Suppose that $f$ is analytic and bounded in the unit disk $D$ and $f(\zeta)\to 0$ as $\zeta \to 1^{-}$ along the upper half of the circle $C:|\zeta - 1/2|=1/2$. Then, $f(x)\to 0$ as $x\to 1^{-1}$ along $[0,1)$.
The hint to this exercise says : For each $n \in \mathbb{N}$, let $f_n(z)=z^nf(z)$, and $F_n(z)=f_n(z)\overline{f_n(\bar{z})}$. Then, $F_n$ is analytic and for $\epsilon>0$ there exists an $n$ such that $|F_n(\zeta)|<\epsilon$ for all $\zeta \in C$. By the Maximum modulus principle, this inequality yields that $|x^{2n}f(x)|\le \epsilon$ for $x\in (0,1)$.
I don't follow the hint given. How do we know that for $\epsilon>0$ there exists an $n$ such that $|F_n(\zeta)|<\epsilon$ for all $\zeta \in C$? I would greatly appreciate any help.
Let $C_\delta : |\zeta -1/2|=1/2,\, |\zeta -1|<\delta.$ $C_\delta $ is a part of $C$ near $1$.
By the condition $f(\zeta )\to 0$ as $\zeta \to 1^{-}$ along the upper half of $C$, $$ |F_n(\zeta )|= |\zeta ^{2n}f(\zeta )\overline{f(\bar{\zeta })}|\le |f(\zeta )\overline{f(\bar{\zeta })}| \le M|f(\zeta )|\,\, (\text{or } M|f(\bar{\zeta })|)<\epsilon$$ for $\zeta \in C_\delta .$
Let $\zeta _\delta $ be a point such that $|\zeta _\delta -1/2|=1/2,$ $|\zeta _\delta -1|=\delta .$
Note that $|\zeta ^{2n}|\le |\zeta _\delta |^{2n}$ for $\zeta \in C,$ $|\zeta -1|\ge \delta $. Since we can take $n$ so that $|\zeta _\delta |^{2n}<\epsilon/M^2 $, we have $$ |F_n(\zeta )|= |\zeta ^{2n}f(\zeta )\overline{f(\bar{\zeta })}|\le M^2|\zeta _\delta |^{2n}<\epsilon. $$ Thus we know that there exists an $n$ such that $|F_n(\zeta)|<\epsilon$ for all $\zeta \in C.$