Example of holomorphic function such that $|F(x)|\leq 1 + |x|^r$ with $0I have just started my course of introduction to complex analysis. During the last week we have been working the concept of holomorphic function. I am now making a list of different kinds of holomorphic functions to better understand this concept: I have found holomorphic functions in only one point, real differentiable 2d fields which are not holomorphic (being holomorphic is stronger than being real differentiable)...
I know that the only examples of bounded holomorphic functions in the entire plane are constant functions (Liouville theorem), but what happens if we modify "a bit" this hypothesis? For example, does anyone know a holomorphic function $F:\mathbb{C}\to \mathbb{C}$ such that $|F(x)|\leq 1+|x|^r$, with $0<r<1$?
 A: $\newcommand{\C}{\mathbb{C}}$I'll give a solution using Cauchy's estimates. For clarity, I first state the theorem.
Notation. For $z \in \C$ and $R > 0$, $B(z, R)$ denotes the closed disc of radius $R$ centred at $z$. That is, $$B(z, R) := \{w \in \C \mid |z - w| \leqslant R\}.$$

(Cauchy's estimate) Let $f : \C \to \C$ be an entire (holomorphic) function. Let $z_{0} \in \C$ and $R > 0$ be arbitrary. Let $M = M(z_{0}, R)$ be the maximum of $|f|$ on $B(z_{0}, R)$. Then,
$$|f'(z_{0})| \leqslant \frac{M}{R}.$$

(This is a very special case of Cauchy's estimate. You don't need the domain to be all of $\C$ but this works for our case.)
Note interestingly that the left-hand side depends only on $z_{0}$ and not on $R$, whereas the right side depends on both. This is what we will exploit.

Now, let $f$ satisfy the hypothesis of your question and $R > 0$, $z_{0} \in \C$ be arbitrary.
How big can the modulus of an element $z \in B(z_{0}, R)$ be? Well, the triangle inequality tells us that
$$|z| \leqslant |z - z_{0}| + |z_{0}| \leqslant R + |z_{0}|.$$
Thus, we see that if $z \in B(z_{0}, R)$, then
$$|f(z)| \leqslant 1 + |z|^{r} \leqslant 1 + (R + |z_{0}|)^{r}.$$
Applying Cauchy's estimate, we get that
$$|f'(z_{0})| \leqslant \frac{1 + (R + |z_{0}|)^{r}}{R}. \tag{$\ast$}$$
Here's something to be careful about if you're seeing this for the first time:
The above inequality holds for all $z_{0} \in \C$ and all $R > 0$. Thus, keeping $z_{0}$ fixed and letting $R \to \infty$, we conclude that $$|f'(z_{0})| = 0.$$
(Here is where we use that $r < 1$ to conclude that the right side of $(\ast)$ goes to $0$ as $R \to \infty$.)
But the above is true for all $z_{0} \in \C$. Thus, we conclude that $f' \equiv 0$ identically on $\C$. This means that $f$ is constant.
I leave it to you to check that the constant can be anything in $B(0, 1)$.
