Exponential decay estimate

Here is a solution more related to Evan's approach in the mentioned textbook. Let $u = u(x,t)$ be a solution of the PDE. Then using integration by parts and that $u_t = \Delta u$ we get

$$\begin{align}\frac{d}{dt} \left(\frac{1}{2} \|u\|_{L^2(U)}^2\right) &= \int_U u_tu\ dx = \int_U u \Delta u\ dx\\&= -\int_U |Du|^2 dx \overset{(\ast)}\leq - \lambda_1 \|u\|^2_{L^2(U)}\end{align}$$

where $(\ast)$ comes from Rayleigh's Formula

$$\lambda_1 = \underset{\substack {u \in H_0^1 (U)\\ u\neq 0}}\min \frac{B[u,u]}{\|u\|^2_{L^2(U)}} = \underset{\substack{u \in H_0^1 (U)\\ u\neq 0}} \min \frac{\int_U |Du|^2 dx}{\|u\|^2_{L^2(U)}} $$

Now let $\eta (s) = \|u(\cdot,s)\|^2_{L^2(U)}$. Then

$$\frac{d}{ds} \left(\eta(s) e^{2\lambda_1 s}\right) = e^{2\lambda_1 s} (\eta'(s) +2\lambda_1 \eta(s)) \leq 0$$ Integrating from $0$ to $t$ w.r.t. $s$ we obtain

$$\eta(t)e^{2\lambda_1 t} \leq \eta (0)$$

Since $\eta (0) = \|u (\cdot, 0)\|^2_{L^2(U)} = \|g\|^2_{L^2(U)}$ the result follows.

Write $$ u(x,t)=\sum_k c_k e^{- \lambda_k t} \varphi_k(x), $$ where $\lambda_k\leq \lambda_{k+1}$ are the eigenvalues of the laplacian with zero Dirichlet boundary conditions, $\varphi_k$ are the corresponding eigenfunctions and $$ c_k=\int_U g(x) \varphi_k(x) dx $$ are the Fourier coefficients of $g$ with respect to the orthonormal basis $(\varphi_k)_k$. Then we see, by Pythagoras' and Parseval's theorems, $$ \| u(\cdot, t)\|_{L^2(U)}^2 = \sum_k c_k^2 e^{-2\lambda_k t} \leq e^{-2 \lambda_1 t}\sum_k c_k^2 = e^{-2\lambda_1 t} \| g\|_{L^2(U)}^2, $$ where we also used the monotonicity of the eigenvalues in the second to last step. This is what you want.