- One can give a neater formulation of the result in Exercise 10 by proving the following fact.
Suppose \(f ( z )\) is an entire function of strict order 2, that is,
\[ f (z) = O (e ^ {c _ {1} | z | ^ {2}}) \]
for some \(c _ { 1 } > 0\) . Suppose also that for x real,
\[ f (x) = O (e ^ {- c _ {2} | x | ^ {2}}) \]
for some \(c _ { 2 } > 0\) . Then
\[ | f (x + i y) | = O (e ^ {- a x ^ {2} + b y ^ {2}}) \]
for some \(a , b > 0\) . The converse is obviously true.
Solution
We prove the nontrivial implication. Assume \[ |f(z)|\le C e^{c_1|z|^2} \qquad(z\in\mathbb C) \] and \[ |f(x)|\le C e^{-c_2x^2} \qquad(x\in\mathbb R), \] after increasing \(C\) if necessary.
It is enough first to treat the sector \[ D=\{z=x+iy: x\ge y\ge0\}. \] On the ray \(z=re^{i\pi/4}\) the global growth bound gives \[ |f(re^{i\pi/4})|\le C e^{c_1r^2}. \tag{1} \] On the positive real axis, \[ |f(r)|\le C e^{-c_2r^2}. \tag{2} \]
We use the following elementary Phragmén–Lindelöf lemma for the sector \(D\): if \(h\) is holomorphic in \(D\), continuous on its boundary, satisfies \(|h(z)|\le e^{K|z|^2}\) in \(D\), and \(|h|\le M\) on both boundary rays, then \(|h|\le M\) in \(D\). Indeed, after rotating \(D\) to \(|\arg z|\le\pi/8\), for \(\delta>0\) consider \[ h_\delta(z)=h(z)e^{-\delta z^3}. \] Since \(\cos(3\arg z)\ge \cos(3\pi/8)>0\) in the rotated sector, \[ |h_\delta(z)|\le \exp(K|z|^2-\delta |z|^3\cos(3\pi/8)), \] so on a sufficiently large circular arc it is at most \(M\). The maximum principle on the truncated sector yields \[ |h(z)|\le M e^{\delta |z|^3}. \] Letting \(\delta\downarrow0\) proves the lemma.
Fix \(\delta>0\) and set \[ H_\delta(z)=f(z)\exp\!\bigl((c_2-\delta+i(c_1+\delta))z^2\bigr). \] It has order at most \(2\). On the positive real axis, (2) gives \[ |H_\delta(r)|\le C e^{-\delta r^2}\le C. \] On the ray \(z=re^{i\pi/4}\), since \(z^2=ir^2\), \[ \left|e^{(c_2-\delta+i(c_1+\delta))z^2}\right| =e^{-(c_1+\delta)r^2}, \] so (1) gives \(|H_\delta(re^{i\pi/4})|\le C e^{-\delta r^2}\le C\). The lemma therefore implies \(|H_\delta(z)|\le C\) throughout \(D\).
Writing \(z=x+iy\) and then letting \(\delta\downarrow0\), we obtain \[ |f(z)|\le C\left|e^{-(c_2+ic_1)z^2}\right| =C e^{-c_2(x^2-y^2)+2c_1xy}. \tag{3} \] For any \(\varepsilon>0\), \[ 2c_1xy\le c_1\varepsilon x^2+c_1\varepsilon^{-1}y^2. \] Choose \(\varepsilon<c_2/c_1\). Then (3) yields \[ |f(x+iy)|\le C e^{-a x^2+b y^2} \] throughout \(D\), where \[ a=c_2-c_1\varepsilon>0, \qquad b=c_2+c_1\varepsilon^{-1}>0. \tag{4} \]
The same argument applies to the reflected sectors in the other quadrants. In the complementary region \(|y|\ge |x|\), the original growth estimate already gives \[ c_1(x^2+y^2)\le -c_1x^2+3c_1y^2, \] so an estimate of the form (4) holds there as well. Enlarging the constants gives a single pair \(a,b>0\) valid for all \(x,y\in\mathbb R\).
Conversely, if \[ |f(x+iy)|\le C e^{-ax^2+by^2}, \] then on the real axis \(|f(x)|\le Ce^{-ax^2}\), while \[ -ax^2+by^2\le b(x^2+y^2)=b|z|^2. \] Hence \(f(z)=O(e^{b|z|^2})\). This proves the converse.