I have a Feller semigroup $(P_t)_{t\geq 0}$. Based on this semigroup I define the linear operator $L = \int_0^tP_s\,ds$ as follows. $$x \mapsto Lu(x) = \int_{0}^t\int u(y) p_s(x,dy)\,ds$$ where $p_s$ is the unique kernel for $P_s$. What I would like to know is why the following is true. $$P_tu - u = \int_0^t\frac{d}{ds}(P_su) ds$$ I am self-studying this subject so my apologies if this is too trivial. I wrote the following but I am not happy with it. \begin{align} \int_0^t\frac{d}{ds}(P_su) ds &= \int_0^t\frac{d}{ds}\left(\int u(y)p_s(x,dy)\right) ds \\ &= \frac{d}{ds}\int_0^t\int u(y)p_s(x,dy) ds \\ & = \frac{d}{ds}\int_0^tP_su(x) ds \\ & = P_tu(x) - P_0u(x)\\ & = P_tu(x) - u(x) \end{align} The reason why I am not convinced of this reasoning is because I am treating these new differentiation and integration operators as if they behave like ordinary differentiation and integration operators. While that may be the case I haven't proven this yet so I am looking for a proof of the fact above in terms of first principles, so to speak.

  • $\begingroup$ I don't really understand your question... what is the difference between an "ordinary" and a "new" differentiation/integration operator? $\endgroup$
    – saz
    Apr 9, 2018 at 15:18
  • $\begingroup$ @saz For instance, when I go from the third equation to the fourth, I use the fundamental theorem of calculus but $P_s$ is not some function I can integrate. But I guess the right way to look at it is that $P_su(x)$ is some function of $s$ and $x$ and I am integrating along the $s$ dimension. I am also not comfortable with the first equation even though I don't know what is wrong with it. Differentiating an integral with respect to the index of a sequence of kernels sounds strange. I guess the kind of answer I am looking for is some explanation of the steps I took without full justification. $\endgroup$
    – Calculon
    Apr 9, 2018 at 17:37
  • $\begingroup$ Well, it's not that strange ... but before proving the identity you should actually ensure that all expressions are well-defined, in particular you need to show that $s \mapsto P_s u$ is differentiable. Do you know that this is true or is this an assumption in your (not so rigorous) proof? $\endgroup$
    – saz
    Apr 11, 2018 at 17:45
  • $\begingroup$ @saz I am following the book of Schilling and Partzsch and the derivative of the mapping $s \mapsto P_su$ is defined as $\lim_{s\to 0}\frac{P_s u - u}{s}$ with the understanding that the limit is in the uniform sense. On the other hand, $P_su(x) = \int u(y) p_s(x,dy)$ with $p_s$ being the kernel. But I don't feel comfortable with writing $\frac{d}{ds}\int u(y) p_s(x,dy)$ because this makes it look like I am differentiating with respect to the index of a family of kernels, which is different from what we had above. Sorry I haven't been able to articulate the issue I am having very well. $\endgroup$
    – Calculon
    Apr 12, 2018 at 7:31
  • $\begingroup$ Not sure whether my answer is really answering your question. Feel free to comment ... $\endgroup$
    – saz
    Apr 16, 2018 at 19:44

1 Answer 1


Let us first the recall the following (standard) definition of the infinitesimal generator

Let $(P_t)_{t \geq 0}$ be a Feller semigroup. Then the infinitesimal generator $A: \mathcal{D}(A) \to C_{\infty}(\mathbb{R}^d)$ is defined by $$Af := \lim_{t \to 0} \frac{P_t f-f}{t}$$ for any $f \in C_{\infty}(\mathbb{R}^d)$ where the limit exists with respect to the uniform norm.

There is the following well-known statement.

Let $(P_t)_{t \geq 0}$ be a Feller semigroup. For any $u \in \mathcal{D}(A)$ and $t>0$ it holds that $P_t u \in \mathcal{D}(A)$ and $$\frac{d}{dt} P_t u = AP_t u = P_t Au.$$

The diffentiability of the mapping $t \mapsto P_t u$ can be shown for a larger class of functions; however, the derivative fails, in general, to be integrable. Consequently, we we cannot expect to prove the identity

$$P_t u-u = \int_0^t \frac{d}{ds} P_s u \, ds \tag{1}$$

for any $u$; we have to restrict ourselves to a smaller class of functions, e.g. $u \in C_{\infty}(\mathbb{R}^d)$ such that $\frac{d}{dt} P_t u(x)$ is Riemann-integrable on $(0,T)$ for any $T>0$ and $x \in \mathbb{R}^d$ (which holds, by the above statement, for instance for any $u \in \mathcal{D}(A)$).

Now, given such a mapping $u$, the identity $(1)$ is a direct consequence of the fundamental theorem of calculus. For fixed $x \in \mathbb{R}^d$ we know that the mapping

$$t \mapsto F(t) := P_t u(x)$$

is differentiable and its derivative is Riemann integrable. By the fundamental theorem of calculus, we find

$$F(t)-F(0)= \int_0^t \frac{d}{ds} F(s) \,ds,$$


$$P_t f(x)- \underbrace{P_ 0f(x)}_{=f(x)} = \int_0^t \frac{d}{ds} P_s f(x) \, ds.$$


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