Heat equation: Show $\int\limits_{\mathbb{R}^n} u(x,t)\,dx=\int\limits_{\mathbb{R}^n} g(x)\,dx$ for $t>0$ Assume $g\in C(\mathbb{R}^n)$, $g\in L^1(\mathbb{R}^n)$, $|g|\leq M$. Let $u$ be the bounded solution to 
\begin{gather*}
\Delta u -u_t=0 \text{ for } t>0, x\in\mathbb{R}^n, \\
u(x,0)=g(x) \text{ for } x\in\mathbb{R}^n
\end{gather*}
Show


*

*$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)|=0$

*$\int\limits_{\mathbb{R}^n} u(x,t)\,dx=\int\limits_{\mathbb{R}^n}
    g(x)\,dx$ for $t>0$


Proof of (a): Assume $g\in C(\mathbb{R}^n)$, $g\in L^1(\mathbb{R}^n)$, $|g|\leq M$. Let $u$ be a solution to the PDE above. Recall the solution to this PDE is
$$u(x,t)=\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy,$$
where $x\in \mathbb{R}^n$ and $t>0$. So,
\begin{equation*}
\begin{aligned}
|u(x,t)| 
& =\left|\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy\right| \\
& \leq \dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \left|\exp\left[ -\dfrac{|x-y|^2}{4t}\right]\right| \cdot \left|g(y)\right|\,dy \\
& \leq \dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right]\cdot M\,dy
\end{aligned}
\end{equation*}
Recall also the formula for the integral of a Gaussian function
$$\int_{\mathbb{R}} a\exp\left[-\dfrac{(x-b)^2}{2c^2}\right]\,dx=\sqrt{2a} \cdot |c|\cdot \sqrt{\pi}.$$
We can extend this result to $\mathbb{R}^n$ and use it in our inequality above.
\begin{equation*}
\begin{aligned}
|u(x,t)| 
& \leq \dfrac{M}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|y-x|^2}{2(\sqrt{2t})^2}\right]\,dy \\
& = \dfrac{M}{(4\pi t)^{n/2}} \cdot \sqrt{2\pi} \cdot|\sqrt{2t}| 
= \dfrac{M}{(4\pi t)^{n/2}} \cdot \sqrt{4\pi t} 
= \dfrac{M}{(4\pi t)^{(n-1)/2}} 
\end{aligned}
\end{equation*}
Hence 
$$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)| \leq 
\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} \dfrac{M}{(4\pi t)^{(n-1)/2}}
= \lim\limits_{t\to\infty} \dfrac{M}{(4\pi t)^{(n-1)/2}} =0
$$
i.e.
$$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)| =0
$$
Proof of (b):
\begin{equation*}
\begin{aligned}
\int\limits_{\mathbb{R}^n} u(x,t)\,dx 
& = \int\limits_{\mathbb{R}^n}\left[\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy \right]\,dx \\
& = \dfrac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}\left[ \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] \,dx \right]g(y)\,dy  \\
& = \dfrac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}\left[ \sqrt{2\pi} \cdot|\sqrt{2t}|\right]g(y)\,dy  \\
& = \dfrac{1}{(4\pi t)^{(n-1)/2}} \int\limits_{\mathbb{R}^n}
g(y)\,dy  \\
& = \dfrac{1}{(4\pi t)^{(n-1)/2}} \int\limits_{\mathbb{R}^n}
g(x)\,dx  \\
\end{aligned}
\end{equation*}
Questions:


*

*Is my proof of (a) correct?

*I know I am very close to finishing the proof of b. What am I missing?

 A: If $g(x) = M$ then you can easily check that the solution is $u(x,t) \equiv M$ for which $u\not\to 0$ as $t\to\infty$ which contradicts your result. This does not contradict the original claim since $g\not\in L^1$, but this assumption is not used in your derivation. Your mistake is in computing the integral, and if you do the calculation right you will find
$$|u| \leq \int K(x,y,t)|g(y)|\,{\rm d}y \leq M\int K(x,y,t)\,{\rm d}y = M$$
since the integral of the Heat kernel is unity, i.e. $\int K(x,y,t){\rm d}y = 1$. Thus this approach is not good enough to show $u \to 0$, only that $u$ remains bounded.
Since both $g\in L^1$ and the properties of the heat kernel is critical for the claim to hold true it's natural to apply some integral inequality to try to bound the integral. Applying Cauchy-Schwarz we find
$$|u|^2 = \left(\int K(x,y,t)g(y)\,{\rm d}y\right)^2\leq \left(\int K^2(x,y,t)\,{\rm d}y\right)\left(\int g^2(y)\,{\rm d}y\right)$$
Now you can check that $K^2(x,y,t)= \frac{1}{\sqrt{8\pi t}^{n/2}}K(x,y,t/2)$ which gives you the first term and for the second term note that $g^2(y) \leq M|g(y)|$ and use the fact that $g\in L^1$.
A: Hint: For both parts, you are forgetting the fact that the integration is taking place over $\mathbb{R}^n.$ This is why you end up having the extra exponent $n-1$ on the bottom. Thus, the Gaussian formula in $\mathbb{R}^n$ takes a slightly different form. Other than that, your proof has no flaw. 
Also, for part $(b)$, an easier way of doing it would be integrating the LHS with respect to $t$ and use integration under the integral. That might need some justification but usually trivial with the help of Lebesgue's differentiating under integral theorem. 
