Let $f$ be a function on $\Bbb R$ that satisfies
$$f(x) = O(e^{-\pi x^2})\quad\text{and}\quad\hat{f}(\xi) = O(e^{-\pi\xi^2}).$$
Then if $f$ is even then $\hat{f}$ extends to an even entire function. Moreover, if $g(z) = \hat{f}(z^{1/2})$, then g satisfies
$$|g(x)|\leq ce^{-\pi x}\quad\text{and}\quad |g(z)|\leq ce^{\pi R\sin^2(\theta/2)}\leq ce^{\pi|z|}$$
when $x\in\Bbb R$ and $z = Re^{i\theta}$ with $R\geq 0$ and $\theta\in\Bbb R$.
Question. Apply the Phragmen-Lindelof principle to the function
$$F(z) = g(z)e^{\gamma z}\quad\text{where}\ \gamma = i\pi{e^{-i\pi/(2\beta)}\over \sin\pi/(2\beta)}$$
and the sector $0\leq\theta\leq\pi/\beta<\pi$, and let $\beta\to\color{red}{1}$ to deduce that $e^{\pi z}g(z)$ is bounded in the closed upper half-plane.
My attempt: We apply the Phragmen-Lindelof on the given sector (we don't rotate). First, if $z = re^{i\theta}$ then
\begin{align*}
e^{\gamma z} &= \exp\left(i\pi{e^{-i\pi/(2\beta)}\over \sin\pi/(2\beta)} z\right)\\
i\pi{e^{-i\pi/(2\beta)}\over \sin\pi/(2\beta)}r(\cos\theta+i\sin\theta) & = i\pi{\cos(\pi/2\beta)-i\sin(\pi/2\beta)\over\sin(\pi/2\beta)}r(\cos\theta+i\sin\theta)\\
& \overset{\mathrm{\operatorname{Re}}}{=} {-\pi r\over\sin(\pi/2\beta)}(\cos(\pi/2\beta)\sin\theta-\sin(\pi/2\beta)\cos\theta)\\
& = -\pi r{\sin(\theta-\pi/2\beta)\over\sin(\pi/2\beta)}\\
\end{align*}
Note that $0<\theta<\pi/\beta$ so
$$-{\pi\over 2}<-{\pi\over 2\beta}<\theta-{\pi\over 2\beta}<{\pi\over 2\beta}<{\pi\over 2}$$
and $\sin$ is strictly increasing function on $(-\pi/2,\pi/2)$. This implies ${\sin(\theta-\pi/2\beta)\over\sin(\pi/2\beta)}<1$. But this is not good.
$$|F(z)| = |g(z)e^{\gamma z}|=|g(z)|e^{\operatorname{Re}(\gamma z)} = ce^{\pi r(1-\epsilon)},\quad\epsilon<1,$$
by the above argument. So we can't apply P-L principle. Even if I can apply, I'm not sure how the function is bounded on the boundary of the given sector. I'm stuck here. Could you help?
Best Answer
I looked up this problem again today and I found that it's not a very hard question. I think I'm misunderstanding something when I asked this question.
From the post, I concluded that $$|F(z)| = |g(z)e^{\gamma z}| = |g(z)|e^{\operatorname{Re}\gamma z} = ce^{\pi|z|(1-\epsilon_{\theta})},\quad \epsilon_\theta = {\sin(\theta-\pi/2\beta)\over\sin(\pi/2\beta)}.$$ On the boundary of $S = \left\{z:0\leq\arg z\leq{\pi\over\beta}\right\}$ (here, we choose principal branch), \begin{align*} |g(x)|\leq ce^{-\pi x}\Rightarrow |F(x)| & \leq ce^{-\pi x}e^{\pi x} = c,\quad x\in\Bbb R\\ |g(z)|\leq c^{\pi|z|}\Rightarrow |F(z)| & \leq ce^{\pi r}e^{-\pi r} = c,\quad z = re^{i\pi/\beta}\\ \end{align*} Hence, $|F(z)|\leq c$ on the boundary of $S$.
Note that $\epsilon_{\theta}$ is bounded for $0\leq\theta\leq{\pi\over\beta}$ so $$|F(z)|\leq ce^{a|z|},\quad\text{for some}\ a>0.$$ Now we can apply the Phragmen-Lindelof theorem: Choose $1<\alpha<\beta$ and set $$F_\epsilon(z) = F(z)\exp\left(-\left(e^{-{\pi\over 2\beta} i}z\right)^\alpha\right)$$ Then if $z = re^{i\theta}$, $$-\left(e^{-{\pi\over 2\beta} i}re^{i\theta}\right)^\alpha = -r^\alpha e^{i(\theta-\pi/2\beta)\alpha},\quad -{\pi\over 2}<-{\alpha\over 2\beta}\pi<\left(\theta-{\pi\over 2\beta}\right)\alpha<{\alpha\over 2\beta}\pi<{\pi\over 2}.$$ So, $\cos(\theta-\pi/2\beta)\alpha>0$ on $S$. hence, $$|F_{\epsilon}(z)|\leq ce^{ar}e^{-r^\alpha\cos((\theta-\pi/2\beta)\alpha)}\to 0,\quad r\to\infty$$ Hence, $F_\epsilon$ is bounded and by the MMP, $|F_\epsilon(z)|\leq 1$ on $S$ and letting $\epsilon\to 0$, we conclude $|F(z)|\leq c$ on $S$.
If $\beta\to 1$ then $S\to\overline{\Bbb H^+}$ and $\gamma\to \pi$ so $$|g(z)e^{\pi z}|\leq c,\quad z\in\overline{\Bbb H^+}.$$
N.B. The same result holds in the lower half-plane, so by Liouville's theorem $e^{\pi z}g(z)$ is constant.
So far, the whole statement is about even function. But if we assume $f$ is an odd function, then a similar statement holds. Indeed, if $f$ is an odd function, then $f$ extends to an odd entire function $\hat{f}$. Since $\hat{f}(0) =0$ in this case, $\hat{f}(z)/z$ is an entire function which is even. Now, define $\tilde{g}(z) = \hat{f}(z^{1/2})/z^{1/2}$ then we get the same bound and the question statement in this post equally holds for $\tilde{g}$. So, $\hat{f}(z)/z = \text{constant}$ for all $z\in\Bbb C$. Since $\hat{f}(0) =0$, we conclude $\hat{f}\equiv 0$. Applying inverse Fourier transform, $f\equiv 0$.
Now, arbitrary $f$ can be written as a sum of odd and even function $$f(z) = {f(z)+f(-z)\over 2}+{f(z)-f(-z)\over 2}$$ Combining the odd and even function results, we conclude $f$ is a constant multiple of $e^{-\pi x^2}$.