Let $(T_t)_{t \geq 0}$ a Feller semigroup and define a linear operator $(A,\mathcal{D}(A))$ by $$\mathcal{D}(A) := \left\{u \in C_{\infty}(\mathbb{R}^d); \exists f \in C_{\infty} \forall x \in \mathbb{R}^d: f(x) = \lim_{t \to 0} \frac{T_t u(x)-u(x)}{t} \right\} \\ Au(x) := \lim_{t \to 0} \frac{T_t u(x)-u(x)}{t} \qquad (u \in \mathcal{D}(A))$$

($A$ is called weak generator of the semigroup).

Now I want to show that this generator is the generator in the sense of the weak topology on $C_{\infty}(\mathbb{R}^d)$, i.e. that the convergence is bounded pointwise convergence.

Let $u \in \mathcal{D}(A)$. Since (by definition) the sequence is pointwise convergent, the only remaining thing is to show the boundedness, i.e.

$$\sup_{t>0} \left\| \frac{T_t u-u}{t} \right\|_{\infty} < \infty$$

Well, since the sequence is pointwise convergent we have $$\sup_{t > 0} \left|\frac{T_t u(x)-u(x)}{t} \right| < \infty$$ for fixed $x \in \mathbb{R}^d$. A hint says that one should apply the Banach-Steinhaus theorem, but I don't see how to apply this theorem here, because there are not even linear operators (note that $u$ is fixed). Some hint...?

Remark A Feller semigroup is a positivity preserving, conservative, strongly continuous semigroup satisfying the sub-markov property.

  • 1
    $\begingroup$ Not true, the $L_t : u \mapsto \dfrac{T_t\,u - u}{t}$ are linear operators. $\endgroup$
    – Siméon
    Jan 16, 2013 at 18:21
  • $\begingroup$ @Ju'x Yes, but $u$ is fixed. (I have to prove $\frac{T_t u- u}{t} \to Au$ weakly as $t \to 0$ for (fixed!) $u \in \mathcal{D}(A)$.) $\endgroup$
    – saz
    Jan 16, 2013 at 18:29
  • $\begingroup$ What is the a strong generator? $\endgroup$
    – SBF
    Jan 16, 2013 at 21:10
  • $\begingroup$ @Ilya The strong generator is the generator in the sense of the topology generated by the supremum norm, i.e. $\frac{T_t u-u}{t} \to Au$ uniformly. In this case, the weak and strong generator coincide - and that's exactly what I want to prove using the result from above. $\endgroup$
    – saz
    Jan 17, 2013 at 7:56
  • $\begingroup$ I see now, thanks $\endgroup$
    – SBF
    Jan 17, 2013 at 8:47

1 Answer 1


Here is a sketch how to prove the result using the so-called (strong) generator of the semigroup $(T_t)_{t \geq 0}$: Let $$\begin{align} \mathcal{D}(A_s) &:= \left\{u \in C_{\infty}(\mathbb{R}^d); \exists f \in C_{\infty}(\mathbb{R}^d): \lim_{t \to 0} \left\| \frac{T_t u-u}{t} -f \right\|_{\infty} = 0 \right\} \\ A_s u &:= \lim_{t \to 0} \frac{T_t u-u}{t} \qquad (u \in \mathcal{D}(A_s)) \end{align}$$ (The limit is taken w.r.t to the sup-norm.) Then $(A_s,\mathcal{D}(A_s))$ is called the (strong) generator of $(T_t)_{t \geq 0}$. The idea is to show that the weak and the strong generator coincide. The following theorem is quite helpful:

Theorem Let $(A_s,\mathcal{D}(A_s))$ the generator of a Feller semigroup $(T_t)_{t \geq 0}$. Let $(A,\mathcal{D}(A))$ an extension of $(A_s,\mathcal{D}(A_s))$ such that $$Au = u \Rightarrow u=0 \quad (u \in \mathcal{D}(A)) \tag{1}$$ Then $(A,\mathcal{D}(A))= (A_s,\mathcal{D}(A_s))$.

If we would be able to show that the weak generator $(A,\mathcal{D}(A))$ satisfies $(1)$ we would be finished: For $u \in \mathcal{D}(A) = \mathcal{D}(A_s)$ we have $$\frac{T_t u-u}{t} \to Au \quad \text{uniformly}$$ hence in particular $$\sup_{t>0} \left\| \frac{T_t u-u}{t} \right\|_{\infty} < \infty$$

So - that's it. Here is the remaining part of the proof:

Lemma Let $(A,\mathcal{D}(A))$ the weak generator of a Feller semigroup. Let $u \in \mathcal{D}(A)$, $x_0 \in \mathbb{R}^d$ such that $u(x_0)=\sup_{x \in \mathbb{R}^d} u(x) \geq 0$. Then $$Au(x_0) \leq 0$$ (i.e. $A$ satisfies the maximum principle). In particular, $A$ is dissipative, i.e. $$\forall \lambda>0: \|\lambda \cdot u-Au\|_{\infty} \geq \lambda \cdot \|u\|_{\infty} \tag{2}$$

This shows that the weak generator $(A,\mathcal{D}(A))$ satisfies $(1)$ (put $\lambda=1$ in $(2)$).

Literature René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes (Chapter 7).

Remark The given hint (i.e. applying Banach Steinhaus theorem) was not correct - or at least much more difficult than the creator of this exercise was expecting. One would have to apply Banach Steinhaus to the dual space of $C_{\infty}$ (using the fact that for the dirac measures $\delta_x$ the boundedness is given by the pointwise convergence ... and some more considerations about the (vague) density of dirac measures.)

  • $\begingroup$ How did you prove that $(\mathcal D(A),A)$ is dissipative? Schilling is picking an $x_0\in\mathbb R$ with $|u(x_0)|=\sup_{x\in\mathbb R}|u(x)|$ in his argumentation; but that would mean that $|u|$ attains its supremum. That clearly doesn't need to be the case for a general $u\in C_0(\mathbb R)$. $\endgroup$
    – 0xbadf00d
    Feb 4, 2019 at 20:00
  • $\begingroup$ @0xbadf00d Why do you think so? Any non-negative function $v \in C_0$ attains its supremum, right? ... and $v:= |u|$ satisfies these assumptions. $\endgroup$
    – saz
    Feb 4, 2019 at 20:07
  • $\begingroup$ Oops, I never thought about that, but sure, it follows from the compactness of $\left\{|u|\ge\varepsilon\right\},\varepsilon>0$. $\endgroup$
    – 0xbadf00d
    Feb 4, 2019 at 21:53
  • $\begingroup$ So, given such an $u$, we use that $\tilde u:=\operatorname{sgn}(u(x_0))u\in\mathcal D(A)$ and $\tilde u(x_0)=\sup_{x\in\mathbb R}\tilde u(x)$ and then that $(\mathcal D(A),A)$ satisfies the nonnegative maximum principle, right? $\endgroup$
    – 0xbadf00d
    Feb 4, 2019 at 23:52
  • $\begingroup$ @0xbadf00d Well, yes, that's exactly the reasoning in the book... $\endgroup$
    – saz
    Feb 5, 2019 at 10:25

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