Given f holomorphic on $B_1(0)$ and $max_{z \in C_r(0)} |f(z)| \to 0$ as $r \to 1$. Show f is identically 0.

Question

I have tried to use cauchy integral formula and the deformation theorem but that got nowhere: I got $f(z_0) = max_{z \in C_r'(z_0)}|f(z)|$ where $r' > 0$ such that $B_r'(z_0) \subset B_1(0)$. Note: $C_r(x)$ is notation for a circle of radius $r$ and centre $x$ in the complex plane

Let me outline my attempt further. I have attempted to show that $\forall \varepsilon > 0: \forall w \in B_1(0): |f(w)|<\varepsilon$. I let $\varepsilon$ and $w$ be free, I evaluate f(w) at some $C_r(0)$ such that $B_{|w|}(0) \subset \mathrm{Int}(C_r(0))$. From the Cauchy integral formula, I have $$ f(w) = \frac{1}{2i\pi}\int_{C_r(0)}\frac{f(z)}{z-w}dz $$ From which I tried to deform into something that would cancel out the denominator but that led me nowhere. I have also tried just straight up bashing it but that went nowhere too.

Answer 1

Let $\epsilon >0$ there exists $r_0\in (0,1)$ such that $|f(z)| <\epsilon$ whenever $|z|=r$ and $r >r_0$. By Maximum Modulus Principle, $|f(z)| \le \epsilon$ whenever $|z|\le r$. Since this holds for all $r >r_0$ it follows that $|f(z)| \le \epsilon$ for all $z$. Since $\epsilon >0$ is arbitrary it follows that $f(z)=0$ for all $z$.

Answer 2

For $0<r<1$ and $z_0\in B_r(0)$, Cauchy's integral formula yields $$f(z_0)=\frac{1}{2\pi i}\int\limits_{C_r(0)}\frac{f(z)}{z-z_0}\,dz$$ Taking absolute values and using the triangle inequality for integrals, we obtain \begin{align*} |f(z_0)|&\leq\frac{1}{2\pi }\int\limits_{C_r(0)}\frac{|f(z)|}{|z-z_0|}\,dz\\ &\leq \frac{1}{2\pi }\int\limits_{C_r(0)}\frac{\max\limits_{\zeta\in C_r(0)}|f(\zeta)|}{r-|z_0|}\,dz\\ &=\frac{r}{r-|z_0|}\max\limits_{\zeta\in C_r(0)}|f(\zeta)| \end{align*} Taking $r\to 1^-$ now proves the desired result.

Edit: Changed $r$ in denominator to $r-|z_0|$.