# The infinitesimal generator of gaussian is half the Laplacian.

This is an example from Davies' One-Parameter Semigroups.

Define the bounded operators $T_t$ on $L^p(\mathbb{R}^n)$ by $$(T_tf)(x)=(2\pi t)^{-n/2} \int_{\mathbb{R}^n} e^{-(x-y)^2/2t}f(y)dy$$ then one may easily show that $T_t$ is a one-parameter semigroup on $L^p(\mathbb{R}^n)$. By using the invariance of $\mathcal{S}$ under Fourier transforms one sees that $T_t(\mathcal{S})\subset \mathcal{S}$ for all $t\ge 0$, where $\mathcal{S}$ is the Schwartz Space. Then it says that $$(Zf)(x)=\frac{1}{2}\Delta f(x)$$ for all $f\in \mathcal{S}$.

I don't understand how to get this last statement. How can we show that the infinitesimal generator of $f$ becomes half the Lapalcian? I would greatly appreciate any help.

• Do you want a purely analytic answer? There is a very simple probabilistic answer: $T_t$ is the Markov semigroup of Brownian motion, so that $Z$ is the infinitesimal generator of Brownian motion, whose identity is revealed by Ito's formula. – Ian May 22 '17 at 2:01
• @Ian I am looking for a purely analytic answer. This example came up at the very first section on semigroup of operators, but I don't know how to use $Z=\lim_{t\to 0} \frac{1}{t} (T_t -I)$ to show the identity. – takecare May 22 '17 at 2:06
• Well, then you have to calculate a limit of an integral: $\lim_{t \to 0^+} (2 \pi)^{-n/2} t^{-n/2-1} \int_{\mathbb{R}^n} e^{-|x-y|^2/(2t)} (f(y)-f(x)) dy$. For that you could consider splitting the integral between a $\epsilon$ ball around $x$ and the rest of $\mathbb{R}^n$, Taylor expand on the ball out to second order, then bound the rest (using the fact that $f \in \mathcal{S}$). – Ian May 22 '17 at 2:08
• @Ian Yes that's the limit I don't know how to calculate. Can you give me some help on how to compute this limit? – takecare May 22 '17 at 2:10