Given an entire function $f(z)$, define $\displaystyle m(r):=\inf_{\vert z\vert=r}\vert f(z)\vert$. If $f$ satisfies $$\limsup_{r\to+\infty} m(r)=+\infty,$$ can we assert that $f$ is a polynomial?
This question comes to my mind after going through the discussion about the entire functions on Ahlfors.
I tried to prove this by showing $\infty$ is a pole of $f$ but got stuck since $f$ may have infinitely many zeros. Now I tend to believe that the statement is not true, but neither can I find a counterexample.
Let me put my comment as an answer. Any nonconstant entire function of order less than $1/2$ (more precisely satisfying $f(z)=O_{\epsilon}(e^{\epsilon |z|^{1/2}})$ for every $\epsilon >$ so order $1/2$ and type $0$ are allowed too) satisfies $$\limsup_{r\to+\infty} m(r)=+\infty$$
If $f$ has order at most $1/2$ and type $0$ (or order strictly less than $1/2$) by Hadamard factorization theorem we have:
$$f(z)=cz^m\Pi_{k \ge 1}(1-z/z_k)$$ where the $z_k \ne 0$ are the nonzero zeroes taken with appropriate multiplicity of $f$ arranged say in increasing modulus order - the product converges absolutely as the zeroes are sparse as their convergent exponent is $1/2$ and their number up to modulus $R$ is $o(\sqrt R)$ for example
Then considering $$g(z)=cz^m\Pi_{k \ge 1}(1-z/|z_k|)$$ one clearly has $m_f(r) \ge m_g(r)=g(-r)$ since if $m_f(r)=|f(w)|, |w|=r$ then $|1-w/z_n| \ge |1-r/|z_n||$ by the triangle inequality, so $|f(w)| \ge |g(-r)|$
In particular if $\limsup m_f(r) < \infty$ then $g$ is bounded on the negative axis, so $h(z)=g(z^2)$ is bounded on the imaginary axis since $|h(ix)|=|g(-x^2)| \le M$.
But $f,g$ have same order and type by the Hadamard factorization theorem again and related results about the order and type vs the zeroes (for example the order is the convergence exponent of the roots so depends only $|z_k|$ when nonintegral and zero type is characterized by the number of roots of modulus at most $r$ being $o(r^{\rho})$ again for nonintegral order $\rho$) so $g(z)=O_{\epsilon}(e^{\epsilon |z|^{1/2}})$ for all $\epsilon >0$ hence $h(z)= O_{\epsilon}e^{\epsilon |z|}$ and $|h(ix)| \le M, x \in \mathbb R$.
But Phragmen Lindelof theorem implies $h$ constant (for arbitrary $a>0$ take $e^{-az}h(z), \Re z>0$ and $e^{az}h(z), \Re z <0$ and deduce that $|h(z)| \le Me^{a|z|}$ and let $ a \to 0$), so $g,f$ must be constant which is a contradiction.
For example $f(z)=\sum_{n \ge 0}\frac{z^n}{(3n)!}$ has order $1/3$ by Stirling hence $\limsup m_f(r)=\infty$ but is not a polynomial