Prove that $\{f_n\}$ is uniformly continuous.

Question

Let $\{f_n\}$ a sequence of functions continuous on $\overline{D}$ and analytic on $D$, where $D=\{z\in \Bbb C: |z|<1\}$.

1. If $f_n\to f$ pointwise on $\partial D$ and $\int_{|z|=1}|f_n-f||dz|\to 0$, then $\{f_n\}$ converges uniformly on $D$.
2. If $\{f_n\}$ converges uniformly on $\partial D$, then $\{f_n\}$ converges on $D$.

Try: For second part there is a theorem that says that this function with these hypotheses reaches the maximum at the boundary, then it can be easily shown with this theorem that it is cauchy uniform and implies that it is uniformly continuous in $D$.

For part 1, I don't really know how to do it. I am trying to prove with these hypotheses precisely that the hypothesis of part 2 is fulfilled and use it but I don't know how. Some help.

Answer 1

This is a partial answer showing uniform convergence on concentric discs rather than the whole of $D$, I might come back to it if I figure out how to get uniform convergence on $D$.

You should not need to use part $2$ to prove part $1$, as I believe in whoever wrote the question to you that they ordered the questions so that either they are independent or so that you can use them but going in the chronological order. Though still, part $1$ might be provable using part $2$, I haven't looked into it.

As said by León in the comments above, you have the inequality $$\int_{|z|=1}|f_n-f_m||dz|\le \int_{|z|=1}|f_n-f||dz|+\int_{|z|=1}|f-f_m||dz|$$ You can obtain this by integrating the inequality $$|f_n(e^{it}) -f_m(e^{it})||ie^{it}| \le |f_n(e^{it})-f(e^{it})||ie^{it}| + |f(e^{it})-f_m(e^{it})||ie^{it}|$$ w.r.t. $t \in [0, 2\pi]$.
Now, let $z \in D$. By Cauchy's integral formula, we have: $$f_n(z) = \int_{|\xi| = 1} \frac{f_n(\xi)}{\xi - z} \mathrm{d}\xi$$ Hence, for all $n$ and $m$: $$|f_n(z) - f_m(z)| = \left|\int_{|\xi| = 1} \frac{f_n(\xi) - f_m(\xi)}{\xi - z} \mathrm{d}\xi\right| \leq \int_{|\xi| = 1} \left|\frac{f_n(\xi) - f_m(\xi)}{\xi - z}\right| |\mathrm{d}\xi|$$ But, knowing that $\min_{|\xi| = 1} |\xi - z| = 1 - |z|$, we get: $$\begin{split}\int_{|\xi| = 1} \left|\frac{f_n(\xi) - f_m(\xi)}{\xi - z}\right| |\mathrm{d}\xi| &= \int_{0}^{2\pi} \left|\frac{f_n\left(e^{it}\right) - f_m\left(e^{it}\right)}{e^{it} - z}\right| |ie^{it}|\mathrm{d}t\\ &\leq \int_{0}^{2\pi} \frac{\left|f_n\left(e^{it}\right) - f_m\left(e^{it}\right)\right|}{1 - |z|} |ie^{it}|\mathrm{d}t\\ &\leq \frac{1}{1 - |z|}\int_{|\xi| = 1} \left|f_n(\xi) - f_m(\xi)\right| |\mathrm{d}\xi\end{split}$$

Since the RHS only depends on $|z|$, and using León's inequality, $(f_n)_n$ is Cauchy for the uniform convergence on $D(0,r)$ for all $0< r < 1$, and hence uniformly converges to a (holomorphic) function on all the $D(0,r)$s.

The issue with my reasoning is that $\frac{1}{1 - |z|}$ tends to infinity as $|z|$ tends to $1$, which is why I cannot get uniform convergence on all of $D$ at once for now.