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 $s\in[0,\infty)$.


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.


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