To show a integrable function which satisfies some conditions is identically zero.

Question

Let $f\in L^1([0,\infty))$ satisfying $$\int_0^\infty e^{itz}f(t)dt=0, \quad \text{ for all } z \text{ in upper half complex plane.}$$ Then to show $f\equiv 0.$

Answer 1

Just put $z=is$ where $s>0$. You get $\int e^{-st} f(t)dt=0$ for all $s >0$. This means that the Laplace transform of $f$ is $0$ and this implies that $f=0$ a.e.