If $f$ is an entire function such that for each $\theta$, $|f(re^{i\theta})|\rightarrow \infty$ as $r\rightarrow \infty$

Question

Let $f: \mathbb{C}\mapsto \mathbb{C}$ be an entire function such that for each $\theta$, $|f(re^{i\theta})|\rightarrow \infty$ as $r\rightarrow \infty$.
a) Does this imply that $|f(z)|\rightarrow \infty$ as $|z|\rightarrow\infty$?
b) Does this imply that $f(z)$ is a polynomial?

I know that if a) is true, then $b)$ is true since a) implies that $\infty$ is a pole for $f(z)$ which is equivalent to $f(z)$ is a polynomial.
However, I don't know how to show that a) is true or construct a counter example.

Edited 5/11:
I am also considering a closely related question: Let $f: \mathbb{R}^2\mapsto \mathbb{R}$ be a continuous function. Suppose for all $k\in \mathbb{R}$, we have $|f(x,kx)|\rightarrow \infty$ as $x\rightarrow \infty$.
a) Does this imply that $|f(x,y)|\rightarrow \infty$ as $|(x,y)|\rightarrow\infty$
b) If we require $f$ to be differentiable, does this imply that $|f(x,y)|\rightarrow \infty$ as $|(x,y)|\rightarrow\infty$

Answer 1

Here is a simpler example of a transcendental entire function $f : \mathbb{C} \rightarrow \mathbb{C}$ such that $f(r e^{i \theta}) \underset{r \rightarrow +\infty}{\longrightarrow} \infty$ for every $\theta \in \mathbb{R}$.

Consider the entire function $f : z \mapsto e^{z} +e^{-z} +z^{2}$.

It is obvious that $f$ is not polynomial - one can easily write its power series expansion or check that none of its derivatives is identically zero.

Furthermore, one can check that for all $\theta \in \mathbb{R}$, $f(r e^{i \theta}) \underset{r \rightarrow +\infty}{\longrightarrow} \infty$.
Indeed, let $\theta \in \mathbb{R}$.
Notice that since for all $z \in \mathbb{C}$, $f(z) = f(-z)$, we can assume that $\theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

• If $\theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, then $$\left|f(r e^{i \theta})\right| \geq e^{r \cos(\theta)} - e^{-r \cos(\theta)} -r^{2} \underset{r \rightarrow +\infty}{\longrightarrow} +\infty$$

• If $\theta = \frac{\pi}{2}$, then $$\left|f(r e^{i \theta})\right| = |f(i r)| = \left|2 \cos(r) -r^{2}\right| \geq r^{2} -2 \underset{r \rightarrow +\infty}{\longrightarrow} +\infty$$

Answer 2

You are right about (a) implies (b). If we show that (b) is false, then we can show that (a) is also false.

Indeed, (b) is false.

Consider the same function $H(z)$ in the link: Radial Limits for Holomorphic Functions

$$ H(z)=\int_0^{\infty} t^{-t} e^{tz} dt, $$ $$F(z)=H(z+i\pi)H(iz+i\pi).$$ Then $F(z)\rightarrow 0$ as $z\rightarrow \infty$ along every line through $0$.

Now, let $f(z)=F(z)+1$, then $f(z)\rightarrow 1$ as $z\rightarrow \infty$ along every line through $0$. We consider $$ G(z)=\int_0^z f(w)dw. $$ The function $G$ is well-defined since $f$ is entire and integral is path independent, and then $G$ is also entire with $G'=f$.

Let $z\rightarrow\infty$ along a line through $0$. We take the path of the integral to be also a line through $0$, from $0$ to $z$. Then, we can see that $G(z)\rightarrow\infty$ as $z\rightarrow\infty$ along every line through $0$.

To clarify this, let $L_{\theta}(a,b)$ be the line segment from $a$ to $b$ on the line $L_{\theta}=\{re^{i\theta}|r\geq 0\}$. Then we have $$ f(w)\rightarrow 1 \ \textrm{as} \ \ w\rightarrow\infty \ \textrm{along} \ \ L_{\theta}. $$ $$ \int_{L_{\theta}(0,z)} f(w)dw = z +o(|z|) \ \textrm{as} \ \ z\rightarrow\infty \ \textrm{along} \ \ L_{\theta}. $$ However, $G$ is not a polynomial since $f$ is not a polynomial.

By this argument with repeated constant multiplication and integration, we have further:

Let $P(z)$ be a nonconstant polynomial. Then there is an entire function $G_P(z)$ which is not a polynomial such that $$ \frac{G_P(z)}{P(z)} \rightarrow 1 \ \textrm{as} \ \ z\rightarrow\infty \ \textrm{along each} \ \ L_{\theta}. $$