Theorem 2.5.2 ([Theorem 4.31, Bau14]).label Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$, and $\bracs{\bp_t|t \ge 0}$ be its heat semigroup. For each $p \in [1, \infty]$ and $t \ge 0$, $\bp_{t}$ extends uniquely to a mapping $L^{p}(\mu) \to L^{p}(\mu)$ with
\[\norm{\bp_t f}_{L^p(\mu)}\le \norm{f}_{L^p(\mu)}\]
Proof. Let $f, g \in L^{1}(\mu) \cap L^{\infty}(\mu)$, then
\begin{align*}\dpn{\bp_tf, g}{L^2(\mu)}&= \dpn{f, \bp_tg}{L^2(\mu)}\le \norm{f}_{L^1(\mu)}\norm{\bp_tg}_{L^\infty(\mu)}\\&\le \norm{f}_{L^1(\mu)}\norm{g}_{L^\infty(\mu)}\end{align*}
so $\norm{\bp_t f}_{L^1(\mu)}\le \norm{f}_{L^1(\mu)}$.$\square$