First, $g$ can only have a finite number of zeros say $a_1,\ldots, a_n$. More importantly $g(z)=(z-a_1)\cdots(z-a_n)\cdot h$ where $h$ is some entire nonzero function. But then $|\frac{f(z)}{h(z)}|<|(z-a_1)\cdots(z-a_n)|$ for every $|z|>1$. That is $\frac{f}{h}$ is bounded by a polynomial. Thus, $\frac{f}{h}$ is a polynomial, which shows both that $h$ must be a constant and that $f$ is a polynomial. Thus both $f$ and $g$ are polynomials.
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Theorem. Let $f$ be an entire function and $n\in\mathbb{N}$ such that $|f|\le |z|^n$. Then $f$ must be a polynomial.
Proof. Let $f(z)=\sum_0^\infty a_k z^k $, we will show that the $a_k$'s must be zero eventually; they will in fact be zero when $k>n$. We perform the following calculation ($n$ is fixed, $m$ is arbitrary):
$\displaystyle |a_{n+m}|=\left|\frac{f^{(n+m)}(0)}{(n+m)!}\right|=\lim_{R\to\infty} \left|\frac{1}{2\pi i} \int_{B(0,R)} \frac{f(z)}{z^{n+m+1}} dz\right|\le\lim_{R\to\infty} \frac{1}{2\pi} \underset{z\in B(0,R)}{max}\ \left|\frac{f(z)}{z^{n+m+1}} \right|\cdot 2\pi R \le\lim_{R\to\infty} \underset{z\in B(0,R)}{max}\frac{|z|^n}{|z|^{n+m+1}} \cdot R \le\lim_{R\to\infty} \frac{R^n}{R^{n+m+1}} \cdot R = 0$
Thus, $a_k=0$ for $k>n$ and so $f$ must be a polynomial.
For the general result notice that any polynomial $|p|\le M\cdot |z|^n$ for some $M\in\mathbb{R}$ and $n\in\mathbb{N}$.
Interestingly, using the real or imaginary version of cauchy integral formula for taylor coefficients, we can extend this result to the real or imaginary part of a function. That is if $f=u+vi$ and $|u|\le |p|$ for some polynomial $p$ then $f$ must a polynomial. We can even (shockingly) drop the absolute values.
Theorem (Markushevich - Volume 2 - Page 265) Let $f=u+vi$. Suppose that $u(z)\le|z|^n$ for some $n\in\mathbb{N}$, then $f$ must be a polynomial.
I'd like to attempt an answer. I will have no problem with down-votes or with deleting it, since I am not 100% confident in it.
Let $A$ be the annulus $\frac{2}{3} \le |z| \le \frac{3}{2}$. For a sequence $a_n \in \mathbb{C}$, define $f_{a_n}(z) = f(a_n z)$ and $B_{a_n} = \{a_n z : z \in A \}$.
$B_{a_n}$ is an annulus with inner radius $|a_n|\frac{2}{3}$ and outer radius $|a_n|\frac{3}{2}$.
Let $a_n = n$, n positive integer. Let $f_{n_k}$ be the subsequence of $f_n$ guaranteed by the statement. Then either $f_{n_k} \to \infty$ uniformly on $A$ or $f_{n_k} \to g$ uniformly on $A$ for an analytic function $g$.
In the second case, $g$ is analytic so $g(A)$ is bounded. Using uniform convergence, you can see that There is an $M$ such that $f_{n_k}(A) < M$ for all $k$ sufficiently large. This means $f(z) < M$ for all $z \in B_{n_k}$, with $k$ sufficiently large. Mimicking the proof of Liouville's theorem that uses the Cauchy Integral Formula, this shows that $f(z)$ is constant.
In the first case, for any $M$ you can achieve $|f_{n_k}(z)| > M$ for $k$ sufficiently large. If in addition, for $k$ sufficiently large, $B_{n_k} \cap B_{n_{k+1}}$ is non-emtpy, then this shows $f(z) \to \infty$ as $z \to \infty$, and therefore $f$ must be a polynomial.
If on the other hand there are infinitely many $k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, then we need to look at what happens in these "gaps". Note that the outer radius of $B_{n_k}$ is in this case strictly less than the inner radius of $B_{n_{k+1}}$. This "gap" is an open annulus, but $f$ is analytic on the closure. Suppose such a gap does not contain a zero of $f$. Then the Minimum Modulus Theorem says that the modulus of $f$ in the gap is at least as great as the minimum modulus on the boundary, and so in the gap $f|(z)| > M$.
Therefore if only finitely many gaps contain zeros, we again see that $f(z) \to \infty$ as $z \to \infty$.
Finally, suppose there are infinitely $n_k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, and infinitely many of these "gaps" contain a zero of $f$. For the $m$-th gap pick a zero $b_m$, with $|b_m| < |b_{m+1}|$ and $|b_m| \to \infty$. These $b_m$ define a new sequence $f_{b_m}$ of functions on $A$, a subsequence of which converges uniformly. But in this case it cannot converge uniformly to $\infty$, because each function in the sequence has a zero at $z = 1$. Then as in the proof of the second case, $f$ must be constant.
Best Answer
Yes. Cauchy's formula states that $$f(z) = \frac{1}{2\pi} \int_{\theta} f(z + re^{i \theta}) d\theta$$ which gives an expression for $f(z)$ in terms of the average around a circle. By integration it follows that $$ f(z) = \frac{1}{r_0^2 \pi} \int_{r \leq r_0, \theta} f(z+re^{i \theta}) r dr d\theta$$ which implies that if the square integral of $f$ is bounded, then (by Cauchy-Schwarz) $f$ is locally bounded by a bound depending on the square integral of $f$ and the circle in question. From this your claim follows.
Basically, the point is that $f$ is the average of its values in a neighborhood. Note that this implies that the space of holomorphic functions in an open set $U$ such that $\int_{U} |f|^2$ is finite is actually a Hilbert space (i.e., complete) under the usual inner product.