8
$\begingroup$

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 $a,b>0$.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
+50
$\begingroup$

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 $\delta \to 0$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.