Let $f$ be a analytic in Domain $D$ which is an open disk of radius 1 centered at $z=0$ , it is given that in Domain $D$ , $|f(z)|\leq 1-|z|$ then prove that $f(z)=0$ in Domain $D$
It is asked that we have to prove $f(z) = 0$ using Cauchy's Integral Formula
I have attached my solution here My solution Please verify this!!
Your solution is correct. A few notes: the second page could be substituted with a reductio ad absurdum: if $f(0)\neq 0$ and $f$ is constant, there is a $z:1-|z|<|f(0)|=|f(z)\le 1-|z|$.
In general, there's a faster approach to this kind of problems, by means of the cauchy inequality (which is an easy corollary of Cauchy integral formula) : given a function $f$ which is holomorphic in $B(z_0,R)$, we have that, $\forall r<R$:
$$|f^{(n)}(z_0)|\le n! r^{-n}\underset{|z-z_0|=r}{\sup}|f|$$
By Cauchy inequality and the hypotesis,
$|f^{(n)}(0)|\le n!r^{-n}\underset{|z|=r}{\sup}|f|\le n!r^{-n}(1-r)$
Letting $r\to 1^-$, we obtain
$$f(0)=0\\ \forall_{n>0}f^{(n)}(0)=0$$
Thus we have proved that $f$ is zero on the open unit disc