Theorem 2.4.5 ([Theorem 4.27, Bau14]).label Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$, then for any $f \in L^{2}(\mu) \cap L^{\infty}(\mu)$,
\[\norm{\bp_t f}_{\infty} \le \norm{f}_{\infty}\]
Proof. Similar to the previous theorem, it is sufficient to show that $(1 - \lambda L)^{-1}$ satisfies the bound for all $\lambda > 0$.
To this end, let $\lambda > 0$, $g \in L^{2}(\mu) \cap L^{\infty}(\mu)$ with $0 \le g \le 1$ (a.e.), $f = R_{\lambda} g$, and $h = 1 \wedge f$, then
\begin{align*}0&\le \norm{f - h}_{\lambda}^{2} = \norm{f}_{\lambda}^{2} - 2\dpn{f, h}{\lambda}+ \norm{h}_{\lambda}^{2} \\&\le \dpn{f, g}{L^2(\mu)}- 2\dpn{f, h}{\lambda}+ \norm{h}_{L^2(\mu)}^{2} - \dpn{h, \lambda Lh}{L^2(\mu)}\\&= \dpn{f, g}{L^2(\mu)}+ \norm{g - h}_{L^2(\mu)}^{2} - \norm{g}_{L^2(\mu)}^{2}- \dpn{h, \lambda Lh}{L^2(\mu)}\\&\le \dpn{f, g}{L^2(\mu)}+ \norm{g - f}_{L^2(\mu)}^{2} - \norm{g}_{L^2(\mu)}^{2}- \dpn{f, \lambda Lf}{L^2(\mu)}\\&\le \norm{f}_{L^2(\mu)}^{2} - \dpn{f, g}{L^2(\mu)}- \dpn{f, \lambda Lf}{L^2(\mu)}\\&= \norm{f}_{L^2(\mu)}^{2} - \dpn{f, f}{\lambda}- \dpn{f, \lambda Lf}{L^2(\mu)}\\&= \norm{f}_{L^2(\mu)}- \norm{f}_{L^2(\mu)}= 0\end{align*}
so $f = h$ a.e., and $0 \le f \le 1$ (a.e.).$\square$