I'm trying to solve an exercise on Folland's book of real analysis on the part of fourier transforms and its applications on PDE's, when working with the heat equation one may show that given the heat kernel $$G_t(x)={(4\pi t)}^{-n/2}\exp\left( - \frac{\mid x \mid ^2}{4t}\right)$$ and some function $f \in L^1_{loc}(\mathbb{R}^n)$ s.t.
$$|f(x)| \leq C_\epsilon \exp(\epsilon \mid x \mid^2) $$ for every $\epsilon > 0$.
I've shown that the convolution $u(x,t) = f \ast G_t (x)$ is well defined and trying to show now that this function solves the equation $(\partial_t - \Delta)u = 0 $ on $\mathbb{R}^n \times (0, \infty)$.
It is easy to show that $G_t$ satisfies this equation, but is it possible to pass with the derivative trough the convolution s.t.
$$ (\partial_t - \Delta)u(x)= (\partial_t - \Delta)\int f(y)G_t(x - y) dy = \int (\partial_t - \Delta)f(y)G_t(x - y) dy = f \ast (\partial_t - \Delta)G_t (x) = 0$$
If so this part of the exercise will be done, but i don't have this kind of result as $G_t$ does not have compact support, so how can i show this?
Yes, it is possible to take the derivatives inside the integral. Compact support is not really necessary for this type of proof - rapid spatial decay of $G$ and its derivatives is good enough. To justify the operation, do it manually by taking difference quotients: for instance, $$ \partial_t \int f(y)G_t(x-y)~dy = \lim_{h\to 0} \frac{1}{h}\left[ \int f(y)G_{t+h}(x-y)~dy - \int f(y)G_t(x-y)~dy\right]. $$ You want to show that the limit on the right is well-defined and equal to $$ \int f(y)\partial_t G_t(x-y)~dy. $$ This can be done using dominated convergence, using the fact that $f\in L_{\operatorname{loc}}^1$ and $\partial_t G_t$ decays rapidly in space to find an appropriate integrable dominating function. You would do similarly for the Laplacian, taking one derivative at a time.