I am trying to prove the following:
Let $g \in C[-1,1]$. Then the function $G(z) = \int_{-1}^1 e^{itz}g(t)dt$ has infinitely many zeros.
I know that $G(z)$ is entire and $\lim_{x \to \pm \infty} G(x) = 0$. I have tried the following. Assume the contrary, that is, that $G(z)$ has only $n \in \mathbb{N}$ zeros. Then we can write it as $G(z) = e^{h(z)}P(z)$ where $P(z)$ is a polynomial of degree $n$ and $h(z)$ is entire. The limit above implies that $h(x) + \log |P(x)| \to -\infty$, i.e. that $h(x) \to -\infty$ (on the real axis) faster than some asymptotically logarithmic positive function.
Unfortunately the above does not seem to solve the problem, or at least I do not know how to continue.
Asking for your guidance.
EDIT: Other thought were:
Approximate $g(t)$ with a step function $h_n(t)$ with $2^n$ steps. Define $H_n(z)$ as the transform of $h_n(t)$, show that $|G(z) - H_n(z)| < |H_n(z)|$ and apply Rouche's theorem. One problem is that $H_n(z)$ is also small on the boundary, and even if I could prove the inequality it is still unclear how infinity of zeros follows for $G(z)$.
Show that $G(z)$ is of fractional order and apply Hadamard's theorem. This is clearly false since I can show that the order of $G(z)$ is bounded by $1$ from above, and, at least for some $g(t)$, the bound is achieved.