I apologice in advance if this question is too trivial. Let $\nu$ be the standard gaussian measure and $f\in L^2_{\nu}(\mathbb R)$.
Let \begin{equation} g(z)=\int_{\mathbb R} e^{i x z} f(x)\nu(d x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb R} e^{i x z} f(x) e^{-x^2/2} dx \end{equation}
How can I see the following implication ?
All derivatives of the holomorphic function $g(z)$ vanish at $z=0$ implies that $g\equiv 0$.
At first I tried thinking of $g$ as the characteristic function of a r.v. whose distribution has density $f(x)e^{-x^2/2}$ wrt the Lebesgue measure, but we don't know whether $f$ is positive or not.
Can you give me some hint?
Thanks in advance!
If $g$ is holomorphic in a neigborhood of $0$, then $$g(z)=\sum_{n=1}^\infty a_nz^n,$$ and $a_n=\frac{g^{(n)}(0)}{n!}.$ If all derivatives vanished at $0$, then $a_n=0$ for all $n$. Therefore $g=0$ in a neigborhood of $0$. Using the analytic continuation theorem yields that $g=0$ on $\mathbb C$.