Limit of the supremum norm of the convolution of a Gauss kernel with a certain function Let $f: \mathbb R^n \to \mathbb R$ be a continuous function with $\lim_{|x| \to \infty} f(x) = 0$, and let $g_t: \mathbb R^n \to \mathbb R, g_t(x) = \frac 1{(4 \pi t)^{\frac n 2}} e^{- \frac{|x|^2}{4 t}}$ be the Gauß kernel (for $t > 0$).
I now want to show that 
$$\lim_{t \to \infty} \|g_t * f \|_\infty = 0$$
where $*$ is the convolusion.
Now I know some basic properties about the Gauß kernel like $g_t(x) = \frac 1{t^{\frac n 2}} g_1 \left( \frac x{\sqrt t} \right)$, or $\int_{\mathbb R^n} g_t(x) d x = 1$. Using this perspective, it seems plausible that the above result holds, but I'm a bit lost on how to formally show it. I tried to write out the convolution 
$$g_t * f(x) = \int_{\mathbb R^n} g_t(x - y) f(y) d y = \int_{\mathbb R^n} \frac 1{(4 \pi t)^{\frac n 2}} e^{- \frac{|x - y|^2}{4 t}} f(y) d y$$
but I'm not sure how I can get that the $\|\cdot\|_\infty$-norm of that expression now becomes arbitrarily small.
I found this question which seems similar which might be of relevance, but since the function family that we take the limit of (and also the desired result) is different, I wasn't able to see how it could help me with my question (but I figured it might not hurt to link the thread nevertheless).
 A: The two relevant facts about $g_t$ are: 
$$
\int_{\mathbb{R}^n} g_t(x)\,dx = 1 \tag1
$$
$$
\text{For every } R, \quad  \sup_{y\in \mathbb{R}^n}\int_{B_R(y)} g_t(x)\,dx \to 0 \quad \text{as } t\to\infty  
\tag2
$$
Here $B_R(y)=\{x : |x-y|\le R\}$.  
To prove (2), first note that 
$$
\text{For every } R, \quad  \int_{B_R(0)} g_t(x)\,dx \to 0 \quad \text{as } t\to\infty  
\tag3
$$
which is proved by changing the variable so that $g_t$ becomes the standard Gaussian; then the radius of the domain of integration becomes a function of $t$, and shrinks as $t\to\infty$. 
To show (2), by rotational symmetry of Gaussian we may assume $y=(t, 0, 0, \dots, 0)$ with $t>0$. If $t\le R$, then $B_R(y) \subset B_{2R}(0)$, and the result follows from (3). Otherwise, note that $B_R(y)$ is contained in half-space $x_1>0$, and that $g_t(x_1, x_2, \dots, x_n)$ is a decreasing function of $x_1$ for $x_1>0$. These facts imply 
$$
\int_{B_R(y)} g_t(x)\,dx  \le  \int_{B_R(y')} g_t(x)\,dx 
$$ 
where $y'=(R, 0, \dots, 0)$. Since $B_R(y')\subset B_{2R}(0)$, the result again follows from (3). 

Now that we have (1) and (2), the proof is easy. Given $\epsilon>0$, fix $R$ such that $|f(x)|<\epsilon$ when $|x|>R$. By Property (1),
$$
\left| \int_{\mathbb{R}^n \setminus B_R} f(z) g_t(y-z) \,dz \right| 
\le \epsilon \int_{\mathbb{R}^n \setminus B_R} g_t(y-z) \,dz \le \epsilon
$$
By Property (2), 
$$
\left| \int_{B_R} f(z) g_t(y-z) \,dz \right| 
\le \sup_{\mathbb{R}^n}|f| \int_{B_R(y)} g_t(x) \,dx \le \epsilon
$$
if $t$ is sufficiently large, independently of $y$. Combining the two, we get 
$$
|(f*g_t)(y)|\le 2\epsilon
$$
for all $y$, for all sufficiently large $t$.
