Show that $f$ is identically zero in $\mathbb{C}$

Question

Let $f$ be holomorphic in $\mathbb{C}$. Prove that if $|f(z)| \leq M|z|^{\alpha}$ with $0 <\alpha <1$, then $f$ is identically zero in$\mathbb{C}$

I know that $f$ has a Taylor expansion $f(z)=\sum_{n=0}^{\infty}c_nz^n$. And that if there exist positive constants $M$ and $K$ and a positive integer $k$ such that for $|z|\geq K$

$$|f(z)|\leq M|z|^k$$ then $f$ is a polynomial of degre at most $k$. So if I could modify somehow this result and conclude that $f$ is a polynomial of degree at most $\alpha$, then $f$ would be identically zero in $\mathbb{C}$, but I'm not able to do that (it might even be false), so I'd like some hint that points me towards a good path to solve it. Thanks in advance!

EDIT: As indicated in the comments, if $|z|\geq 1$, then $|z^{\alpha}| < |z|$.

So let's consider $R>0$, $R>r\geq1$ and the circle $\gamma(0,r)$. Then the constants $c_n$ of the Taylor expansion are given by

$$c_n=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z^{n+1}} dz$$ and applying the Estimation Theorem gives

$$|c_n|\leq \frac{1}{r^n}M(r)$$ where $M(r)= \sup \{|f(z)|:|z|=r \}$. Then, since $|z|^{\alpha}<|z|=r$

$$M(r) < Mr$$ we have

$$|c_n|\leq \frac{M}{r^{n-1}}$$ And since $r>1$ can be made arbitrarly large, we must have $c_n=0$ for all $n>1$,so $f$ is constant. Since $f(0)=0$ from the hypothesis, $f$ is identically zero in each disc $D(0;R)$, so $f$ is identically zero in $\mathbb{C}$.

Is this correct?

Answer 1

The Cauchy formula for the derivative tells us that $$f'(w)=\oint_{C_R} \frac{f(z)}{(z-w)^2} \frac{dz}{2\pi i}$$ where $C_R$ is a circle centered at zero, of radius $R$ and enclosing $w$. But then: $$ |f'(w)| \leq 2 \pi R \frac{M R^\alpha} {2 \pi(R-|w|)^2} \rightarrow 0$$ as $R\rightarrow +\infty$. So $f$ is a constant. From your hypotheses, $f(0)=0$, so the constant must be zero.

Answer 2

Remark that $f(0)=0$, then $f(z)/z$ is holomorphic and $|f(z)/z|<M$ by Liouville, $f$ is constant, deduce that the constant is zero.

Answer 3

An approach: let $\gamma_R$ be the boundary of the disk of radius $R$ ($R > |z|$, we may always find such an $R$). Use the Cauchy integral formula: $$f(z) = \frac{1}{2\pi i} \int_{\gamma_R} \frac{f(\xi)}{z - \xi} d\xi$$ to obtain a bound for $|f(z)|$ depending on $R$ and $f(\xi)$ where $\xi \in \gamma_R$. Then use your hypothesis, take $R \to \infty$, and $|f(z)|$ should go to $0$.