Suppose $u:\overline{\mathbb{D}}\to \mathbb{C}$ is continuous in $\mathbb{D}$ and that $u(re^{it})\to u(e^{it})$ uniformly as $r\to 1^{-}$. Show that $u$ is continuous in $\overline{\mathbb{D}}$.
I need help for this question.
I think I only need to show $u $ is continuous on $\partial(\mathbb{D})$.
I may need to define a harmonic function $f$ ( using this function $u$) and then using the mean value property for this harmonic function might be useful, but I don't have any vigorous thought and I also don't know how to define this harmonic function. I appreciate if you help me.
Harmonic functions are not needed.
Step 1: the restriction of $u$ to $\partial\mathbb{D}$ is continuous; that is, the function $t\mapsto u(e^{it})$ is continuous. This follows from $u(e^{it})$ being a uniform limit of continuous functions $t\mapsto u(re^{it})$ (recall that uniform convergence preserves continuity).
Step 2: given $a\in \partial \mathbb{D}$ and $\epsilon>0$, use step 1 to conclude there is $\delta>0$ such that $|u(e^{it})-u(a)|<\epsilon/2$ whenever $|e^{it}-a|<\delta$. Then use the uniform convergence assumption: there is $r_0$ such that $|u(e^{it})-u(re^{it})|<\epsilon/2$ whenever $r_0<r<1$. Combine both with the triangle inequality to conclude that $$ |u(re^{it})-u(a)| < \epsilon\quad \text{whenever } r>r_0 \ \text{ and } \ |e^{it}-a|<\delta $$ This proves the continuity at an arbitrary point $a\in\partial \mathbb{D}$.
The continuity at the points of $\mathbb{D}$ is given.