# On a property of resolvents associated with holomorphic semigroups

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).

• The more useful formula seems to be $(\lambda-\mathcal{L})^{-1}f=\int_0^\infty e^{-\lambda t}T_t f\,dt$ for $\operatorname{Re}\lambda>0$, which shows that shows the assertions is true for $\operatorname{Re}\lambda>0$. Apr 21, 2021 at 10:04
• If you allow to change $L^\infty$ to $L^p$, then the answer is 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. Apr 21, 2021 at 16:59
• @MateuszKwaśnicki Thank you for your comment and the reference. Apr 21, 2021 at 23:36

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 $$L^\infty$$.