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).

  • 1
    $\begingroup$ 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 $\operatorname{Re}\lambda>0$. $\endgroup$
    – MaoWao
    Apr 21, 2021 at 10:04
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    $\begingroup$ 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. $\endgroup$ Apr 21, 2021 at 16:59
  • $\begingroup$ @MateuszKwaśnicki Thank you for your comment and the reference. $\endgroup$
    – sharpe
    Apr 21, 2021 at 23:36

1 Answer 1


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 $L^\infty$.

  • $\begingroup$ @Thank you for your comment. This is a clear counter example. $\endgroup$
    – sharpe
    Apr 21, 2021 at 23:37

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