Convolution with Gaussian, without distribution theory, part 2 I only know basic $L^p$ theory (nothing about distributions) and am trying to prove the following:
Let $t>0$, $f\in L^{p}(\mathbb{R}^n,m)$, $\Gamma(t,x)=(4\pi t)^{-n/2}e^{-|x|^2/4t}$ and
$$
u(t,x)=\int_{\mathbb{R}^n}\Gamma(t,x-y)f(y)dy.
$$
If $1< p<\infty$, then $u(t,\cdot)\to 0$ in $L^p$ as $t\to \infty$.
I've shown that $||u(t,\cdot)||_p\leq ||f||_p$ for $1\leq p\le\infty$ but haven't gotten anywhere from here. Could someone please give some hints or suggest a reference (these seem to be very well-known and widely applicable results).
(This is the first, and this is the third part of my question.) 
 A: As the solution available here:
Let $f \in L^1\cap L^p$ at first. Then, by the Young's inequality, 
$$ \|u(t,\cdot)\|_{\infty} \le \|\Gamma(t,\cdot)\|_{p'} \|f\|_p $$
And
$$ \|u(t,\cdot)\|_{1} \le \|f\|_1 $$
By $L^p$-Interpolation, we've that
$$ \|u(t,\cdot)\|_p \le \|u(t,\cdot)\|_1^{1/p}\|u(t,\cdot)\|_{\infty}^{(p-1)/p}\le C(f,p)\|\Gamma(t,\cdot)\|_{p'} ^{(p-1)/p} $$
Where $p'^{-1} + p^{-1} = 1$. But 
$$\|\Gamma(t,\cdot)\|_{r}^r = \frac{1}{(4\pi t)^{nr/2}}\int e^{\frac{-r|x|^2}{4t}}dx = \frac{1}{(4\pi t)^{nr/2}} c(n) \int_{0}^{\infty} s^{n-1}e^{-\frac{rs^2}{4t}}ds = \\ \frac{c(n)}{(4\pi t)^{nr/2}} \sqrt{\frac{4t}{r}}^n \int_0^{\infty} w^{n-1}e^{-w^2} dw $$
Which shows that, for $r>1$, $\|\Gamma(t,\cdot)\|_r \rightarrow 0$ as $t \rightarrow \infty$. Thus, we've proved the Theorem for $f \in L^1 \cap L^p$. 
For the general case, let $g \in L^1\cap L^p$ be such that $\|g-f\|_p \le \varepsilon$, and then 
$$ \|u_f(t,\cdot)\|_p \le \|u_g(t,\cdot)-u_f(t,\cdot)\|_p + \|u_g(t,\cdot)\|_p \le \varepsilon + \varepsilon$$ 
If $t$ is big enough, where we've used once again that $\|u(t,\cdot)\|_p\le \|f\|_p$. So, the Theorem is proved. 
