Uniform convergence of complex analytic function on unit circle

Question

I'm trying to solve the following problem. Some guidance would be greatly appreciated!

Let $F(z)$ be continuous on the closed disc $\overline{\mathbb{D}} = \{z: |z| \leq 1\}$ and analytic on the open disc $\mathbb{D} = \{z: |z|< 1\}$. Prove $$\lim_{\lambda \uparrow 1}F(\lambda z) = F(z)$$ uniformly on $\overline{\mathbb{D}}$

I want to show that $\forall \epsilon> 0, \exists \delta > 0$ such that $|\lambda - 1| < \delta \implies \forall z: |F(\lambda z) - F(z)| < \epsilon$. Since $F$ is analytic in the open disc, I thought to write $F(z) = \sum_{k=0}^\infty a_k z^k$ where I didn't include negative $k$ since otherwise the serious would have a pole at $z = 0$. It follows that $$|F(\lambda z) - F(z)| = \left\vert \sum_{k=1}^\infty a_k z^k(\lambda^k - 1)\right\vert \leq |a_1||z||1 - \lambda| + \left\vert \sum_{k=2}^\infty a_k z^k(1-\lambda^k)\right\vert$$

where it's not clear how to deal with the summation term. I don't think this is the right way to appraoch it anyways since this summation is only guaranteed to converge on the open disk not the closed one.

Answer 1

A continuous function on a compact set is uniformly continuous. So $F$ is uniformly continuous on $\overline{\Bbb D}$, and for all $\epsilon > 0$ exists a $\delta > 0$ such that $$ |z-w| < \delta \implies |F(z) - F(w)| < \epsilon \, . $$ It follows that in particular for $1 - \delta < \lambda < 1$ and all $z \in \overline{\Bbb D}$ $$ |\lambda z - z| = (1 - \lambda)|z| \le 1 - \lambda < \delta \implies |F(\lambda z) - F(z)| < \epsilon \, , $$ which proves that $\lim_{\lambda \to 1} F(\lambda z) = F(z)$, uniformly for $z \in \overline{\Bbb D}$.

The fact that $F$ is analytic is not needed here. Also (as pointed out in the comments) the Taylor series expansion may not be valid on the boundary of the unit disk, so that approach does not work, unfortunately.