Assume we have $u \in C^3(\mathbb{R}^n \times (0,\infty))$ satisfying the heat equation $$ \Delta u(x,t) = \partial_t u(x,t)$$ and a function $u_0:\mathbb{R}^n \to \mathbb{R}$ with unknown regularity (at least two times differentiable if necessary) such that $$ \sup_{x \in B(0,R)} \left\lvert u(x,t) – u_0(x) \right\rvert \xrightarrow{t \to \infty} 0 \quad \forall R>0.$$
I need to show that $\Delta u_0 = 0$ on the whole $\mathbb{R}^n$.
My first approach was the mean value theorem for functions satisfying the heat equation. Then, it is enough to show that $$ \frac{1}{4r^n} \int_{E(x,t;r)} u(y,s) \frac{\left| x-y \right|^2}{(s-t)^2} \, dy \, dt \xrightarrow{t \to \infty} \frac{1}{\left| B(x,R) \right|} \int_{B(x,R)} u_0(y) \, dy$$
for every $R$ and some $r$ (may be the same), where $$E(x,t;r) = \{ (y, s) \in \mathbb{R^{n+1}}\ :\ s \le t,\ \Phi(x-y,t-s)\ge\frac{1}{r^n}\}$$ are the famous heat balls and $\Phi$ the fundamental solution of the heat equation.
I'm not sure whether this approach leads to the desired result. Even though, I have no idea how to proceed with the heat balls here.
Best Answer
Let us fix $R>0$ first. Let $\phi \in C_c^\infty(B(0,R))$ be an arbitrary function. Then multiplying both sides of the equation, integrating in $B(0,R)$ and using one of Green's formulas, one has $$ \int_{B(0,R)}(u(x,t)-u_0(x))\Delta\phi(x)dx+\int_{B(0,R)}\Delta u_0(x)\phi(x)dx=\partial_t\int_{B(0,R)}(u(x,t)-u_0(x))\phi(x)dx. $$ Integrating from $s$ to $2s$, one has $$ \int_s^{2s}\int_{B(0,R)}(u(x,t)-u_0(x))\Delta\phi(x)dxdt+s\int_{B(0,R)}\Delta u_0(x)\phi(x)dx=\int_{B(0,R)}(u(x,2s)-u_0(x))\phi(x)dx-\int_{B(0,R)}(u(x,s)-u_0(x))\phi(x)dx. $$ Letting $s\to\infty$, one can easily derive $$ \int_{B(0,R)}\Delta u_0(x)\phi(x)dx=0. $$ By the arbitraryness of $\phi$, one has $$ \Delta u_0(x)=0 \text{ in }B(0,R) $$ for any $R>0$. So $\Delta u_0(x)=0$ for all $x$ in $\mathbb{R}^n$