Let $d$ be an integer. We denote by $m$ the Lebesgue measure on $\mathbb{R}^d$. We define $BL(\mathbb{R}^d)$ by
\begin{align\*}
BL(\mathbb{R}^d)=\{f \in L^2\_{\rm loc}(\mathbb{R}^d,m) \mid |\nabla f|\in L^2(\mathbb{R}^d,m)\},
\end{align\*}
where $L^2\_{\rm loc}(\mathbb{R}^d,m)$ denotes the space of locally square integrable function on $\mathbb{R}^d$. For $f \in L^2\_{\rm loc}(\mathbb{R}^d,m)$, we write $\nabla f$ for the distributional derivative. For $f,g \in BL(\mathbb{R}^d)$, we define $\mathcal{E}(f,g)=\int\_{\mathbb{R}^d}\nabla f \cdot \nabla g\,dm$, where $\cdot$ denotes the standard inner product on $\mathbb{R}^d$.


We take a positive continuous function $V\colon \mathbb{R}^d \to \mathbb{R}$,which may be unbounded. We set 
\begin{align\*}
\mathcal{F}=\left\{f \in BL(\mathbb{R}^d) : \int\_{\mathbb{R}^d}f^2\,Vdm<\infty \right\} 
\end{align\*}


We can show that $(\mathcal{E},\mathcal{F})$ becomes a Dirichlet form on $L^2(\mathbb{R}^d,V\,dm)$. Therefore, $(\mathcal{E},\mathcal{F})$ generates a strongly continuous contraction semigroup $\{T\_t\}\_{t>0}$ on $L^2(\mathbb{R}^d,V\,dm)$, which is extended to a contraction semigroup on $L^{\infty}(\mathbb{R}^d,V\,dm)$. The extension is still denoted as $\{T\_t\}\_{t>0}$. In fact, $\{T\_t\}\_{t>0}$ is identified with the semigroup of a time-changed Brownian motion, and we can show that $\{T\_t\}\_{t>0}$ is a strongly continuous contraction semigroup on $C\_{0}(\mathbb{R}^d)$. Here, $C\_{0}(\mathbb{R}^d)$ stands for the space of continuous functions vanishing at infinity.


**My question**


Can we show that $\{T\_t\}\_{t>0}$ is extended to an bounded analytic semigroup on $C\_{0}(\mathbb{R}^d)$? The generator of $\{T\_t\}\_{t>0}$ is given by $\frac{1}{V}\Delta$, where $\Delta$ is the Laplacian on $\mathbb{R}^d$.


I am interested in whether analyticity of semigroups are stable under time change, which is one of the most fundamental transformations of stochastic processes.



Analyticity of the semigroup generated by a time-changed Brownian motion
I find that the answer is no. Let us work on the half line $(0,\infty)$ with Dirichlet boundary conditions at $0$; however the problems come from $\infty$. Let $L=a(x)D^2$ where $a=1/V$ is supposed to be smooth, positive and $a(0)=1$ and consider the change of variable $s=\phi(x)=\int\_0^x \frac{1}{\sqrt{a(t)}} dt$. If $u$ is a $C^2$-function, then $$a(x)u\_{xx}=u\_{ss}-D\_x (\sqrt{a})u\_s.$$ To be more precise, the change of variable above induces an isometry $T: C\_0([0,\infty[) \to C\_0([0,\ell]$, $\ell=\int\_0^\infty\frac{1}{\sqrt{a(t)}} dt$ given by $Tu(s)=u(\phi^{-1} s)$ and such that $TLT^{-1}=M$, where $$M=D\_{ss}-D\_x(\sqrt{a})D\_s.$$ This chanhe of variable simplifies the diffusion but adds a drift $D\_x(\sqrt a)$ which, however, should be written in the variable $s$. Next we choose $a$ in such a way that $\ell=\infty$ and $D\_x(\sqrt {a})=s$. Letting $b=\sqrt {a}$ this leads to the Cauchy problem $$b''=\frac{1}{b}, \quad b'(0)=0, \quad b(0)=1.$$
This equation can be solved almost explicitely by multiplying by $b'$ and using the initial values, thus leading to $$\int\_0^{\sqrt{\log b(x)}} e^{t^2}dt =\frac{\sqrt {2}}{2} x.$$ However, one can see directly from the equation that $b$ is globally defined for $x \ge 0$, positive, increasing and convex. Finally
$$
\ell=\int\_0^\infty \frac{1}{b}=\int\_0^\infty b''=\lim\_{x \to \infty}b'(x).$$
If $\ell <\infty$, then $b(x) \le 1+\ell x$ and again $1/b$ is not integrable near $\infty$. With this choice of $b=\sqrt{a}$ the operator $M=D\_{ss}-sD\_s$ is the Ornstein-Uhlenbeck in the half-line which is known not to be the generator of an analytic semigroup. By similarity, the same happens for $L$.


Hoping it is correct. I find very interesting the question and let me point out that it seems that the counterexample cannot be obtained by using powers: if $a(x)=x^\alpha$, then $D\_x(\sqrt{a})\approx 1/s$ with the above notation and the semigroup is analytic (also the singularity for small $s$ can be treated by using Bessel functions). It is not very clear to me what is behind.


