Suppose $f$ is an entire function such that $|f(z)| \le 2 \ln (|z|+1)$ for all $z \in \Bbb{C}$. Prove that $f(z) = 0$ for all $z \in \Bbb{C}$.
Here is my solution which I am hoping someone could check:
First, it easily follows from the hypothesis that $f(0)= 0$, so if we could prove $f$ is constant, then it would automatically follow that $f=0$ on $\Bbb{C}$. Indeed,
$$|f(0)| \le 2 \ln (|0|+1) = 2 \ln (1) = 0.$$ Now, let us prove that $f$ is constant. Let $z_0 \in \Bbb{C}$ and $r > 0$ arbitrary. Then, because $f$ is entire, we know it is holomorphic on a neighborhood of $\overline{B(z_0,r)}$, so Cauchy estimates tell us that
$$|f'(z_0)| \le \frac{\sup_{|z-z_0|=r} |f(z)|}{r} \le 2 \frac{\sup_{|z-z_0|=r} \ln (|z|+1)}{r}.$$ Let $z$ be any complex number satisfying $|z-z_0| = r$. Then there exists a unit complex number $w_z$ such that $z = rw_z + z_0$. Since $x \mapsto \ln x$ is an increasing function on $(0,\infty)$, we have
$$\ln(|z| + 1) = \ln(|rw_z + z_0| + 1) \le \ln (|rw_{z}| + |z_0| + 1) = \ln (r + z_0 + 1)$$ and hence,
$$|f'(z_0)| \le 2 \frac{\ln (r + z_0 + 1)}{r}.$$ As $r \to \infty$, the numerator and denominator of the fraction on the RHS go to $\infty$. But L'Hôpital's rule says
$$|f'(z_0)| \le 2 \lim_{r \to \infty} \frac{\ln (r + z_0 + 1)}{r} = 2 \lim_{r \to \infty} \frac{1}{r + |z_0| + 1} = 0,$$ so $f'(z_0)=0$. Since $z_0$ was arbitrary, it follows that $f'=0$ and therefore $f$ is constant. But, by our initial remark, we know this means $f=0$ on $\Bbb{C}$.
Yes, your solution looks nice. I just want to add a similar solution using a Taylor series.
Since $f$ is entire, $$f(z) = \sum_{n=0}^\infty a_n z^n$$ on $\mathbb{C}$ and $$a_n = \frac{f^{n}(0)}{n!} = \frac{1}{2\pi i}\int_{C_R}\frac{f(z)}{z^{n+1}}dz,$$ where $C_R$ is circle of radius $R>0$ centered at $0$.
Note that on $C_R$, $$\left|\frac{f(z)}{z^{n+1}}\right| \leq \frac{2 \ln (R+1)}{R^{n+1}}.$$ Thus, $$|a_n|= \left| \frac{1}{2\pi i} \int_{C_R} \frac{f(z)}{z^{n+1}}dz\right| \leq \frac{1}{2\pi} \frac{2 \ln (R+1)}{R^{n+1}} \cdot 2 \pi R = \frac{2 \ln (R+1)}{R^n} \to 0$$ as $R \to \infty$ for all $n \geq 1$. This implies that $a_n=0$ for all $n\geq 1$. Thus, $f(z) \equiv a_0$ for all $z \in \mathbb{C}$.
As you said, $f(0) = 0$, so $f(z) \equiv 0$ for all $z \in \mathbb{C}$.