# Zeros of Fourier transform of a function in $C[-1,1]$

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.

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.

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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. $\blacksquare$

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This is great. Nicely done. – Antonio Vargas Dec 23 '12 at 3:31

Because $G$ is the Fourier transform of a function with compact support, you probably must use the inequality from the Paley-Wiener theorem somehow. In our case, the inequality is $|G(z)| \leq C e^{|\operatorname{Im} z|}$ with $C=2\max_{[-1,1]}|g(t)|$.

Below, I try to use this inequality together with your assumption $G(z)=e^{h(z)}P(z)$ to derive a contradiction. Because this is homework, there is probably a much simpler solution. I wasn't even aware of the Borel–Carathéodory theorem, which is used to prove the missing part in the answer to the follow up question.

With $G(z)=e^{h(z)}P(z)$, we get $\operatorname{Re} h(z) + \log|P(z)/C| \leq |\operatorname{Im} z|$. The term $\log|P(z)/C|$ is bounded from below for $|z| \geq R$, if all zeros of $P(z)$ are contained within $|z| < R$. So we have $$\operatorname{Re} h(z) \leq |\operatorname{Im} z|+c \qquad \forall|z|\geq R$$

Now some strong theorem (like the Big Picard Theorem) is needed to show that this implies that $h(z)$ can't have an essential singularity at infinity. The answer to this follow up question shows that we could conclude this from $$\operatorname{Re}(f(z))\leq |z|^n\quad \forall z\in \mathbb{C}$$ With $c':=\max_{|z|\leq R}|h(z)|$, the function $f(z):=h(z)-c-c'$ satisfies this inequality, so neither $f(z)$ nor $h(z)$ can have an essential singularity at infinity. Because $h(z)$ can't have an essential singularity at infinity, it must be of the form $h(z)=i\alpha z + b$ with $\alpha \in [-1,1]$. But then $G(z)$ doesn't satisfy $\lim_{x \to \pm \infty} G(x) = 0$, so we have a contradiction.

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Everything is fine except for the missing part ... Except for the intuition that near essential singularities the function goes crazy enough, I cannot see why $\frac{h(z)}{z}$ cannot maintain an almost imaginary ratio with $\frac{z}{|z|}$ so that its image would still be almost all of $\mathbb{C}$ yet the real part of the product would remain close to zero. – ybungalobill Dec 20 '12 at 7:18
OK. Actually what I said (except that for ratio I meant product) is equivalent to the question of whether $\Im h(z)$ may grow faster than $\Re h(z)$. I'll see if I can apply Cauchy-Riemann equations. Does it even sound correct? – ybungalobill Dec 20 '12 at 7:46
@ybungalobill I asked the missing part as a separate question. The Cauchy-Riemann equations are the reason why I believe that there can't be an essential singularity. However, the ... was more meant to symbolize a laborious cornering of the function instead of an elegant argument. I'm curious whether somebody will present an elegant argument. – Thomas Klimpel Dec 20 '12 at 8:10
@ybungalobill Based on the answer to the other question, the proof should now be complete. But since this is homework, I would also be interested to learn about the "official" solution, i.e. a solution which doesn't need to reference some strong or unknown theorem on the way. – Thomas Klimpel Dec 20 '12 at 17:51
+1 for solving this in an interesting way :). See the "intended" solution posted by me. – ybungalobill Dec 22 '12 at 11:58