Let $$u(x,t) = \frac{1}{\pi^{N/2}}\int_{\mathbb{R}^N} u_0(x-2\sqrt{t}y)e^{-|y|^2}\ dy$$
Show that if there are constants $M, \delta>0$ such that $$|u_0(x)|\le Me^{-\delta|x|^2}, x\in\mathbb{R}^N$$ then $$|u(x,t)|\le M(1+4\delta t)^{-N/2}e^{-\delta|x|^2/(1+4\delta t)}$$ Conclude that if $u_0$ has compact support, then $\lim_{t\to \infty} u(x,t) = 0$ uniformly
$$|u(x,t)| = \frac{1}{\pi^{N/2}}\left|\int_{\mathbb{R}^N} u_0(x-2\sqrt{t}y)e^{-|y|^2}\ dy\right|\le \frac{1}{\pi^{N/2}}\int_{\mathbb{R}^N} \left|u_0(x-2\sqrt{t}y)e^{-|y|^2}\right|\ dy $$
So I need to find a bound for $\left|u_0(x-2\sqrt{t}y)e^{-|y|^2}\right|$ using $|u_0(x)|\le Me^{-\delta|x|^2}$
Ok,
$$\left|u_0(x-2\sqrt{t}y)e^{-|y|^2}\right|\le Me^{-\delta|x|^2}e^{-|y|^2} = Me^{-\delta|x|^2-|y|^2}$$
I tried working with $-\delta|x|^2-|y|^2$ and using $|a|^2 -|b|^2 \le |a-b|^2$ or something like that but it looks like it won't work.
UPDATE:
the right bound is $|u_{0}(x-2\sqrt{t}y)| \leq M e^{-\delta |x-2\sqrt{t}y|^2}$ so we get
$$|u(x,t)|\le \frac{1}{\pi^{N/2}}\int_{\mathbb{R}^N}|u_{0}(x-2\sqrt{t}y)e^{-|y|^2}|dy \leq \frac{1}{\pi^{N/2}}\int_{\mathbb{R}^N}M e^{-\delta |x-2\sqrt{t}y|^2} e^{-|y|^2}dy =\\ \frac{1}{\pi^{N/2}}\int_{\mathbb{R}^N}M e^{-\delta |x-2\sqrt{t}y|^2 -|y|^2}dy $$