uniform limit of integral of heat kernel For any $(x,t)\in \Bbb R^n\times(0,+\infty)$. Let $$K(x,t)=\frac 1{{(4\pi t)}^{\frac n2}}e^{-\frac {|x|^2}{4t}}$$ be the heat kernel and consider $$u(x,t)=\int_{\Bbb R^n}K(x-y,t)u_0(y)dy$$. 
Suppose $u_0$ is continuous in $\Bbb R^n$ and that $u_0(x)\to0$ uniformly as $x\to+\infty$
Prove that $t\to+ \infty$ $ u(x,t)=0$, uniformly in x.
I am preparing for qualifying exam so could you please answer this.
 A: Let $\epsilon >0$. As $u_0$ goes to zero uniformly as $x \to \infty$, there is $r= r(\epsilon)$ such that 
$$|u_0(x)| \leq \epsilon/2$$
for all $|x|\geq r$. Then 
$$u(x, t) =\int_{B_{r}} K(x-y, t) u_0(y) dy + \int_{\mathbb R^n \setminus B_{r}}K(x-y, t) u_0(y) dy$$
For the first term, as 
$$|K(x-y, t)| = \frac{1}{(4\pi t)^{\frac{n}{2}}} e^{-\frac{|x-y|^2}{4t}} \leq  \frac{1}{(4\pi t)^{\frac{n}{2}}} $$
Let $T$ be large so that 
$$ \int_{B_r} u_0(y)dy \leq (4\pi T)^{\frac{n}{2}} \frac{\epsilon}{2}$$
Thus $T$ depends only on $\epsilon$. The for all $t\geq T$, 
$$\bigg|\int_{B_{r}} K(x-y, t) u_0(y) dy \bigg| \leq  \epsilon/2\ .$$
For the second term, 
$$\bigg|\int_{\mathbb R^n \setminus B_{r}}K(x-y, t) u_0(y) dy \bigg| \leq \epsilon /2 \bigg| \int_{\mathbb R^n \setminus B_{r}}K(x-y, t) dy\bigg| \leq \epsilon/2$$
as 
$$ \int_{\mathbb R^n}K(x-y, t) dy = 1\ .$$
That is, for any $\epsilon>0$ there is $T(\epsilon)$ such that for all $x\in \mathbb R^n$, 
$$|u(x, t)| \leq \epsilon$$
whenever $t\geq T$. Wish you good luck with your qualifying exam!
