$L^2$ decay of solutions of heat equation Consider heat equation: in $\Omega \subset \mathbb R^n:$
\begin{equation*}
\begin{cases}
\begin{aligned}
u_t - \Delta u &=0 \\
u(0,x) &= u_0 \\
u &= 0\quad\text{on }\partial \Omega
\end{aligned}
\end{cases}
\end{equation*}
I want to show $\|u(t)\|_{L^2}\rightarrow 0$ as $t \rightarrow \infty$.
So far, we know
$$\frac{d}{dt} \int_\Omega u^2 = 2\int uu_t = 2\int u\Delta u = -2\int|\nabla u|^2 \leq 0.$$
I doubt we can conclude from here that $\|u(t)\|_{L^2}\rightarrow 0$ as $t \rightarrow \infty$. The derivative of the function is negative and the function $\int_\Omega u^2$ is non-negative.
 A: You're right, this won't happen unless $u_0=0$ a.e. Here is a simple example.
Consider the domain $[0,1]\subset \mathbb{R}$, with $u_0=\sin(\pi x)$. The solution to the heat equation is easily calculated as
$$u=\sin(\pi x)e^{-\pi^2t}\;.$$
Since $\sin(\pi x)>0$ on the domain, we calculate the $L^2$-norm as
$$\int_0^1\sin^2(\pi x)e^{-2\pi^2t}dx=\frac{1}{2}e^{-2\pi^2t}.$$
As $t\to 0$ this gives you $1/2$.
This is intuitively clear in general if you think about it. The heat equation models the distribution of heat in the domain over time. If you ask for the $L^2$-norm of the solution to approach zero as $t\to 0$, you're really asking for the initial distribution to be identically zero.
Solution for edited question:
For the limit as $t\to \infty$, your calculation 
$$\frac{d}{dt}\int_{\Omega}u^2=-2\int \vert \nabla u \vert^2\leq 0$$
along with @Hans Engler's comment is enough to prove the claim. By Poincare's inequality, we have 
$$\int_{\Omega}\vert \nabla u \vert^2\geq C\int_{\Omega}u^2\;,$$
for some constant $C$ (depending only on $\Omega$). We therefore have
$$f^{\prime}(t)\leq -2Cf(t)\leq 0\;,$$
for $f(t)=\int_{\Omega}u^2>0$. Since $f(t)$ is decreasing and positive, it is bounded below. If there was a constant $A>0$ such that $f(t)\geq A$ then we would have
$$\infty<\lim_{t\to \infty}\int_0^{t}f(s)\leq -\lim_{t\to \infty}\frac{1}{2C}f(t)+f(0)\;,$$
which contradicts the fact that $f$ is bounded below. This implies $f(t)\to 0$ as $t\to \infty$
