There is a long history on constructing entire functions bounded along every direction. For example, we refer to Burckel's math review on Newman (Amer. Math. Monthly 1976 MR0387593) or this mathoverflow post
There is a Liouville theorem on such functions (first discovered by Carleman)
Theorem 1:[Carleman 1926] Let $f(z)$ be an entire function so that $$ |f(re^{i\theta})| \leq L(\theta), \ \ \theta \in [0, 2\pi). $$ where $L(\theta)$ is a measurable function on $[0, 2\pi)$ with $$ \int_{0}^{2\pi} \log^{+} \log^{+} L(\theta) d\theta<+\infty, \ \ \text{where for any given}\ x \in \mathbb{R}, \ \log^{+}x=\begin{cases} \ln x \ \ x \geq 1\\ 0 \ \ x \leq 1\\\end{cases}. $$ Then $f$ is constant.
Carleman, T.. Extension d'un th'{e}or`{e}me de liouville. Acta Math. 48, 3-4 (1926), 363--366.
Theorem 2: [Levinson and many people] Consider a rectangle $$ \Omega=\{z=x+yi\ |\ x \in (a_1, a_2), y \in (-b, b)\} $$ and $L(y)$ is a measurable function on $(-b, b)$ with $$ \int_{-b}^b \log^{+} \log^{+} L(y) dy<+\infty. $$ Then for any compact subset $K \subset \Omega$, there exists a constant $C$ depending only on $L(y)$ and $\operatorname{dist}(K, \partial \Omega)$ so that for any holomorphic function in $\Omega$ with $|f(x+yi)| \leq L(y)$ we always have $$ |f(z)| \leq C, \ \ \text{for any}\ z \in K. $$ Moreover, $C$ is decreasing with respect to $\operatorname{dist}(K, \partial \Omega)$.
Obviously Theorem 1 is a special case of Theorem 2 (like the connection between three-circle theorem and three-line theorem). Theorem 2 is proved by establishing similar result on $\ln |f|$, we may refer to Domar's beautiful proof (also presented in Koosis's book) of Theorem 2.
Koosis, P. The logarithmic integral. I., vol. 12 of Cambridge Studies in Advanced Mathematics, Cambridge University Press, Cambridge, 1998.
From a historical perspective, I am interested in Carleman's original proof. His 1926 paper was written in French. I have some difficulty in understanding arguments there. Let me summarize in below, following notations in his original paper.
Notations: $$ M(r)=\sup_{\varphi \in[0, 2\pi)} |f(re^{i\varphi}), \ \ln M(r)=v(r), \ \ln |f(\rho e^{i\varphi})|=U(\rho, \varphi) $$ $$ |f(re^{i\theta})| \leq L(\theta) \doteq e^{e^{\omega(\theta)}}, \ \int_0^{2\pi} \omega(\theta)d\theta<\infty. $$
For any given $r>0$, we pick a corresponding $\xi_{r}=r e^{i \theta_r}$ so that $|f(re^{i\theta_r})|=M(r)$. Moreover, we choose $\xi_{r-\delta}=(r-\delta)e^{i\theta_{r-\delta}}$ so that $|f(\xi_{r-\delta})|=M(r-\delta)$ and $\lim_{\delta \rightarrow 0} \xi_{r-\delta}=\xi_r$. Roughly speaking, we need a continuous path in the set of maximum modulus.
Now a fixed $r>0$, we pick a subdomain $D=\{z=\rho e^{i\varphi}, |\rho|<r, U(\rho, \varphi)>\frac{1}{2}v(r)\}$ so that $\xi_r \in \partial D$. In the meantime, we make sure $\partial D$ consists of two parts $\gamma \cup L$ where $U(\rho, \varphi) \geq \frac{1}{2}v(r)$ on $\gamma \subset \partial D(0, r)$, and $U(\rho, \varphi) = \frac{1}{2}v(r)$ on $L \subset D(0, r)$ respectively, Next we use $W(\rho, \varphi)$ to denote the harmonic measure of $\gamma$ with respect to $\partial D(0, r)$ at $\rho e^{i\varphi}$.
Pick $r$ large enough so that $v(r)>0$. Now we have $U(\rho, \varphi)-\frac{1}{2}v(r)-\frac{1}{2}v(r) W(\rho, \varphi) \leq 0$ on $\partial D$. By maximum principle (for subharmonic functions) we have inside $D$, $$ U(\rho, \varphi)-\frac{1}{2}v(r)-\frac{1}{2}v(r) W(\rho, \varphi) \leq 0. $$ Plug into $\rho=r-\delta$ and $\varphi=\theta_{r-\delta}$ and we get $$ v(\rho-\delta)-\frac{1}{2}v(r)-\frac{1}{2}v(r) W(r-\delta, \theta_{r-\delta}) \leq 0. $$ If we assume the total length of $\gamma$ is $2\pi \theta$, it is well known that $$ W(\rho, \varphi) \leq h(\rho) \doteq \frac{2}{\pi} \arctan \Big( \frac{r+\rho}{r-\rho} \tan (\frac{\theta}{2})\Big) $$ Then $$ v(\rho-\delta)-\frac{1}{2}v(r)-\frac{1}{2}v(r) h(r-\delta) \leq 0. $$ which means $$ v(r)-v(\rho-\delta) \geq \frac{1}{2}v(r)(h(r)-h(r-\delta)) $$ where we use $h(r)=1$. Therefore whenever $v(r)$ is differentiable, we have $$ v'(r) \geq \frac{1}{2}v(r)h'(r)=\frac{1}{2}v(r) \frac{\cot(\frac{\theta}{2})}{\pi r}. $$
Now we reparametrize and introduce $s=\ln r$ and $g(s)=\ln{\frac{v(r)}{2}}$, then the above differential inequality becomes $$ \frac{dg(s)}{ds} \geq \frac{1}{2\pi \tan(\frac{\theta}{2})}. $$ If we introduce $$ \lambda(s)=\text{Measure} \{ \varphi \in [0, 2\pi], \ |\ \omega(\varphi) \geq g(s)\}, $$ Obviously $\lambda(s) \geq Length(\gamma)=2\pi \theta$ as any point $re^{i\varphi} \in \gamma$ we have $$ \omega(\varphi) \geq \ln U(r, \varphi) \geq \ln (\frac{v(r)}{2})=g(s). $$
Claim 1: $$ \frac{dg(s)}{ds} \geq \frac{1}{2\pi \tan(\frac{\theta}{2})} \geq \frac{k}{\lambda(s)} $$ for some constant $k>0$.
Once Claim 1 holds, we have $\lambda dg \geq k ds$. After integration $$ \int_{g_0}^{\infty} \lambda dg \geq \lim_{s \rightarrow \infty}k(s-s_0)=\infty $$
On the other hand,
Claim 2: $$ \int_{g_0}^{\infty} \lambda dg =-\int_0^{\infty} g d\lambda \leq \int_0^{2\pi} \omega(\varphi)d\varphi<\infty $$ Here we note $\lambda(s)$ is non-increasing on $s$.
Once Claim 1 and 2 hold, Theorem 1 follows.
Question 1: How do we prove Claim 1? It seems we need to show $\theta(r)$ is uniformly bounded by some angle strictly smaller than $\pi$, so that $\cos{\frac{\theta}{2}}$ is bounded from below. This point is unclear to me.
Question 2: How do we prove Claim 2? Namely why does the integration by parts hold and how do we compare compare $d\lambda$ and $d\varphi$?
Another minor issue is that in the proof it is assumed $v(r)=\ln M(r)$ is differentiable with respect to $r$, and there is a continuous path in the set of maximum modulus from the inside.
Thanks for any answers or suggestions!
