• $b\in C^1(\mathbb R)$ be Lipschitz continuous
  • $\sigma\in C^2(\mathbb R)$ be Lipschitz continuous with $\sigma(\mathbb R)\subseteq\mathbb R\setminus\left\{0\right\}$ and $\sigma''$ being bounded
  • $Lf:=bf'+\frac12\sigma^2f''$ for $f\in C^2(\mathbb R)$ and $L^\ast g:=\frac12(\sigma^2g)''-(bg)'$ for $g\in C^2(\mathbb R)$
  • $(\kappa_t)_{t\ge0}$ be a semigroup of Markov kernels on $(\mathbb R,\mathcal B(\mathbb R))$ and $$\kappa_t f:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)\tag1$$ for bounded Borel measurable $f:\mathbb R\to\mathbb R$.

Assume $(\kappa_t)_{t\ge0}$ is a strongly continuous semigroup (via $(1)$) on $C_0(\mathbb R)$ (continuous functions on $\mathbb R$ vanishing at infinity equipped with the supremum norm) with infinitesimal generator $(\mathcal D(A),A)$ given by the restriction $A$ of $L$ to $$\mathcal D(A):=\left\{f\in C_0(\mathbb R)\cap C^2(\mathbb R):Lf\in C_0(\mathbb R)\right\}.$$ Moreover, assume there is a Borel measurable $p_t:\mathbb R\times\mathbb R\to[0,\infty)$ with $$\kappa_t(x,B)=\int_Bp_t(x,y)\:\lambda({\rm d}y)\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\tag2,$$ where $\lambda$ denotes the Lebesgue measure, for all $t\ge0$ with

  1. $[0,\infty)\ni t\mapsto p_t(x,y)$ is differentiable for all $x,y\in\mathbb R$ and $$[0,\infty)\times\mathbb R\ni(t,y)\mapsto\frac{\partial p}{\partial t}(t,x,y)\tag3$$ is locally bounded for all $x\in\mathbb R$
  2. $\mathbb R\ni y\mapsto p_t(x,y)$ is twice continuously differentiable for all $(t,x)\in[0,\infty)\times\mathbb R$.

We can observe the following: Fix $f\in C_c^2(\mathbb R)$. Then, $$\frac{\kappa_{t+h}f-\kappa_tf}h\xrightarrow{h\to0}A(\kappa_tf)=\kappa_t(Af)\tag4,$$ where the convergence is with respect to the supremum norm, for all $t\ge0$. Fix $(t,x)\in[0,\infty)\times\mathbb R$. On the one hand, $$\frac{(\kappa_{t+h}f)(x)-(\kappa_tf)(x)}h\xrightarrow{h\to0}\int\frac{\partial p}{\partial t}(t,x,y)f(y)\:{\rm d}y\tag5$$ by 1. and the dominated convergence theorem. On the other hand, $$\left(\kappa_t\left(Af\right)\right)(x)=\int f(y)\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\:{\rm d}y\tag6$$ by 2. and integration by parts. Thus, $$\int\frac{\partial p}{\partial t}(t,x,y)f(y)\:{\rm d}y=\int f(y)\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\:{\rm d}y\tag7.$$ Are we able to conclude $$\frac{\partial p}{\partial t}(t,x,y)=\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\tag8$$ (since $f$ was arbitrary)?


1 Answer 1


Essentially your question is whether $C_c^2(\mathbb{R})$ is measure-determining and the answer is "yes".

Definition Let $\mu$ and $\nu$ be (non-negative) measures on a measurable space $(X,\mathcal{A})$. A family $\mathcal{D} \subseteq L^1(\mu) \cap L^1(\nu)$ is called determining if $$\left\{ \forall f \in \mathcal{D}: \quad \int f \, d \mu = \int f \, d\nu \right\} \implies \mu = \nu.$$

It is known that $C_c^{\infty}(\mathbb{R}^n)$ is determining for all measures which are finite on compact sets, see e.g. Theorem 17.12 in $\{1\}$. The idea is to use the fact that $C_c^{\infty}(\mathbb{R}^n)$ is dense in $L^1(\mu)$ for any measure $\mu$ which is finite on compacts. As a consequence of this, we get that $\mathcal{D}=C_c^2(\mathbb{R}^n)$ is determining, i.e.

$$\left\{\forall f \in C_c^2(\mathbb{R}^n): \quad \int f \, d\mu = \int f \, d\nu \right\} \implies \mu=\nu \tag{1}$$

for any (non-negative) measures $\mu$, $\nu$ which are finite on compacts. Alternatively, $(1)$ can be proved by noting that it suffices to show $\mu(K)=\nu(K)$ for any compact set $K$ and applying Urysohn's lemma to approximate the indicator function $1_K$ by (sufficiently) smooth functions.

In your case we are not dealing with non-negative measures but that's not much of a bother. If $p$, $q$ are two locally bounded functions such that $$\int f(x) p(x) \, dx = \int f(x) q(x) \, dx, \qquad f \in C_c^2(\mathbb{R}^n)$$ then $$\int f(x) (p^+(x)+q^-(x)) \, dx = \int f(x) (p^-(x)+q^+(x)) \, dx$$ where $\pm$ denotes the positive and negative part, respectively. Applying $(1)$ we get

$$\int_B (p^+(x)+q^-(x)) \, dx = \int_B (p^-(x)+q^+(x)) \, dx$$

for any Borel set $B$, i.e.

$$\int_B p(x) \, dx= \int q(x) \, dx.$$

Choosing $B=\{p>q\}$ and $B=\{q<p\}$ this yields $p=q$ (Lebesgue)-almost everywhere. If $p$ and $q$ are continuous, then this, in turn, entails that $p=q$.

Hence, $(7)$ gives $(8)$ provided that $\partial_t p(t,x,\cdot)$ and $L^* p_t(x,\cdot)$ are locally bounded and continuous.


$\{1\}$ R. Schilling: Measures, integrals and martingales, Cambridge, 2nd edition.

  • $\begingroup$ I've only got a copy of the 1st edition. The relevant chapter is completely new, right? $\endgroup$
    – 0xbadf00d
    Jan 25, 2019 at 17:30
  • $\begingroup$ @0xbadf00d Seems so, yes. You can take a look at this question which gives two references (though not exactly for the statement which I mentioned above; since I don't have copies of the books I can't check what material they contain about determining families). $\endgroup$
    – saz
    Jan 25, 2019 at 17:40
  • 1
    $\begingroup$ Which version of Urysohn's lemma do you've got in mind? (Surely not the one from Wikipedia.) $\endgroup$
    – 0xbadf00d
    Jan 26, 2019 at 0:34
  • $\begingroup$ @0xbadf00d For instance something like this $\endgroup$
    – saz
    Jan 26, 2019 at 7:42

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