1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.