# Analytic function is identically zero on open unit disk if $|f(z)|\leq 1-|z|$. [duplicate]

Let $$f$$ be an analytic function defined on open unit disk with $$|f(z)|\leq 1-|z|$$. I need to establish $$f$$ is zero on disk .

If I'm able to prove that $$f(0)=0$$ and $$f(0)\leq f(z)$$ then $$f$$ is identically zero, I'm not able to achieve either of them. Any hint would be sufficient.

## marked as duplicate by Martin R, rtybase, Delta-u, Xander Henderson, BrahadeeshOct 12 '18 at 12:41

• – Martin R Oct 11 '18 at 18:15

We have a Taylor expansion $$f(z)=\sum_{n=0}^\infty a_nz^n$$ , valid on open unit disc. Now if $$0≤r<1$$ we have parseval's identity: $$\frac{1}{2\pi} \int_0^{2\pi}|f(re^{\iota\theta})|\ d\theta=\sum_{n=0}^\infty|a_n|^2r^{2n}$$

But for $$0≤r<1$$ we have $$\frac{1}{2\pi} \int_0^{2\pi}|f(re^{\iota\theta})|\ d\theta≤\frac{1}{2\pi} \int_0^{2\pi}(1-|re^{\iota \theta}|)\ d\theta=1-r^2$$

Letting $$r\rightarrow 1$$ we can say for the function $$g:(-1,1)\rightarrow \Bbb R$$ defined by $$g(x)=\sum_{n=0}^\infty|a_n|^2x^{2n},-1,the limit $$\lim_{x\rightarrow 1-}g(x)$$ exists and equals to $$0$$. Hence by Able's limit theorem we have $$0=\sum_{n=0}^\infty|a_n|^2$$ i.e. $$a_n=0$$ for each $$n≥0$$. Therefore $$f$$ is identically zero in open unit disc.

The unit circle satisfies $$|z|=1$$, therefore you have $$|f(z)|\le 1-1=0$$ on the boundary of the unit disk.

Since for the analytic function $$f(z)$$, the maximum of |f(z)| on the unit disk is attained on the boundary and this maximum is $$0$$, we have $$f(z)=0$$ on the entire unit disk. The above proof works only if the function is analytic on the closed unit disk. For the case where $$f$$ is defined on the open disk see the proof given by UserS.

• $f$ is only defined in the unit disk, not on the unit circle. – Martin R Oct 11 '18 at 18:14
• Thanks Martin R, I have edited my answer to address the problem with my proof. – Mohammad Riazi-Kermani Oct 11 '18 at 18:39

Fix $$z\in \mathbb D.$$ Then for $$|z|\le r <1,$$ the maximum modulus theorem and the given condition show

$$|f(z)|\le |f(re^{it})| \le 1-r.$$

As $$r\to 1^-$$ the right side $$\to 0.$$ Therefore $$|f(z)|\le 0.$$ This implies $$f(z)=0.$$ Since $$z$$ was an arbitrary point in $$\mathbb D,$$ we have $$f\equiv 0.$$