The problem is
Suppose $\Omega\subset\mathbb{R}^{n}$ is a bounded open set. Consider the initial-boundary value problem $$\begin{cases} \partial_{t}u(x,t)-\Delta u(x,t)=0 & \text{in}\ \Omega\times(0,\infty)\\ u=u_{0} & \text{on}\ \Omega\times\{0\}\\ \dfrac{\partial}{\partial\nu}u(x,t)=0 & \text{on}\ \partial\Omega\times(0,\infty).\\ \end{cases}$$ And assume the first and second order derivatives of its solution $u$ exist in $\overline{\Omega}\times[0,\infty)$. Then use Arzelà-Ascoli to show $$\lim_{t\rightarrow\infty}\lvert\nabla u(x,t)\rvert=0\quad \forall\,x\in\Omega.$$
I tried using the energy method. Let $$E(t)=\int_{\Omega}u^{2}(x,t)\,dx.$$It is easy to see $E’(t)\leq0$ and $\lim\limits_{t\to\infty}E’(t)=0$. But I’m confused how to find a sequence to use the theorem. Any suggestion is appreciated.
If $\lim_{t\rightarrow\infty}\lvert\nabla u(x,t)\rvert\neq0$, then there exists a sequence $(t_{n})_{n}$ and a constant $\alpha>0$ such that $t_{n}\rightarrow\infty$ and $\lVert\nabla u(x,t_{n})\rVert_{L^{\infty}}\geq\alpha,\ n\rightarrow\infty$. Set $u_{n}(x)=\lvert\nabla u(x,t_{n})\rvert$. Since the second order derivative of $u$ exists on $\overline{\Omega}\times[0,\infty)$, we know $u_{n}$ is continuous on the compact set $\overline{\Omega}$. Moreover, $\lvert\Delta u(x,t_n)\rvert=\lvert\partial_t u(x,t_n)\rvert$ is also uniformly bounded on $\overline{\Omega}$, thus $(u_n)$ is equi-continuous. Consequently, by Arzelar-Ascoli, there exists a subsequence $(t_{n_{k}})_{k}$ converges to a continuous function $g$ uniformly on $\overline{\Omega}$. But$$\int_{\Omega}g^{2}(x)\,dx=\lim_{k\rightarrow\infty}\int_{\Omega}\lvert\nabla u(x,t_{n_{k}})\rvert^{2}\,dx=0.$$So $g\equiv0$, a contradiction.
Is it right?