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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.

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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.) –  Willie Wong Feb 3 '11 at 12:47
    
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 math.stackexchange.com/questions/11104/… for example. Do you perhaps want $g\in L^1$? –  Willie Wong Feb 3 '11 at 12:53
    
Yes, sorry I forgot to write that $g$ is also $L^1$ function, –  user6574 Feb 3 '11 at 13:11
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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. –  Willie Wong Feb 3 '11 at 17:24
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1 Answer 1

If $a$ tends to zero at infinity, then $T$ is compact (e.g., by a little result in an old paper of mine). In that case, I rather doubt that there is any simple characterization of the eigenvalues in terms of the functions $a$ and $g$.

However, as you conjecture, it is true that $T$ has positive spectrum if $a$ and the Fourier transform of $g$ are positive!

Proof: Suppose $\lambda$ is in the spectrum of $T$. Then either

(i) there is a sequence $f_n$ in $L^2$ with $\|f_n\|=1$ for all $n$ and $\|(T-\lambda)f_n\|\to0$, or

(ii) the range of $T-\lambda$ is closed but not dense in $L^2$.

In case (i) we have $$ ((T-\lambda)f_n,af_n) = \int_{\mathbb R^n} \hat g|\widehat{af}_n|^2 - \lambda \int_{\mathbb R^n} a|f_n|^2 \to 0. $$ If either $\lambda$ is negative or has nonzero imaginary part, it follows $\|\sqrt{a}f_n\|\to0$, hence $\|Tf_n\|\to0$, and hence $|\lambda|=\|\lambda f_n\|\to0$, a contradiction.

In case (ii) there must exist some nonzero $u\in L^2$ orthogonal to the range of $T-\lambda$: $$ 0=((T-\lambda)f,u) \quad\mbox{for all $f\in L^2$} $$ Note that the convolution operator given by $Gf=g*f$ is self-adjoint due to the positivity of $\hat g$. Taking $f=Gu$ we find $$ 0= ((T-\lambda)Gu,u) = \int_{\mathbb R^n}a|Gu|^2 -\lambda\int_{\mathbb R^n}\hat g|\hat u|^2. $$ If either $\lambda$ is negative or has nonzero imaginary part, it follows $\hat g\hat u=0$ a.e., so $Gu=0$. But then for any $f$ we have $(Tf,u)=(af,Gu)=0=\lambda(f,u)$ and so $u=0$, a contradiction.

Consequently $\lambda\ge0$.


Actually, this argument easily generalizes to show that for any product $T=AB$ of two non-negative bounded self-adjoint operators in a Hilbert space, the spectrum of $T$ is non-negative. Probably this is well-known --- e.g., it follows from results described in

Hladnik, Milan, Omladic, Matjaz, Spectrum of the product of operators. Proc. Amer. Math. Soc. 102 (1988), no. 2, 300–302.

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