Consider the heat equation defined as
$$\begin{cases} \partial_{t}u(x,t)=\Delta_{x}u(x,t) &\mbox{ } t>0\mbox{, } u\in U \\ u(x,t)=0 &\mbox{ } t\ge 0\mbox{, } x\in\partial U \\ u(x,0)=g(x) &\mbox{ } x\in U \end{cases}$$
posed on a bounded connected open subset $U\subset\mathbb{R}^{n}$ that has a $C^{2}$-smooth boundary $\partial U$.
We know that $\exists$ a $C_{0}$-semigroup $(T(t))_{t\ge 0}$ on $L^{2}(U)$ generated by the linear operator
$$A:D(A)=H^{1}_{0}(U)\cap H^{2}(U)\to L^{2}(U) \mbox{, with } Au=\Delta u.$$
Let $\lambda_{1}>0$ be the principle eigenvalue of $-\Delta$ on $U$ with zero boundary conditions. I want to show that
$$\|T(t)\|_{\mathcal{L}(L^{2}(U),L^{2}(U))}\le e^{-\lambda_{1}t}$$
We know $\lambda_{1}=\min{\{B[u,u]|u\in H^{1}_{0}(U), \|u\|_{L^{2}(U)}=1\}}$, and from Rayleigh's formula, we have
$$\lambda_{1}=\min_{\substack{u\in H^{1}_{0}(U) \\ u\neq 0}}\frac{B[u,u]}{\|u\|^{2}_{L^{2}(U)}}$$
How do I tackle the problem from here? I assume I must exploit the semigroup property somehow.
Because $(\Delta f,f) \le -\lambda_1\|f\|^{2}$ for $f\in\mathcal{D}(-\Delta)$, then \begin{align} \frac{d}{dt}\|T(t)f\|^{2}& =(\Delta T(t)f,T(t)f)+(T(t)f,\Delta T(t)f) \\ & \le -2\lambda_1(T(t)f,T(t)f)= -2\lambda_1\|T(t)f\|^{2},\;\; f \in\mathcal{D}(-\Delta) \end{align} Hence, $$ \frac{d}{dt}\left( e^{2\lambda_1 t}\|T(t)f\|^{2}\right) \le 0\\ \|T(t)f\|^{2} \le e^{-2\lambda_1 t}\|T(0)f\|^{2} \\ \|T(t)f\| \le e^{-\lambda_1 t}\|f\|. $$ This easily extends to all $f$ because $\mathcal{D}(\Delta)$ is dense.