Let $f: \mathbb{R^n} \to \mathbb{R}$ be continuous and of compact support. For $x \in \mathbb{R^n}$ and $t>0$ consider,
$$u(x,t) = (4 \pi t)^{\frac{-n}{2}} \int_{\mathbb{R}^{n}} e^{\frac{-\|x-y\|^2}{4t}} f(y)dy.$$
I need to show that $$\displaystyle\lim_{t\rightarrow \infty}u(x,t)=0$$ uniformly in $x$.
Since $(4 \pi t)^{\frac{-n}{2}} \rightarrow 0 $ as $ t \rightarrow \infty$, it is enough to integrate into the support of function f and use the fact that the exponential is bounded?
Since, f is be continuous and of compact support we have that $f\in L^1(\Bbb R^n)$. Now observe that for all $x,y\in\Bbb R^n $ we have, $$0\le e^\frac {-\|x-y\|^2}{4t} \le 1$$ Thus, $$\color{red}{\lim_{t\to\infty}\sup_{x\in\Bbb R^n}|u(x,t)|}\le \lim_{t\to\infty} \frac 1{(4 \pi t)^{\frac{-n}{2}}} \int_\Bbb R e^\frac {-\|x-y\|^2}{4kt}|f(y)|dy \\=\lim_{t\to\infty} \frac 1{(4 \pi t)^{\frac{-n}{2}}} \int_\Bbb R|f(y)|dy = \lim_{t\to\infty} \frac {\|f\|_1}{(4 \pi t)^{\frac{-n}{2}}} =0 $$
Which prove the uniform limit.