Let $f:\mathbb{C}\to \mathbb{C}$ be an entire function such that $Re(f)\le |p(z)|$ for some polynomial, can we derive that $f(z)$ is a polynomial.
If $p(z)$ is constant, then this can be shown by considering $e^f$. If we instead consider $|u(z)|\le |p(z)|$, then it can also be shown. But if we do not establish the lowerbound, then I cannot figure out how to generlize the proof.
The condition is equivalent to $\operatorname{Re} f\le K|z|^m$ for some $m$ and $K$.
Under this condition we can conclude that $f(z)$ is a polynomial of order less than or equal to $m$.
Let $f(z)=u+iv=\sum_{k=0}^\infty a_kz^k$ and $A(r)=\max _{|z|=r} u(z)$.
It is well-known that for $k\ge 1$ $$ a_kr^k=\frac{1}{\pi}\int_0^{2\pi} u(re^{i\theta })e^{-ik\theta }d\theta . $$ This leads $$ |a_k|r^k+2u(0)\le \frac{1}{\pi}\int_0^{2\pi} \left(|u(re^{i\theta })|+u(re^{i\theta })\right)d\theta \tag{1} $$ since $u(0)=\frac{1}{2\pi}\int_0^{2\pi} u(re^{i\theta })d\theta .$
If $A(r)\le 0$, then $|u|+u=0$ and we have $|a_k|r^k+2u(0)\le 0$ from $(1)$. If $A(r)>0$, then we have $$ |a_k|r^k+2u(0)\le 4A(r),$$ since $|u|+u\le 2A(r)$. In both cases we have $|a_k|r^k\le \max\{4A(r),\, 0\}-2u(0).$
Now suppose that $\operatorname{Re} f \le K|z|^m$. Then we have $$ |a_n|\le 4Kr^{m-n}-\frac{2u(0)}{r^n}\to 0 \quad (r\to \infty) $$ for $n>m$.