## On the Euclidean Growth of Entire Functions

In Mapping properties of analytic functions on the unit disk, Proceedings of the American Mathematical Society, Vol. 135, N. 9 (2007), 2893-2898. I show that there is a universal constant $0 such that whenever $f$ is analytic in the unit disk $\mathbb{D}$ and whenever a disk  $D$ centered at $f(0)$ has the property that the Euclidean area counting multiplicity of $f$ over $D$ is strictly less than the Euclidean area of $D$, then $f$ must necessarily send the smaller disk $r_0\mathbb{D}$ into $D.$ In formulas,

$\displaystyle \int_{\{z\in\mathbb{D}: f(z)\in D\}}|f^{\prime}(z)|^2 dA(z)< A(D) \Longrightarrow f(r_0\mathbb{D})\subset D.$

At the time I did not think of rephrasing this for entire functions. It goes like this. Suppose $g$ is an entire function (say with $g(0)=0$). We measure the growth of $g$ in two different ways. Given a radius $r>0$, one quantity we can measure is the maximum modulus

$\displaystyle M(r)=\max_{|z|=r}|g(z)|.$

The other quantity we will be concerned with is $E(r)$ the Euclidean area counting multiplicity covered by $g$ when restricted to the disk $\{|z|, i.e.,

$\displaystyle E(r)=\int_{|z|

The claim then is, that under this notations and conditions,  there is an absolute  constant $\theta_0>1$ so that

$\displaystyle M(r)\leq \sqrt{E(\theta_0 r)/\pi}.$

This follows from the result in the unit disk mentioned above by picking $\theta_0>1/r_0$. Indeed, suppose that for some radius $r>0$, we have $M(r)>\sqrt{E(\theta_0 r)/\pi}$. Then,  $E(\theta_0 r)<\pi M(r)^2$. Now apply the result mentioned above to the function $f(\zeta)=g(\theta_0 r\zeta)$ and to the image disk  $D=\{|w|. The Euclidean area over $D$ covered by $f$ is less than the total Euclidean area covered by $f$, which is $E(\theta_0 r)$, and is hence strictly less than $A(D)$. So we find that $g$ must necessarily send the disk $\{|z| into $\{|w|, but this  contradicts the maximum principle because we chose $r_0\theta_0>1$.

This post is listed under “Curiosity”, because the classical and much more important way to measure the growth of an entire function is to use the Spherical area in the image. Namely,

$\displaystyle S(r)=\int_{|z|

Moreover, $S(r)$ is related to the growth of $\log M(r)$ instead of $M(r)$.