# Analyticity of the semigroup generated by a time-changed Brownian motion

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.

• If I understand correctly, the question is about analyticity of the semigroup generated by $\frac{1}{V} \Delta$ in spaces of continuous functions. May 6, 2020 at 21:24
• @GiorgioMetafune Thank you for your comment. That's right. Certainly, this is a question for the analyticity of $\Delta/V$. May 7, 2020 at 0:27
• I notew that in the functional analysis community it is also usual to denote this space with $C_0$, your $C_\infty$ might be a source of confusion. See en.wikipedia.org/wiki/Function_space#Functional_analysis May 7, 2020 at 6:13
• @AndrásBátkai Thank you for your comment. I would like to write $C_{0}$ in stead of $C_{\infty}$. May 7, 2020 at 6:41

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.

• Thank you very much for your very kind reply. I feel there is some relationship between the analyticity of Laplacian with unbounded drift and that of the generator of a time-changed Brownian motion. Your answer is very interesting! May 7, 2020 at 10:38