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:
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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 $G(z)$.
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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.
Best Answer
OK, what I missed is that since the order of $G(z)$ is bounded by $1$, we know that $h(z) = \alpha z$. The rest is straightforward. Here I fill in the details using the above fact.
Proof. Let $M = \max_{t \in [-1,1]} |g(t)|$. Then for $y = \Im z$, for large enough $z$, we have $$|G(z)| \leq M \int_{-1}^{1} e^{-ty}dt = M \frac{e^{|y|} - e^{-|y|}}{|y|} \leq Me^{|y|} \leq Me^{|z|}$$ So the order $\rho_G \leq 1$ and accordingly the genus is $0$ or $1$.
Now, assume that $G(z)$ has finitely many roots and denote the non-zero roots by $a_k$. Since the genus of $G$ is bounded by $1$ we can write either $$G(z) = Cz^m e^{\alpha z} \prod_k \left(1 - \frac{z}{a_k}\right)$$ or $$G(z) = Cz^m e^{\alpha z} \prod_k \left(1 - \frac{z}{a_k}\right) e^{\frac{z}{a_k}}.$$ Because the products are finite, we can absorb the exponents in the later product into the outer exponent and rewrite $G(z)$, in both cases, as $$G(z) = e^{\alpha z} P(z)$$ for some $\alpha \in \mathbb{C}$, where $P(z)$ is a non-zero polynomial.
From real analysis we know that $\int_{-1}^1 g(t) \sin xt dt$ and $\int_{-1}^1 g(t) \cos xt dt$ tend to zero as $x$ tends to $\pm \infty$. This implies that $\lim_{x\to\pm\infty}G(x) = 0$, or $$\lim_{x\to\pm\infty} e^{\Re\alpha x} |P(x)| = 0.$$ Considering $x \to +\infty$ we must have $\Re\alpha < 0$. Considering $x \to -\infty$, $\Re\alpha > 0$. Contradiction. $\blacksquare$