Here is a problem from Stein and Shakarchi Complex Analysis, can somebody help me to solve it? I guess we can use Phragmen-Lindelof theorem but I don't know the exact way.

Suppose $f(z)$ is an entire function s.t. $f(z)=O(e^{c_1|z|^2})$ for some $c_1>0$, and for $x$ real $f(x)=O(e^{-c_2|x|^2})$ for some $c_2>0$. Then $f(x+iy)=O(e^{-ax^2+by^2})$ for some $a,b>0$.

  • $\begingroup$ Hint: Note $f(iy) = O(e^{c_1 y^2})$, and apply the theorem to $f(z) e^{C z^2}$ for $c_1 < C < c_2$ to the upper right quadrant $S = \{z\ |\ 0 < \arg z < \pi/2\}$. $\endgroup$ – Evan Aug 10 '13 at 17:36
  • $\begingroup$ Yes, that's what I was trying to do, but for this region, we should require $|f(z)|\le Ce^{c_1|z|^a}$ with $a<2$. $\endgroup$ – Danqing He Aug 10 '13 at 18:33
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    $\begingroup$ Apply Theorem 2 from B. Ya. Levin, Lectures on Entire Functions. Here is its screenshot. $\endgroup$ – user64494 Aug 10 '13 at 20:24
  • $\begingroup$ Oh yikes, and I made an extra assumption on the constants. So the idea is to apply the theorem twice? Once to get up to angle $\pi/4$ and again to finish? $\endgroup$ – Evan Aug 10 '13 at 23:24
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    $\begingroup$ I just realized that we could multiply by $g(z) = e^{c_2 z^2 + iBz^2}$ to handle the $c_1 < c_2$ issue. this would have magnitude $|g(x,y)| = \exp(c_2(x^2 - y^2) - 2B(xy))$. Inverting, $|g(x,y)|^{-1} = \exp(-c_2 x^2 + c_2 y^2 + 2Bxy)$, but you can bound $2xy \leq \epsilon^2 x^2 + y^2/\epsilon^2$ for any $\epsilon$. You can't exactly use the imaginary axis but you can probably work out pushing it a little bit inside? $\endgroup$ – Evan Aug 11 '13 at 16:09

The credit belongs to @user64494 and @Evan (and Phragmén–Lindelöf).

Write $z = x + iy$. It is enough to consider $x$ such that (say) $x^2 > y^2$, since otherwise $$ c_1 |z|^2 = c_1(x^2+y^2) \le 2c_1y^2 \le 3c_1 y^2 - c_1 x^2, $$ and so by the assumption we have $$|f(z)|\le O(e^{c_1|z|^2}) = O(\exp(- c_1 x^2 + 3c_1 y^2)).$$

By symmetry we may consider the case $x \ge y \ge 0$ (the sector $D = \{ z : \arg z \in [0,\frac14 \pi]\}$).

Lemma 1: (baby Phragmén–Lindelöf) Let $M>0$. If $g(z)$ is analytic inside the sector $D$, $|g(z)|\le M$ on the sides of $D$, and $|g(z)|\le e^{C|z|^2}$ inside $D$ for some $C>0$, then $|g(z)|\le M$ inside $D$.

Assuming Lemma 1 is correct we can prove:

Lemma 2: Let $A,B>0$ and $C\ge 1$. Suppose $g(z)$ is analytic inside the sector $D$, that $|g(z)| \le \exp(C |z|^2)$ for $z\in D$, and that $|g(r)| \le C \exp(-A r^2)$, $|g(r e^{\frac14 \pi i})| \le C \exp(B r^2)$, for $r\ge 0$, then $$ |g(z)| \le C \exp(-A(x^2-y^2) + 2B xy), \,\, \forall z = x + iy \in D. $$

Assuming Lemma 2 is correct we have for some $C, \varepsilon>0$: $$ \begin{gather*} |f(z)| & \le & C \exp(-c_2(x^2-y^2) + 2c_1 xy) \le C \exp\left(-c_2(x^2-y^2) + c_1 \varepsilon x^2 + \frac{c_1}{\varepsilon} y^2\right)\\ & = & C \exp\left(-(c_2 - \varepsilon c_1) x^2 + (c_2 + \frac{c_1}{\varepsilon}) y^2\right), \end{gather*}$$ which is what we wanted to proved.

Proof of Lemma 1: By rotating the function $g(z)$ we may work in the sector $D^\prime = \{ z : | \arg z | \le \frac18 \pi \}$. For $\delta > 0$, we introduce $$ \varphi_\delta(z) = g(z) e^{-\delta z^3}. $$ We have $$ |\varphi_\delta(z)| \le \exp\left(C|z|^2 - \delta |z|^3 \cos\left(\frac{3 \pi}{8}\right) \right), \,\, z\in D^\prime. $$ This means there exists $R_\delta > 0$, such that $$ |\varphi_\delta(R e^{i \theta})| \le M , \,\, \forall R>R_\delta, |\theta| \le \frac{\pi}{8}.$$ By the maximum principle we have $|g(z)| \le M e^{\delta |z|^3}$ for any $z \in D^\prime$. Since $\delta > 0$ can be made arbitrarily small, we proved the lemma.

Proof of Lemma 2: The idea is similar. For $\delta > 0$, consider now the function $$\phi_\delta(z) = g(z) e^{(A-\delta+i(B+\delta))z^2}.$$ Cleary $|\phi_\delta(z)| \le \exp(C^\prime |z|^2),\, z \in D$ for some $C^\prime>0$. In addition, for $r\ge 0$, $$\begin{gather*} |\phi_\delta(r)| & \le & C \exp(-A r^2 + (A-\delta) r^2) = C \exp (-\delta r^2), \\ |\phi_\delta(re^{\frac14 \pi i})| & \le & C \exp(B r^2 -(B+\delta) r^2 ) = C \exp (-\delta r^2). \end{gather*}$$ Thus, by Lemma 1, $$ |g(z)| \le C | e^{-(A-\delta+i(B+\delta))z^2} | =C e^{-(A-\delta)(x^2-y^2)+2 (B+\delta) xy}, \, z = x + iy \in D. $$ Again we can take $\delta \to 0$.

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    $\begingroup$ Wonderful proof. +1 I would upvote more to all the people contributed to the proof if I could. $\endgroup$ – Hans Jul 22 '16 at 19:59
  • $\begingroup$ Great solution +1. There is a minor typo at the end, you missed writing $C$. I hope you don't mind my edit. $\endgroup$ – Sungjin Kim Mar 23 '17 at 2:37

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