This question is about semigroup theory.

Let $E$ be a locally compact metric space, and $X=(X_t,t\ge 0;\,P_x,x\in E)$ be a Markov process on $E$. We assume that $X$ is symmetric with respect to $m$, a Radon measure on $E$. The semigroup $\{T_t\}_{t \ge 0}$ of $X$ is extended to a strongly continuous contraction semigroup on $L^2(E,\mu)$, which is denoted by the same symbol. By the symmetry, the semigroup $\{T_t\}_{t \ge 0}$ is extended to a holomorphic semigroup on (the complexification of) $L^2(E,\mu)$. We write $(\mathcal{L},\mathcal{D}(\mathcal{L}))$ for the generator of $\{T_t\}_{t \ge 0}$. Then, the resolvent set $\rho(\mathcal{L})$ contains a sector $S_\theta$ of angle $\theta \in (\pi/2,\pi)$. We moreover obtain that \begin{align} (1)\quad T_t=\frac{1}{2\pi i}\int_{\gamma}e^{\lambda t}(\lambda-\mathcal{L})^{-1}\,d\lambda,\quad t \in (0,\infty), \end{align} where $i=\sqrt{-1}$, and $\gamma$ denotes a curve in the sector.

Let $f \in L^2(E,\mu )\cap L^\infty(E,\mu)$. Then, can we show that $(\lambda-\mathcal{L})^{-1}f \in L^\infty(E,\mu)$ for every $\lambda \in \gamma$ ? Since $\{T_t\}_{t \ge 0}$ is generated by the Markov process, it is trivial that $T_tf \in L^\infty$ for every $t>0$. Therefore, it should not be so unnatural to expect such a thing from formula (1).

yes, given that $\theta$ is sufficiently close to $\tfrac\pi2$. My guess is that the result does not extend to $L^\infty$, but I do not know a counterexample. I would start by looking at: V. A. Liskevich, M. A. Perelmuter,Analyticity of Submarkovian Semigroups. Proc. Amer. Math. Soc. 123(4) (1995): 1097–1104, DOI:10.1090/S0002-9939-1995-1224619-1 and the references therein. $\endgroup$