Entire rigidity under Gaussian growth and real-axis decay

The critical function. Put $\mathrm{\Phi }(z):=e^{\pi z^2\mathrm{/}a}F(z)$. The map $z\mapsto \pi z^2\mathrm{/}a$ is a polynomial, hence complex differentiable everywhere, and composing it with the entire exponential and multiplying by the entire $F$ shows that $\mathrm{\Phi }$ is entire ([F1, F2]). For real $x$ and $y$, using $\mathrm{\mid }{e}^{\pi z^2\mathrm{/}a}\mathrm{\mid }=e^{\pi \mathrm{Re}(z^2)\mathrm{/}a}$ and the two hypotheses, $\mathrm{\mid }\mathrm{\Phi }(x)\mathrm{\mid }=e^{\pi x^2\mathrm{/}a}\mathrm{\mid }F(x)\mathrm{\mid }\le C_2{e}^{-\pi (b-1\mathrm{/}a)x^2}\le C_2,\mathrm{\mid }\mathrm{\Phi }(iy)\mathrm{\mid }=e^{-\pi y^2\mathrm{/}a}\mathrm{\mid }F(iy)\mathrm{\mid }\le C_1,$ and, since $\mathrm{Re}(z^2)=(\mathrm{Re}z{)}^2-(\mathrm{Im}z{)}^2$, $\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }=e^{\pi ((\mathrm{Re}z{)}^2-(\mathrm{Im}z{)}^2)\mathrm{/}a}\mathrm{\mid }F(z)\mathrm{\mid }\le C_1{e}^{\pi (\mathrm{Re}z{)}^2\mathrm{/}a}\mathrm{.}$ In particular $\mathrm{\mid }\mathrm{\Phi }(0)\mathrm{\mid }=\mathrm{\mid }F(0)\mathrm{\mid }\le C_2\le C$.

Sector data. Fix $\delta >0$ and any $\theta$ with $\arctan \frac{\pi }{2a\delta }<\theta <\frac{\pi }{2}$; equivalently $\pi {\cos }^2\theta \mathrm{/}a<\delta \sin 2\theta$. Fix also $\epsilon >0$ small enough that $(2+\epsilon )\theta <\pi$, and put ${\sigma }_{\epsilon }:=\cos \frac{(2+\epsilon )\theta }{2}>0,{\eta }_{\epsilon }:=\frac{\pi }{2}-\frac{(2+\epsilon )\theta }{2},$ defining two auxiliary functions on the sectors $S_1=\{0<\arg z<\theta \}$ and $S_2=\{\pi -\theta <\arg z<\pi \}$, both contained in the slit plane $S$ of [F3]: $h_{\epsilon }(z):=\exp \text{\hspace{0.17em}⁣}\left(i\epsilon e^{i\mu }{z}_{\mathrm{p}\mathrm{r}}^{2+\epsilon }\right),q_{\delta }(z):=e^{i\delta z^2}(z\in S_1),q_{\delta }(z):=e^{-i\delta z^2}(z\in S_2),$ with $\mu :={\eta }_{\epsilon }$ on $S_1$ and $\mu :={\eta }_{\epsilon }-(2+\epsilon )(\pi -\theta )$ on $S_2$. Since $z_{\mathrm{p}\mathrm{r}}^{2+\epsilon }$ is holomorphic on $S$ and exponentials and polynomials are entire, $h_{\epsilon }$ and $q_{\delta }$ are holomorphic on each sector, and so is $G_{\epsilon }:=h_{\epsilon }{q}_{\delta }\mathrm{\Phi }$ ([F1, F2, F3]). For $z=r{e}^{i\alpha }$ in the closure of either sector, $\mathrm{\mid }{z}_{\mathrm{p}\mathrm{r}}^{2+\epsilon }\mathrm{\mid }=r^{2+\epsilon }$ and $\mathrm{\mid }{h}_{\epsilon }(z)\mathrm{\mid }=\exp \text{\hspace{0.17em}⁣}(-\epsilon r^{2+\epsilon }\sin (\mu +(2+\epsilon )\alpha ))\le e^{-\epsilon {\sigma }_{\epsilon }{r}^{2+\epsilon }}\le 1,$ because $\mu +(2+\epsilon )\alpha \in [\frac{\pi }{2}-\frac{(2+\epsilon )\theta }{2},\frac{\pi }{2}+\frac{(2+\epsilon )\theta }{2}]$ for $\alpha \in [0,\theta ]$ on $S_1$ and for $\alpha \in [\pi -\theta ,\pi ]$ on $S_2$. Also $\mathrm{\mid }{q}_{\delta }(z)\mathrm{\mid }=e^{\mp \delta \mathrm{Im}(z^2)}$ with the sign making $\mathrm{\mid }{q}_{\delta }\mathrm{\mid }\le 1$ on the sector.

Boundary bounds. On the far ray $\arg z=\theta$ of $S_1$ one has $\mathrm{Re}z=r\cos \theta$ and, by [F6], $\mathrm{\mid }{q}_{\delta }(z)\mathrm{\mid }=e^{-\delta r^2\sin 2\theta }$, so steps 1.1 and 1.2 give $\mathrm{\mid }{G}_{\epsilon }(z)\mathrm{\mid }\le \mathrm{\mid }{q}_{\delta }(z)\mathrm{\mid }\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }\le C_1\exp (r^2(\pi {\cos }^2\theta \mathrm{/}a-\delta \sin 2\theta ))\le C_1\le C$ by the choice of $\theta$. On the far ray $\arg z=\pi -\theta$ of $S_2$ one has $\mathrm{Re}z=-r\cos \theta$ and $\mathrm{\mid }{q}_{\delta }(z)\mathrm{\mid }=e^{-\delta r^2\sin 2\theta }$ as well, so the same computation gives $\mathrm{\mid }{G}_{\epsilon }(z)\mathrm{\mid }\le C_1\le C$ there. On the anchor rays $\arg z=0$ and $\arg z=\pi$ the identity [F6] gives $\mathrm{\mid }{q}_{\delta }(z)\mathrm{\mid }=1$, and the display of step 1.2 gives $\mathrm{\mid }{h}_{\epsilon }\mathrm{\mid }\le 1$; hence $\mathrm{\mid }{G}_{\epsilon }(z)\mathrm{\mid }\le C_2\le C$ there, and at $z=0$ one has $\mathrm{\mid }{G}_{\epsilon }(0)\mathrm{\mid }=\mathrm{\mid }F(0)\mathrm{\mid }\le C$.

Boundedness on the sectors. In the closure of either sector, steps 1.1 and 1.2 give $\mathrm{\mid }{G}_{\epsilon }(z)\mathrm{\mid }\le C_1\exp (r^2(\frac{\pi }{a}+2\delta )-\epsilon {\sigma }_{\epsilon }{r}^{2+\epsilon })⟶0(r\to \mathrm{\infty }),$ because ${\cos }^2\alpha \le 1$ and $\mathrm{\mid }\sin 2\alpha \mathrm{\mid }\le 1$ on the compact angular interval; this is the control at infinity. Step 2.1 gives $\mathrm{\mid }{G}_{\epsilon }\mathrm{\mid }\le C$ on the boundary rays, and $\mathrm{\mid }{G}_{\epsilon }\mathrm{\mid }$ is continuous on the closed sector (the principal power extends continuously from $S$ to the closure of each sector, and $\mathrm{\mid }{G}_{\epsilon }\mathrm{\mid }$ is a finite product of continuous functions), so every boundary point has a neighbourhood $V$ with $\mathrm{\mid }{G}_{\epsilon }\mathrm{\mid }<C+{\epsilon }^{\mathrm{\prime }}$ on $V$ intersected with the sector. The maximum modulus principle [F5], applied to the domains $S_1$ and $S_2$, therefore gives $\mathrm{\mid }{G}_{\epsilon }\mathrm{\mid }\le C$ on both sectors.

Removing the auxiliary and the sector truncation. Fix $\delta >0$. The perturbation $h_{\epsilon }$ tends to $1$ pointwise as $\epsilon \to 0$ along any sequence with $(2+\epsilon )\theta <\pi$, so step 3.1 gives $\mathrm{\mid }{q}_{\delta }\mathrm{\Phi }\mathrm{\mid }\le C$ on each sector $S_1(\theta ),S_2(\theta )$ for every admissible $\theta \in (\arctan \frac{\pi }{2a\delta },\frac{\pi }{2})$. Since $\theta$ was arbitrary in that interval and the sectors with larger $\theta$ contain those with smaller $\theta$, while $q_{\delta }$ does not depend on $\theta$, taking $\theta \to \pi \mathrm{/}2$ yields $\mathrm{\mid }{e}^{i\delta z^2}\mathrm{\Phi }(z)\mathrm{\mid }\le C$ for every $z$ with $0<\arg z<\pi \mathrm{/}2$ and $\mathrm{\mid }{e}^{-i\delta z^2}\mathrm{\Phi }(z)\mathrm{\mid }\le C$ for every $z$ with $\pi \mathrm{/}2<\arg z<\pi$. Letting now $\delta \downarrow 0$ along any sequence, $\mathrm{\mid }{q}_{\delta }(z)\mathrm{\mid }\to 1$ at each fixed $z$ by [F1] (the exponent $i\delta z^2\to 0$), so $\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }\le C(0<\arg z<\pi ,z\mathrm{\ne }0)\mathrm{.}$

The lower half-plane. The function $\tilde{\mathrm{\Phi }}(z):=\mathrm{\Phi }(-z)$ is entire ([F2]) and satisfies the same three estimates as $\mathrm{\Phi }$ in step 1.1, since the bounds $\mathrm{\mid }F(-x)\mathrm{\mid }\le C_2{e}^{-\pi b{x}^2}$, $\mathrm{\mid }F(-iy)\mathrm{\mid }\le C_1{e}^{\pi y^2\mathrm{/}a}$ and the growth estimate only involve absolute values and $(\mathrm{Re}(-z){)}^2=(\mathrm{Re}z{)}^2$. Applying steps 1.1–4.1 to $\tilde{\mathrm{\Phi }}$ gives $\mathrm{\mid }\mathrm{\Phi }(-z)\mathrm{\mid }\le C$ for $0<\arg z<\pi$, that is, $\mathrm{\mid }\mathrm{\Phi }(w)\mathrm{\mid }\le C$ for $-\pi <\arg w<0$.

Conclusion of the critical case and of (i). By steps 4.1 and 5.1 the entire function $\mathrm{\Phi }$ satisfies $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }\le C$ off the coordinate axes, while on the axes step 1.1 gives $\mathrm{\mid }\mathrm{\Phi }(x)\mathrm{\mid }\le C_2\le C$ and $\mathrm{\mid }\mathrm{\Phi }(iy)\mathrm{\mid }\le C_1\le C$ directly. Hence $\mathrm{\Phi }$ is a bounded entire function, so $\mathrm{\Phi }$ is constant by [F4]; the constant is $\mathrm{\Phi }(0)=F(0)$. Therefore, whenever $b\ge 1\mathrm{/}a$, $F(z)=F(0)e^{-\pi z^2\mathrm{/}a}(z\in \mathbb{C})\mathrm{.}$ If $ab>1$ then $b>1\mathrm{/}a$, so the display applies; evaluating at real $x$ gives $\mathrm{\mid }F(0)\mathrm{\mid }{e}^{-\pi x^2\mathrm{/}a}=\mathrm{\mid }F(x)\mathrm{\mid }\le C_2{e}^{-\pi b{x}^2}$, that is, $\mathrm{\mid }F(0)\mathrm{\mid }\le C_2{e}^{-\pi (b-1\mathrm{/}a)x^2}$ for every real $x$, and letting $x\to \mathrm{\infty }$ forces $F(0)=0$ and hence $F\equiv 0$. If $ab=1$ then the display is assertion (ii).