If $h(z)=0$ for all $z\in\mathbb{R}^+$ then $h(z)=0$ in $\mathbb{C}$.

Question

I'm doing a course in probability and while calculating the characteristic function of the Gamma distribution my teacher did the following reasoning: (where $X\sim \mathcal{G}(r,\lambda)$)

By the definition of $\phi_X$: $$\phi_X(\xi)=\int_0^\infty e^{i\xi x}\frac{\lambda^r}{\Gamma(r)}\:x^{r-1}\:e^{-\lambda x}\:\mathrm{d}x=\frac{\lambda^r}{(\lambda-i\xi)^r}\int_0^\infty \frac{(\lambda-i\xi)^r}{\Gamma(r)}\:x^{r-1}\:e^{-(\lambda-i\xi) x}\:\mathrm{d}x.$$ Let $h:\mathbb{C}\to\mathbb{C}$ be defined as $$h(z)=\int_0^\infty \frac{z^r}{\Gamma(r)}\:x^{r-1}\:e^{-z x}\:\mathrm{d}x.$$ (That is, $\phi_X(\xi)=\lambda^r h(\lambda-i\xi)/(\lambda-i\xi)^r$.) If $z\in\mathbb{R}^+$, then $h(z)=1$ since this is the integral of the density of $\mathcal{G}(r,z)$.

After that he told us that, since $h$ is holomorphic, $h(z)=1$ for all $z\in\mathbb{C}$. I don't get how we can do that. $\mathbb{R}^+$ is not an open subset of $\mathbb{C}$ we I can't use any extension results that I know.

Is there some result that proves this?

Answer 1

It follows from the identity theorem and from the fact that $\mathbb{R}^+$ has accumulation points.

Answer 2

This is result on Analytic functions. If the set of zeros of an analytic function $ƒ$ has an accumulation point inside its domain, then $ƒ$ is zero everywhere on the connected component containing the accumulation point.

This is the case here. Any $z \in \mathbb R_+$ is an accumulation point of zeros of $h-1$.