[proofplan] Multiply the heat equation $\partial_t u = \Delta u$ by $u$, integrate over $\Omega$, and apply Green's first identity to convert the $\int u\,\Delta u$ term into $-\int |\nabla u|^2$ plus a boundary integral that vanishes by the Dirichlet condition $u = 0$ on $\partial\Omega$. [/proofplan] [step:Differentiate the energy $E(t)$ under the integral sign] Define $E(t) := \frac{1}{2}\int_\Omega u(x, t)^2 \, d\mathcal{L}^n(x)$. Since $u \in C^2(\overline{\Omega}_T)$, differentiation under the integral sign is justified: \begin{align*} \frac{dE}{dt} = \int_\Omega u(x, t)\,\partial_t u(x, t) \, d\mathcal{L}^n(x). \end{align*} [/step] [step:Substitute $\partial_t u = \Delta u$ and apply Green's first identity] Since $\partial_t u = \Delta u$ in $\Omega_T$: \begin{align*} \frac{dE}{dt} = \int_\Omega u\,\Delta u \, d\mathcal{L}^n. \end{align*} Apply Green's first identity $\int_\Omega u\,\Delta u \, d\mathcal{L}^n = -\int_\Omega |\nabla u|^2 \, d\mathcal{L}^n + \int_{\partial\Omega} u\,\frac{\partial u}{\partial\nu} \, d\mathcal{H}^{n-1}$ to obtain: \begin{align*} \frac{dE}{dt} = -\int_\Omega |\nabla u|^2 \, d\mathcal{L}^n + \int_{\partial\Omega} u\,\frac{\partial u}{\partial\nu} \, d\mathcal{H}^{n-1}. \end{align*} Since $u = 0$ on $\partial\Omega \times [0, T]$, the boundary integral vanishes: \begin{align*} \frac{dE}{dt} = -\int_\Omega |\nabla u|^2 \, d\mathcal{L}^n \leq 0. \end{align*} [/step] [step:Conclude monotonicity and the $L^2$ bound] The energy $E(t)$ is non-increasing since $dE/dt \leq 0$. In particular, $E(t) \leq E(0)$ for all $t \in [0, T]$, which reads: \begin{align*} \frac{1}{2}\int_\Omega u(x, t)^2 \, d\mathcal{L}^n(x) \leq \frac{1}{2}\int_\Omega g(x)^2 \, d\mathcal{L}^n(x), \end{align*} giving $\|u(\cdot, t)\|_{L^2(\Omega)} \leq \|g\|_{L^2(\Omega)}$. The inequality $dE/dt \leq 0$ is strict unless $\nabla u \equiv 0$ on $\Omega$, i.e., unless $u$ is spatially constant and hence identically zero by the boundary condition. [/step]