For $t \geq 0$, we define a family of operators $P_t f(x) = \mathbb{E}_x[f(B_t)]$, where $B_t$ is one-dimensional Brownian Motion and $f \in C_0(\mathbb{R})$, i.e. continuous $\mathbb{R} \rightarrow\mathbb{R}$ functions that vanish at infinity. I am asked to show that this family defines a strongly continuous contraction semi group on $C_0(\mathbb{R})$. Here is the definition:
Let $B_0$ be a Banach space. A family $P_t : B_0 \rightarrow B_0$ of bounded linear operators is called a strongly continuous contraction semi group on $B_0$, if:
- $||P_t|| \leq 1$ for all $t \geq 0$
- $P_t P_s = P_{t+s}$ for all $s,t \geq 0$
- $||P_tf-f|| \rightarrow 0$ if $t\rightarrow 0$ for any $f\in B_0$
Now
$$ P_0f(x) = \mathbb{E}_x[f(B_0)] = \mathbb{E}[f(B_0) : B_0 = x] = f(x)\\ P_t f(x) = \mathbb{E}_x[f(B_t)] = \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi t}} \exp \left( - \frac{(y-x)^2}{2t}\right) f(y) dy $$ is easily seen to be a linear operator, furthermore it is bounded since $$ ||P_t f || \leq ||f|| \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi t}} \exp \left( - \frac{(y-x)^2}{2t}\right) dy = ||f|| $$ and also $||P_t|| \leq 1$. The Chapman Kolmogorov equation can also be verified by a little bit of computation. Now for 3. I am not quite sure. This is what I got so far:
First, we note that $f$ must be bounded, so $||f|| < \infty$. Furthermore, by substituting $z := \frac{y-x}{\sqrt{t}}$, we get $dy = \sqrt{t}dz$, hence
$$ \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi t}} \exp \left( - \frac{(y-x)^2}{2t}\right) f(y) dy = \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi }} \exp \left( - \frac{z^2}{2}\right) f(\sqrt{t}z + x) dz $$ Now, $$ \left|\left|\frac{1}{\sqrt{2 \pi }} \exp \left( - \frac{z^2}{2}\right) f(\sqrt{t}z + x)\right|\right| \leq \frac{1}{\sqrt{2 \pi }} \exp \left( - \frac{z^2}{2}\right) ||f||$$ which is integrable, so by DCT we may pull the limit inside of the integral, i.e.: $$\lim_{t\rightarrow 0} P_t f(x) = \int_{-\infty}^\infty \lim_{t\rightarrow 0}\frac{1}{\sqrt{2 \pi }} \exp \left( - \frac{z^2}{2}\right) f(\sqrt{t}z + x) dz = f(x)$$
Is this correct? I think there has to be a mistake, since it was never used that $f$ is vanishes at infinity, only boundedness was used. As a matter of fact, the next exercise is to show that this fails in the case where we only consider $C_b(\mathbb{R})$, i.e. real functions that are bounded but don't vanish at infinity. I am confused, because I don't see why / where my argumentation fails.
Edit: Okay, thanks to the comment I realized that I have only shown pointwise convergence, however it should be easy to verify convergence w.r.t. the $|| \cdot ||_\infty$ norm. We have: $$ \left| \left| P_t f - f\right| \right| = \sup_{x \in \mathbb{R}}\left| \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi t}} \exp \left( - \frac{(y-x)^2}{2t}\right) (f(y)-f(x)) dy\right| \\ = \sup_{x \in \mathbb{R}}\left| \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \exp \left( - \frac{z^2}{2}\right) (f(\sqrt{t}z+x)-f(x)) dy\right| \\ \leq \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \exp \left( - \frac{z^2}{2}\right) \sup_{x \in \mathbb{R}}\left|f(\sqrt{t}z+x)-f(x) \right| dy $$ Now again, we notice that the integrand on the RHS is bounded by $\frac{2}{\sqrt{2 \pi }} \exp \left( - \frac{z^2}{2}\right) ||f||$, thus by DCT we get $$\lim_{t\rightarrow 0}\left| \left| P_t f - f\right| \right| = 0$$