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 1st factor is probably smooth of slow (at most polynomial) growth whereas the 2nd 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.

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.

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!

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 1st factor is probably smooth of slow (at most polynomial) growth whereas the 2nd 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.