I came across the following remark in my reading.
Remark. If the initial data is in $L^{1}$, then heat equation solutions approach $0$ within the sup-norm. That is, if $K(x,t)$ is the heat kernel and $f \in L^1(\mathbb{R}^n)$, then $$\lim_{t \to \infty} \int_{\mathbb{R}^n} K(x - y, t)f(y)dy = 0,$$ uniformly.
I can picture this working if $f$ is uniformly continuous in $\mathbb{R}^n$ and $\lim_{x \to \infty} f(x) = 0$. Does $f \in L^1(\mathbb{R}^n)$ imply this somehow, and I'm not seeing this? I'm relatively new to $L^{p}$-spaces.
If so, then for $\epsilon > 0$, we can find some radius $r_{\epsilon} > 0$ such that $|f(y)| < \epsilon/2$.
Then we can bound the heat kernel, and use the fact that, $\int K(x-y, t)dy = 1$ to conclude, more or less.
This is quite straightforward from the definition. In $n$ dimensions, the heat kernel is given by $$ K(x-y,t) = \frac{1}{(4\pi t)^{n/2}}e^{-|x-y|^2/4t}. $$ As $e^{-x} \leq 1$ for $x \geq 0$ this implies that $|K(x-y,t)| \leq {(4\pi t)^{-n/2}}$ and therefore $$\left|\int_{\mathbb{R}^n} K(x - y, t)f(y)dy \right| \leq {(4\pi t)^{-n/2}} \int_{\mathbb{R}^n} |f(y)|dy,$$ which converges to $0$ as $t \to \infty$ uniformly in $x$, as soon as $f$ is in $L^1 (\mathbb{R}^n)$.