4
$\begingroup$

Let

  • $(E,\mathcal E)$ be a measurable space
  • $\mathcal M_b(E,\mathcal E):=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$
  • $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$ and $$\kappa_tf:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)\tag1$$ for $f\in\mathcal M_b(E,\mathcal E)$ and $t\ge0$
  • $\mu$ be a probability measure on $(E,\mathcal E)$ subinvariant with respect to $(\kappa_t)_{t\ge0}$

It's easy to see that $(\kappa_t)_{t\ge0}$ is a contraction semigroup on $\left(\mathcal M_b(E,\mathcal E),\left\|\;\cdot\;\right\|_{L^2(\mu)}\right)$ and hence has a unique extension to a contraction semigroup on $L^2(\mu)$. Let $(\mathcal D(A),A)$ denote the generator of that semigroup.

Let $f\in\mathcal D(A)$ such that $f^2\in\mathcal D(A)$. I want to show that $$Af^2\ge 2fAf.\tag2$$

The crucial point might be the following: If $g:E\to\mathbb R$ is $\mathcal E$-measurable and $(\kappa_t|g|)(x)<\infty$ for all $x\in E$, then $$\varphi\left(\left(\kappa_tg\right)(x)\right)\le\left(\kappa_t\left(\varphi(g)\right)\right)(x)\;\;\;\text{for all }x\in E\tag3$$ for all convex $\varphi:\mathbb R\to\mathbb R$ by Jensen's inequality (Clearly, for the question we would take $\varphi(x)=x^2$).

However, it's not clear to me how (and if at all) $(3)$ extends to $g\in L^2(\mu)$.$^1$

Clearly, we know that there is a $(g_n)_{n\in\mathbb N}\subseteq\mathcal M_b(E,\mathcal E)$ with $$|g_n|\le|g|\;\;\;\text{for all }n\in\mathbb N\tag4$$ and $$g_n\xrightarrow{n\to\infty}g\tag5.$$ By the dominated convergence theorem (and construction of $(\kappa_t)_{t\ge0}$), $$\left\|\kappa_tg_n-\kappa_tg\right\|_{L^2(\mu)}\le\left\|g_n-g\right\|_{L^2(\mu)}\xrightarrow{n\to\infty}0\tag6\;\;\;\text{for all }t\ge0.$$ $(3)$ holds for $g=g_n$. Moreover, we could extract a subsequence $\left(g_{n_k}\right)_{k\in\mathbb N}$ with $$g_{n_k}\xrightarrow{k\to\infty}g\;\;\;\mu\text{-almost surely}\tag7.$$ But that doesn't mean (does it?)$^2$ that $$\kappa_tg_{n_k}\xrightarrow{k\to\infty}\kappa_tg\;\;\;\mu\text{-almost surely for all }t\ge0\tag8.$$ So, I'm stuck at this point.


$^1$ One may note that, by subinvariance, $(\kappa_t|g|)(x)<\infty$ for $\mu$-almost all $x\in E$, but I hope that $(3)$ can be proved by a general extension argument.

$^2$ Maybe we can argue that $$\left|\kappa_tg_{n_k}-\kappa_tg_{n_l}\right|\le\kappa_t\left|g_{n_k}-g_{n_l}\right|\xrightarrow{k,\:l\to\infty}0\tag9$$ (pointwise) by the dominated convergence theorem and hence $\left(\left(\kappa_tg_{n_k}\right)(x)\right)_{k\in\mathbb N}$ is Cauchy for all $x\in E$.

$\endgroup$
  • $\begingroup$ What are these functions $f$ and $f_{n_k}$ in Eq. (8)? Anyway, the case $\phi(x)=x^2$ seems easier because you can use monotonicity arguments. $\endgroup$ – MaoWao Feb 6 at 14:16
  • $\begingroup$ @MaoWao They should be $g$ and $g_{n_k}$ instead. Feel free to assume $\varphi(x)=x^2$. (But I'm interested in the general case too) $\endgroup$ – 0xbadf00d Feb 6 at 16:21
2
$\begingroup$

First of all note that for any $g \in L^2(\mu)$ we have

$$(\kappa_t g)^2 \leq \kappa_t(g^2) \quad \text{$\mu$-almost everywhere}\tag{1}$$

where the exceptional null set may depend on $t \geq 0$ and $g$; this follows by a standard approximation procedure, see @MaoWao's answer for details.


Now let $f \in D(A)$ be such that $f^2 \in D(A)$. Set $t_n := 1/n$ for $n \in \mathbb{N}$. Because of $(1)$ there exists a $\mu$-null set $N_0$ such that

$$(\kappa_{t_n}f)^2(x)\leq \kappa_{t_n} (f^2)(x) \quad \text{for all $x \in E \backslash N_0$, $n \in \mathbb{N}$}$$

i.e.

$$\frac{1}{t_n} \big[ \kappa_{t_n} (f^2)(x)-f(x)^2 \big] -\frac{1}{t_n} \big[ (\kappa_{t_n} f)^2(x) -f(x)^2 \big] \geq 0 \quad \text{for all $x \in E \backslash N_0$, $n \in \mathbb{N}$.} \tag{2}$$

Since $f \in D(A)$ we have $Af = \lim_{t \to 0} t^{-1} (\kappa_tf-f)$ in $L^2(\mu)$; in particular, we can choose a subsequence $(t_n')$ of $(t_n)$ such that

$$Af(x) = \lim_{n \to \infty} \frac{\kappa_{t_n'}f(x)-f(x)}{t_n'}, \quad x \in E \backslash N_1 \tag{3}$$ for a $\mu$-null set $N_1$.Note that this implies in particular

$$\kappa_{t_n'} f(x) \xrightarrow[]{n \to \infty} f(x), \qquad x \in E \backslash N_1. \tag{4}$$ Similarly, $f^2 \in D(A)$ implies that there exists a $\mu$-null set $N_2$ and a further subsequence $(t_n'')$ of $(t_n')$ such that

$$A(f^2)(x) = \lim_{n \to \infty} \frac{\kappa_{t_n''}(f^2)(x)-f^2(x)}{t_n''}, \quad x \in E \backslash N_2. \tag{5}$$

Clearly, $(2)$-$(4)$ remain valid with $t_n$ (resp. $t_n'$) replaced by $t_n''$. Set $N := N_0 \cup N_1 \cup N_2$ and fix $x \in E \backslash N$. Writing

$$(\kappa_{t_n''} f)^2(x) -f(x)^2 = (\kappa_{t_n''} f(x)+f(x)) (\kappa_{t_n''}f(x)-f(x))$$

and dividing both sides by $t_n''$ it follows from $(3)$ and $(4)$ that

$$\frac{(\kappa_{t_n''} f)^2(x) -f(x)^2}{t_n''} \to 2f(x) Af(x). \tag{6}$$

Using $(2)$ (for $t_n''$) and letting $n \to \infty$ it now follows from $(5)$ and $(6)$ that

$$A(f^2)(x)-2f(x) Af(x) \geq 0.$$

We have shown this identity for any $x \in E \backslash N$ and since $N$ is a $\mu$-null set this proves the assertion.

$\endgroup$
  • $\begingroup$ The differentiability at $0$ from the orbits is with respect to the $L^2(\mu)$-norm. It seems like you're using that this convergence implies pointwise convergence. What am I missing? $\endgroup$ – 0xbadf00d Feb 6 at 9:57
  • $\begingroup$ @oxbadfood Ah sorry; my mistake. I will fix it when I'm back home. Essentially the idea is that L2 convergence implies that there exists a subsequence which converges a.e. $\endgroup$ – saz Feb 6 at 11:11
  • $\begingroup$ Thank you in advance. $\endgroup$ – 0xbadf00d Feb 6 at 11:12
  • $\begingroup$ @0xbadf00d see my edited answer. $\endgroup$ – saz Feb 7 at 11:18
  • $\begingroup$ Just to be sure: Why do you think we need an approximation argument for $(1)$? By subinvariance, $\mu(\kappa_t|f|)=(\mu\kappa_t)|f|\le\mu|f|<\infty$ and hence $\kappa_t|f|<\infty$ $\mu$-almost surely. So, there is a $\mu$-null set $N$ such that $f$ is integrable with respect to $\kappa_t(x,\;\cdot\;)$ (and hence we can apply Jensen's inequality) for all $x\in E\setminus N$. Am I missing something? $\endgroup$ – 0xbadf00d Feb 7 at 12:57
0
$\begingroup$

Here is an answer for the case $\phi(x)=x^2$. First note that since $\kappa_t$ is positivity-preserving, i.e. $f\geq 0$ implies $\kappa_t f\geq 0$, one has $|\kappa_t g|\leq \kappa_t|g|$. Thus it suffices to prove (3) for positive $g\in L^2$.

Now let $g_n=g\wedge n$. Then $g_n\in\mathcal{M}_b(E,\mathcal{E})\cap L^2(\mu)$, $0\leq g_n\leq g$ and $g_n\to g$ in $L^2$ and a.e. Hence $$ (\kappa_t g_n)^2\leq\kappa_t (g_n^2)\leq \kappa_t (g^2). $$ The left side converges a.e. to $(\kappa_t g)^2$ by monotone convergence.

One more remark: You can always get property (8) from your question by passing to another subsequence (you already have convergence in $L^2$). The problem in the case of general $\phi$ is rather the right side of the inequality. Without any assumptions on $\phi$, the right side might be ill-defined.

$\endgroup$
  • $\begingroup$ (a) I agree that we can pass to a subsequence of $(\kappa_tf_n)$ which converges almost surely, but I think the problem is that we cannot find a common subsequence such that both $(f_{n_k})$ and $(\kappa_tf_{n_k})$ converge almost surely (b) "Ill-defined" since not integrable with respect to $\kappa_t(x,\;\cdot\;)$? Is that what you mean? (c) Just a minor remark: Since $\mu$ is finite, $\mathcal M_b(E,\mathcal E)\subseteq L^2(\mu)$. $\endgroup$ – 0xbadf00d Feb 6 at 17:48
  • $\begingroup$ (a) The trick is not to take a subsequence of $(\kappa_t f_n)$, but of $(\kappa_t f_{n_k})$. The corresponding subsequence of $(f_{n_k})$ will of course still converge a.e. (after all, it's just a subsequence). (b) Yes. You can of course restrict to such $f$ for which $\phi\circ f$ is integrable, but you don't immediately have nice continuity properties of the right side. (c) Yes, but the argument also works for infinite measures $\mu$. $\endgroup$ – MaoWao Feb 6 at 18:01
  • $\begingroup$ We need to be careful. If I'm not missing anything, you're argument for $|\kappa_tf|\le\kappa_t|f|$ is wrong. For $f\in L^2(\mu)$, $\kappa_tf$ is defined in terms of an $L^2(\mu)$-limit. So, you cannot conclude by positivity-preservingness. To be precise: $\kappa_tf$ is a well-defined object in $L^2(\mu)$, while $\int\kappa_t(x,{\rm d}y)f(y)$ is only well-defined for $\mu$-almost all $x\in E$. Outside the corresponding null-set both objects coincide. $\endgroup$ – 0xbadf00d Feb 6 at 21:33
  • $\begingroup$ @0xbadf00d Since you are working with $L^2$-functions you cannot expect that $(3)$ holds for any $x \in E$ but only for ($\mu$)-almost every $x \in E$. So all statements in your questions hold "almost everywhere", for instance also (2) holds only "almost everywhere". $\endgroup$ – saz Feb 7 at 8:33
  • $\begingroup$ @saz Sure, but if $\kappa_tf^2\ge(\kappa_tf)^2$ only holds outside a null set depending on $t$, don't we have a problem when we want to use this in the proof of $(2)$? $\endgroup$ – 0xbadf00d Feb 7 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.