TITLE: $|f(z)|\le 1-|z|\forall z\in D$, we need to show $f\equiv 0$
QUESTION [5 upvotes]: $f$ is analytic function on open unit disk, and $|f(z)|\le 1-|z|\forall z\in D$, we need to show $f\equiv 0$, just a hint please.

REPLY [6 votes]: What value does $1-|z|$ take on $\partial D$? Then apply the maximum modulus theorem appropriately.
Answer:
Let $ r \in (0,1)$, and consider $f$ on $B(0,r)$. Note that $\partial B(0,r) = \{z \, | \, |z|=r\}$, and hence $|f(z)| \leq 1-|z| = 1-r$ for $z \in 
\partial B(0,r)$. By the maximum modulus theorem, this gives $|f(z)| \leq 1-r$ for $z \in B(0,r)$.
Now fix $z \in D$ (hence $|z|<1$). Note that $z \in B(0,r)$ for all $r \in (|z|,1)$. Hence it follows from above that $|f(z)| \leq 1-r$ for all $r \in (|z|,1)$. Hence $f(z) = 0$.