Let $K(x): \mathbb R^d \to \mathbb R^d$, $K \in L^1(\mathbb R^d)$ such that $|K(x)| < M$
We define the integral operator $T$ as such:
$\displaystyle(Tf)(x) = \int_{\mathbb R^d}f(y)K(|x-y|)dy$
Show that the spectrum of $T$ spans $L^2(\mathbb R^d)$
It's very tempting to use the spectral theorem here. The theorem states that if $T$ is compact and self adjoint, then what we want to show, holds.
Thus I want to show that this operator is compact and self adjoint.
What I tried:
Define $\alpha (x,y) = K(|x-y|)$. Obviously we have $\alpha(x,y) = \bar\alpha(y,x)$ and $(Tf) = \int_{\mathbb R^d}f(y)\alpha(x,y)dy$ so this is a self adjoint operator.
I can't show it's bounded though. I need to show that $\int_{\mathbb R^d}\int_{\mathbb R^d}|\alpha(x,y)|^2dxdy < \infty$ right? But all I know is that $|\alpha|^2 < M^2$ and that doesn't help since $\int_{\mathbb R^d}\int_{\mathbb R^d}M^2dxdy = \infty$
What am I missing here?
Answer for the boundedness part (compactness is more involved): let $g(x)=K(|x|)$. Then $Tf(x)=(f*g)(x)$. Hence $\|Tf\|_2 \leq \|f\|_2\|g\|_1$ which shows that $T$ is a bounded opeartor with $\|T\| \leq \|g\|_1$.