Using spectral theorem for integral Hilbert-Schmidt (maybe) operator Let $T: L^2(\mathbb R^d) \to L^2(\mathbb R^d)$ be defined by $\displaystyle (Tf)(x) = \int_{\mathbb R^d}f(y)K(|x-y|)dy$ where $K(|z|) = \frac{1}{1+|z|^\alpha}$ where $z \in \mathbb R^d$ and $\alpha > \frac{d}{2}$.
Show that the spectrum of $T$ (the set of eigenvectors) spans $L^2(\mathbb R^d)$
What I tried (false proof):
If we can show $T$ is a Hilbert-Schmidt integral operator, we can use the spectral theorem.
Notice that $\displaystyle \int \frac{1}{(1+|z|^\alpha)^2}dz = \int_{|z| \leq 1} \frac{1}{(1+|z|^\alpha)^2}dz + \int_{|z| > 1} \frac{1}{(1+|z|^\alpha)^2}dz \leq \\ \displaystyle \int_{|z| \leq 1} \frac{1}{(1+|z|^\alpha)^2}dz + \int_{|z| > 1} \frac{1}{(|z|^\alpha)^2}dz$.
$\displaystyle \int_{|z| \leq 1} \frac{1}{(1+|z|^\alpha)^2}dz \leq \int_{|z| \leq 1}1dz = \pi < \infty$ so this part is not problematic. Let's analyze the other part of the integral.
Define $A_k = \{z: 2^k < |z| \leq 2^{k+1}\}$. Then we have $\displaystyle \{z: |z| > 1\} = \bigcup_{k=0}^{\infty}A_k$, and because $A_k$ is simply every element of $A_0$ but multiplied by $2^k$, from dilation property of measures $m(A_k) = (2^k)^d m(A_0) = 3\cdot \pi \cdot 2^{kd}$
Finally then, \begin{align*} \int_{|z| > 1} \frac{1}{(|z|^\alpha)^2}dz &= \sum_{k=0}^{\infty}\int_{A_k}\frac{1}{(|z|^\alpha)^2}dz \leq \sum_{k=0}^{\infty}\int_{A_k}\frac{1}{(2^k)^{2\alpha}}dz \\
&= \sum_{k=0}^{\infty}2^{-2k\alpha}m(A_k) = 3\pi \sum_{k=0}^{\infty}2^{kd-2k\alpha} \\
&< \infty\end{align*}
This sum converges because $2\alpha > d$. Hence, $K(|z|) \in L^2(\mathbb R^d)$.
Now the false part
Because $K(|z|) \in L^2(\mathbb R^d)$, we can say that $K(|x-y|) \in L^2(\mathbb R^d \times \mathbb R^d)$. [Not really. This is actually false. Infact, $\iint |K(|x-y|)|^2dxdy = \infty]$
So we can say that $T$ is hilbert schmidt, the fact that it's self adjoint is trivial, and by the spectral theorem we are done.
As I said, there's a problem with this proof. Is there a way to make it correct? Is the problem statement even correct?
 A: As MaoWao, I'm inclined to think this operator has no eigenvectors. My intuition is that an operator with a radial kernel must be a function of the Laplacian, that is $T = f(-\Delta)$ for some function $f$. See Theorem 7.22 in Teschl's freely available book here. You can try to compute this $f$ for your $T$. If it is invertible and if $T$ had an eigenvector, then $-\Delta$ would have an eigenvector, which is false since $-\Delta$ has purely absolutely continuous spectrum.
A: As remarked by DisintegratingByParts in the comments, the spectrum contains the eigenvalues (and possibly other spectral values), not the eigenvectors. As suggested by the OP, I'll read the question as: Are the eigenfunctions of $T$ total in $L^2(\mathbb{R}^d)$?
The answer is no, at least for $d=1$ and $\alpha=2$. In this case, the eigenfunction equation $Tf=\lambda f$ corresponds under Fourier transform to
$$
e^{-|\xi|}\hat f(\xi)=\lambda \hat f(\xi).
$$
This equation clearly does not have a non-trivial solution in $L^2$.
I don't expect the answer to be different for other values of $d$ of $\alpha$, but I don't have a proof (there don't seem to be any explicit formulas for the Fourier transform of $K$ in this case). The only multiplication operators with a total set of eigenfunctions are (possibly infinite) linear combinations of indicator functions, so it remains to show that the Fourier transform of $K$ cannot take this form. This seems at least plausible, since you have some decay at $\infty$.
