An entire function is identically zero? I'm preparing for a PhD prelim in Complex Analysis, and I encountered this question from an old PhD prelim:  
Suppose $f(z)$ is an entire function such that $|f(z)| \leq \log(1+|z|) \forall z$. Show that $f \equiv 0$.  
Well, for $z=0$, $|f(0)| \leq 0$. On the other hand, for $z \neq 0$, $\log(1+|z|) > 0$, a positive constant. I'm guessing this would mean that $f$ turns out to be a bounded entire function, so then by Liouville's theorem, $f$ is constant, but this doesn't necessarily mean that $f \equiv 0$, does it? Am I wrong somewhere? Some guidance would be much appreciated!
 A: We will prove a slightly more genaral fact. The proof is based on this answer

Theorem.
  Let $f\in\mathcal{O}(\mathbb{C})$ and for all $z\in\mathbb{C}$ we have $|f(z)|\leq\varphi(|z|)$. Assume that 
  $$
\lim\limits_{R\to+\infty}\frac{\varphi(R)}{R^{p+1}}=0
$$
  then $f$ is a polynimial with $\deg (f)\leq p$.

Proof.
Since $f\in\mathcal{O}(\mathbb{C})$, then
$$
f(z)=\sum\limits_{n=0}^\infty c_n z^n
$$
for all $z\in \mathbb{C}$.
Moreover, for all $R>0$ we have integral representation for coefficients
$$
c_n=\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz
$$
Then, we get an estiamtion
$$
|c_n|\leq
\oint\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz|\leq
\oint\limits_{\partial B(0,R)}\frac{\varphi(|z|)}{|z|^{n+1}}|dz|=
\frac{2\pi R\varphi(R)}{R^{n+1}}=
\frac{2\pi \varphi(R)}{R^{n}}
$$
Hence for $n>p$ we obtain
$$
|c_n|\leq\lim\limits_{R\to+\infty}\frac{2\pi\varphi(R)}{R^n}=
2\pi\lim\limits_{R\to+\infty}\frac{1}{R^{n-p-1}}\lim\limits_{R\to+\infty}\frac{ R\varphi(R)}{R^{p+1}}=0
$$
which implies $c_n=0$ for $n>p$. Finally we get
$$
f(z)=\sum\limits_{n=0}^p c_n z^n+\sum\limits_{n=p+1}^\infty c_n z^n=\sum\limits_{n=0}^p c_n z^n
$$
This means that $f$ is a polynimial with $\deg(f)\leq p$. 
For your particular problem $\varphi(R)=\log(1+R)$, and it is easy to check that
$$
\lim\limits_{R\to+\infty}\frac{\varphi(R)}{R}=0
$$
Hence, $f(z)=c_0$ is a constant function. Moreover,
$$
|c_0|=|f(0)|\leq\log(1+|0|)=0
$$
so $c_0=0$ and $f(z)=0$ for all $z\in\mathbb{C}$.
A: We already know that $f(0) = 0$. Now from $|f(z)| \leq \log(1 + |z|)$ and we get that on every circle of radius $R$ about the origin,
$$|f^{(n)}(0)| \leq  \frac{\log(1  + |R|) \times n!}{R^n} \hspace{5mm} \forall n \geq 1$$
by the Cauchy estimate. Since $R^n$ grows faster than $\log (1+ |R|)$ when $n \geq 1$, we see that $f^{(n)}(0) = 0$ for all $n \geq 1$. But we also know $f(0) = 0$ so that $f^{(n)}(0) = 0$ for all $n \geq 0$. Since $f$ is entire the identity principle implies $f \equiv 0$.
A: Since $f$ is entire it can be expressed as a power series that converges everywhere: $f(z) = \sum_{n=0}^\infty a_n z^n$. From $|f(0)| \leq 0$ we know that $f(0)=0$, hence $a_0 = 0$.
So $g(z) := f(z)/z$ can be continued to an entire function that satisfies
$$|g(z)| \leq \frac{\log(1+|z|)}{|z|} \quad \text{for all } z\neq 0.$$
The right side converges to zero for $|z| \to \infty$, in particular it is bounded. By Liouville's theorem, $g$ is constantly zero and so is $f$.
