# how to prove that a function vanishing at an interval is identically zero?

Let $$\phi(s):=\int_{0}^{\infty}\exp(-st)g(t)dt$$ for $$g\in L_1(0,\infty)$$. Assume that $$\phi(s)=0$$ for $$s\in[0,\frac{1}{2})$$. How to prove that $$\phi(s)=0$$ for every $$s\in[0,\infty)$$.

## 1 Answer

I claim that $$\phi$$ is an entire function.

Indeed let $$\gamma$$ be a simply closed curve in the complex plane.

Then $$\int_{\gamma} \phi(z) dz = \int_{\gamma}\int_0^\infty \exp(-zt)g(t) dz dt = \int_0^\infty \int_\gamma \exp(-zt) g(t)dz dt$$ by Fubini's theorem.

But the function $$\exp(-zt)$$ is entire as a function of $$z$$.

Hence, by the Cauchy-Goursat theorem $$\int_\gamma \exp(-zt) dz$$ vanishes for all $$t \in (0, \infty)$$.

Therefore, $$\int_\gamma \phi(z) dz = 0$$ for every simply closed curve $$\gamma$$ in the complex plane. Moreras theorem, then, implies that $$\phi$$ is entire.

Now since $$\phi$$ is an analytic function that vanishes on a set with accumulation point we get that $$\phi$$ is identically zero by the uniqueness theorem for analytic functions.