1