Analytic function is identically zero on open unit disk if $|f(z)|\leq 1-|z|$. 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.
 A: 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. 
A: 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 $f\equiv 0.$
A: 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 $g(x)=\sum_{n=0}^\infty|a_n|^2x^{2n},-1<x<1$,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.
