Sufficient conditions on integration kernel for continuity of the integral operator Suppose that we have a measure $d\mu(v)=e^{-|v|^2}dv$ on $\Bbb R^d$. We define a linear operator 
$$T[f](u)=\int_{\Bbb R^d} |u-v|^\beta f(v) d\mu(v).$$
I want to establish conditons on $\beta\in\Bbb R$ so the operator $T$ is continuous on $L^2(d\mu)$.
Constant functions belong to $L^2(d\mu)$; for $\beta\le -d $ the operator $T$ is not defined for constant functions, hence, necessarily we need $\beta>-d$.
If $\beta>-d/2$, then the integration kernel $k(u,v)=|u-v|^\beta \in L^2(d\mu(u)\,d\mu(v))$, and hence by Hilbert-Schmidt criterion the operator $T$ is compact and continuous.
I'm interested in the case $\beta\in(-d,d/2]$. Is there a criterion that allows to show continuity/compactness of the operator $T$? I'll be glad to hear all suggestions.
 A: $T$ is bounded for all $\beta > -d$. 
The strategy is to show that $T$ is bounded both as a function $L^1(d\mu) \rightarrow L^1(d\mu)$ and as a function $L^\infty \rightarrow L^\infty$; the Riesz-Thorin theorem then shows that it is bounded $L^p(d\mu) \rightarrow L^p(d\mu)$ for any $p$.
We just need the following estimate; everything else is standard.
Lemma: Fix $\beta$ with $-d < \beta < 0$. Then 
$$\int \lvert u - v \rvert^\beta d\mu(v) \leq C < \infty$$
where the bound $C$ is independent of $u$.
Proof: Let $F(u) = \int \lvert u - v \rvert^\beta d\mu(v)$.
Since $\beta < 0$, we have $\lvert u - v \rvert^\beta \leq 1 + \mathbf 1_{B_1(u)} \lvert u-v\rvert^\beta$, where $\mathbf 1_{B_1(u)}$ is the indicator function for the ball of radius 1 around $u$. Thus
$$F(u) \leq \int_{\mathbb R^d} d\mu(v) + \int_{B_1(u)} \lvert u-v\rvert^\beta e^{-\lvert v\rvert^2} dv\text.$$
The first term is clearly independent of $u$. The second term is bounded above by $\int_{B_1(u)} \lvert u-v\rvert^\beta dv$, which is finite because $\beta > -d$, and does not depend on $u$. This shows that $F(u)$ is bounded, and finishes the proof.
The lemma immediately shows that $T$ is bounded as an operator $L^\infty \rightarrow L^\infty$, with operator norm at most $C$.
Using Fubini's theorem, we have
\begin{align} \|T[f]\|_{L^1(d\mu)} &\leq \int \int \lvert u-v \rvert^\beta \lvert f(v) \rvert d\mu(v) d\mu(u) \\
&= \int \int \lvert u-v \rvert^\beta d\mu(u) \lvert f(v) \rvert d\mu(v) \\
&\leq C \int \lvert f(v) \rvert d\mu(v) \\
&= C \|f\|_{L^1(d\mu)}\text,
\end{align}
so $T$ is also bounded $L^1(d\mu) \rightarrow L^1(d\mu)$.
The Riesz-Thorin Theorem then shows that $T$ is bounded $L^p(d\mu) \rightarrow L^p(d\mu)$ for any $p$ with $1\leq p \leq \infty$, in particular for $p = 2$.
