# Show that the carré du champ operator is nonnegative

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 $$x\in E$$.

• 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. – MaoWao Feb 6 at 14:16
• @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) – 0xbadf00d Feb 6 at 16:21

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.

• 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? – 0xbadf00d Feb 6 at 9:57
• @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. – saz Feb 6 at 11:11
• @0xbadf00d see my edited answer. – saz Feb 7 at 11:18
• 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? – 0xbadf00d Feb 7 at 12:57

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.

• (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, $\mathcal M_b(E,\mathcal E)\subseteq L^2(\mu)$. – 0xbadf00d Feb 6 at 17:48
• (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 $\mu$. – MaoWao Feb 6 at 18:01
• 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. – 0xbadf00d Feb 6 at 21:33
• @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". – saz Feb 7 at 8:33
• @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 $(2)$? – 0xbadf00d Feb 7 at 9:59