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).



On a property of resolvents associated with holomorphic semigroups
As pointed out in the comments, this is true for $Re\, \lambda >0$ but can fail in general. An example is the Ornstein-Uhlenbeck operator $L=D^2-xD$ in $L^2(e^{-x^2/2}\, dx)$. The $L^2$ spectrum consists of the negative integers but the $L^\infty$ spectrum equals the left half plane. If $Re\, \lambda <0$, $\lambda \not \in \mathbb R$, then $\lambda-L$ is injective in $L^\infty$, since this last is contained in $L^2(e^{-x^2/2}\, dx)$, but its $L^2$-inverse cannot preserve $L^\infty$, otherwise it would be the resolvent in $L^\infty$.


