Let $a\in\mathbb{R}$, $f$ be an entire function on $\mathbb{C}$ and the inequality $$\int\limits_0^{2\pi} |f(re^{i\theta})|\text{d}\theta\le r^a$$holds for all $r>0$.
Can we prove that $f\equiv0$?
By Cauchy integral formula, $f(0)=\frac{1}{2\pi}\int\limits_0^{2\pi} f(re^{i\theta})\text{d}\theta,\forall r>0$. So $|f(0)|\le \frac{1}{2\pi}\int\limits_0^{2\pi} |f(re^{i\theta})|\text{d}\theta\le\frac{r^a}{2\pi},\forall r>0$.
Theorem. Let $G$ be a connected open set and let $f:G\rightarrow \mathbb{C}$ be an analytic function. Then the following are equivalent statements:
$(1)f\equiv0$
$(2)$there is a point $s$ in $G$ such that $f^{(n)}(s)=0,\forall n\ge 0$
$(3) \{z\in G:f(z)=0\}$ has a limit point in $G$
To prove that $f\equiv0$, it suffices to show that $f^{(n)}(0)=0,\forall n\ge 0$, but I don't know how to prove it. If we can find a limit point in $\{z\in G:f(z)=0\}$, we can also prove the statement, but I can't find it.
By Liouville theorem, if we can prove that $f$ is bounded, then it's easy to prove the statement. Maybe Cauchy integral formula is helpful to solve this problem.
I'm stuck on this for a while and I'll very much appreciate the help!
Pick $|z|=r$ and let $R=2r$; by Cauchy one has $2\pi if(z)=\oint_{|w|=R}\frac{f(w)dw}{w-z}$, so using that $|dw|=Rd\theta, |w-z| \ge R/2$, one gets that $2\pi |f(z)| \le 2\int\limits_0^{2\pi} |f(Re^{i\theta})|\text{d}\theta\le 2R^a \le 2^{a+1}r^a$ since $R=2r$
This means $|f(z)| \le C|z|^a$ which implies by the usual manipulations that $f$ is zero if $a<0$ and $f$ is a polynomial of degree at most $m=[a]$ if $a \ge 0$, with a zero of order at least $m$ at zero; so $f(z)=Az^m$, but then $\int\limits_0^{2\pi} |f(Re^{i\theta})|\text{d}\theta=2\pi |A|R^m$ and since $a=m+c, 0 \le c<1$, one gets $2\pi A \le R^c, R >0$ which is obviously not true for $0<c<1, A \ne 0$ by letting $R \to 0$. However one clearly can have $f(z)=Az^m, 2\pi |A| \le 1$ for $a=m$ integral.
So the complete answer is that $f=0$ if $a$ is not (nonnegative) integral, but $f$ can be non zero $Az^a, 2\pi |A| \le 1$ if $a$ non negative integral