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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.

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    $\begingroup$ If I understand correctly, the question is about analyticity of the semigroup generated by $\frac{1}{V} \Delta$ in spaces of continuous functions. $\endgroup$ May 6, 2020 at 21:24
  • $\begingroup$ @GiorgioMetafune Thank you for your comment. That's right. Certainly, this is a question for the analyticity of $\Delta/V$. $\endgroup$
    – sharpe
    May 7, 2020 at 0:27
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    $\begingroup$ 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 $\endgroup$ May 7, 2020 at 6:13
  • $\begingroup$ @AndrásBátkai Thank you for your comment. I would like to write $C_{0}$ in stead of $C_{\infty}$. $\endgroup$
    – sharpe
    May 7, 2020 at 6:41

1 Answer 1

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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.

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    $\begingroup$ 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! $\endgroup$
    – sharpe
    May 7, 2020 at 10:38

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