Let $\{f_{n}\}$ be a sequence of holomorphic function on the open unit disk, $D$. Show that $f_{n}$ converge uniformly to $f$ in any compact subset of $D$ if and only if there exists a continuous function $f(z)$ defined on $D$ such that $$\int_{\mid z\mid= \rho}\mid f(z)-f_{n}(z)\mid\mid dz\mid$$ converges to zero as $n$ tend to infinity for all $0<\rho<1$.
I have proved this direction, $\Rightarrow$. Conversely, I am stuck at how to show that $f_{n}$ converge to $f$ on $D$.
Were it done, fixed any compact subset of $D$, I can show that $f(z)$ is holomorphic by using the Cauchy integral formula: $$f(z)=\lim_{n\rightarrow \infty}f_{n}(z)=\frac{1}{2\pi i}\lim_{n\rightarrow \infty}\int_{C}\frac{f_{n}(w)}{w-z}dw.$$ With the suitable choice of contour $C$, we can find a bound of the integral and pass the limit into the integral. Then, again using the Cauchy integral formula: $$f(z)-f_{n}(z)=\frac{1}{2\pi i}\int_{C}\frac{f(w)-f_{n}(w)}{w-z}dw.$$ With the suitable choice of $C$, conclude that it converge to $0$ uniformly. Then, $f_{n}$ converge uniformly to $f$ in any compact subset of $D$.
Thanks in advanced.
For $|z| < 1-2\epsilon$ $$\color{red}{|f(z)-f_k(z)|} =\lim_{n \to \infty} |f_n(z)-f_k(z)|=\lim_{n \to \infty} |\frac{1}{2i\pi}\int_{|s| = 1-\epsilon}\frac{f_{n}(s)-f_k(s)}{s-z} ds| \\ \le \frac{1}{2\pi \epsilon}\lim_{n \to \infty} \int_{|s| = 1-\epsilon}|f_{n}(s)-f_k(s)| |ds|=\color{red}{\frac{1}{2\pi \epsilon}\int_{|s| = 1-\epsilon}|f(s)-f_k(s)| |ds|}$$
Where at first $f$ and what is red is not defined, you only know $f_n$ is a sequence of holomorphic functions such that $\lim_{n> k \to \infty}\int_{|s| = 1-\epsilon}|f_{n}(s)-f_k(s)| |ds|= 0$, from which $f_n$ converges uniformly on $|z| < 1-2\epsilon$