Consider $K(x, y)$, $f(x)$ Schwartz functions and $g(y)$ a tempered distribution. Suppose $$K(x, y) = K(y, x)$$ Define


$$h(t) = \int f(x - t) K(x, y) g(y - t) dx dy$$


It appears to me $h(t)$ is a Schwartz function. I'd be glad if anyone can prove / disprove it.


Thx!


Some thoughts so far:


The assumption of symmetry on K is probably redundant. If we take K to be $$K(x, y) = u(x)v(y)$$ for u, v Schwartz, we get $$h(t) = [\int f(x - t)u(x)dx][\int g(y - t)v(y)dy]$$ Now, the 2nd factor is probably smooth of slow (at most polynomial) growth whereas the 1st factor is Schwartz. $h(t)$ is thus Schwartz. Linear combinations of such "factorized" kernels are dense in the entire Schwartz space, but I still don't see how to deduce the claim from this.



A property of "Schwartz" quadratic forms
Let $H(x,y) = f(-x)\otimes g(-y)$ be a tempered distribution on the product space. Then it is well-known that the function $\tilde{h}(t,s) = H\*K(t,s) = \int H(t-x,s-y)K(x,y)$ is in $C^\infty \cap \mathcal{S}'$. (See, e.g. pg 25 of Stein-Weiss, *Introduction to Fourier Analysis on Euclidean Spaces*). 


Noting that $h(t) = \tilde{h}(t,t)$, it suffices to show that $h$ has rapid decay. This follows from the fact that the convolution of a tempered distribution with a Schwartz function has at most polynomial growth (ibid) so that $\sup\_t |\tilde{h}(t,s)| \leq A (1+ |s|)^M$ for some $A$ and $M$. So a small bit of computation tells you that $\tilde{h}(t,s)$ is bounded by the product of a rapidly decreasing envelope in $t$ and a polynomial in $s$. Therefore on its diagonal it is still rapidly decreasing. 


