Prove that $\,f(z) \equiv 0,\,$ if $\,f(z)\,$ is entire

Question

Let $f(z)$ be an entire analytic function, such that $$ \int_{0}^{2\pi}\lvert\, f(re^{i\theta}) \rvert\,d\theta\le r^{16/5}, \quad \text{for all}\,\,\, r>0 $$ Show that

$$f(z)\equiv 0.$$

Thank you for you help.

Maybe we need to use Cauchy Theorem.

Answer 1

Using Cauchy Integral Formula, for $r\to 0$, we obtain that $f(0)=f'(0)=f''(0)=0$. Thus $f(z)=z^3g(z)$, where $g$ is also entire, and we get for $g$ that $$ \int_0^{2\pi} \lvert g(r\mathrm{e}^{i\vartheta})\rvert\,d\vartheta\le r^{1/5}, \quad r>0. $$ Now, for every $z\in\mathbb C$, let $r>2\lvert z\rvert$, and according to Cauchy Integral Formula $$ g(z)=\frac{1}{2\pi i}\int_{|w|=r}\frac{g(w)\,dw}{w-z}, $$ and hence $$ \lvert g(z)\rvert \le \frac{1}{2\pi}\max_{|w|=r}\lvert w-z\rvert^{-1}\int_0^{2\pi}\lvert g(r\mathrm{e}^{i\vartheta})\rvert \,d\vartheta\le \frac{1}{2\pi}\cdot \frac{2}{r}\cdot r^{1/5} =\frac{1}{\pi r^{4/5}}\to 0, $$ as $r\to \infty$.

Thus $g\equiv 0$, and so is $f$.