The continuity of this two variable function. Is this continuous? I have come across with this problem in studying partial differential equation.
Let $f$ be a continuous and periodic function on $\mathbb R$ with period $2\pi$, and denote $D=\{(x,y)\mid x^2+y^2<1\}.$
And let $u:D\to \mathbb R$ be continuous and $C^2$ class on $D$, and satisfy $$\lim_{\delta \to 0}\underset{1-\delta\leqq r<1}{\sup_{|a-b|\leqq \delta}}|u(r\cos a,r\sin a)-f(b)|=0,$$ where I use the polar coodinates $x=r\cos a, y=r\sin a$ for $u(x,y)$.
Then, if I define $v: \overline D \to \mathbb R$ by $$v(x,y)=v(r\cos a, r\sin a):=\begin{cases}u(r\cos a, r\sin a) & \mathrm{if} \ (x,y)\in D\\ f(a) &\mathrm{if} \ (x,y)\in \overline D \setminus D=\partial D\end{cases}$$
, is $v$ continuous on $\overline D$ ?
On $D$, $v$ is continuous from the supposition of continuity of $u$, so the problem is the continuity at $\partial D.$
I expect this holds because
$$\lim_{\delta \to 0}\underset{1-\delta\leqq r<1}{\sup_{|a-b|\leqq \delta}}|u(r\cos a,r\sin a)-f(b)|=0$$
means that if $a\fallingdotseq b$ and $r\fallingdotseq 1$, then $u(r\cos a, r\sin a)\fallingdotseq f(b)\fallingdotseq f(a)$. ($f(b)\fallingdotseq f(a)$ follows from the continuity of $f$.)
So I tried to prove but it doesn't seem to work.

Let $(c,d)\in \overline D$.
If $(c,d)\in D$, $v$ is continuous at $(c,d)$ due to the continuity of $u$.
So I'll consider the case $(c,d)\in \partial D.$
Let $\epsilon>0.$
$(c,d)$ is on the unit circle so I can write $$c=\cos \xi, d=\sin \xi.$$
From $$\lim_{\delta \to 0}\underset{1-\delta\leqq r<1}{\sup_{|a-b|\leqq \delta}}|u(r\cos a,r\sin a)-f(b)|=0,$$
there is $\delta_1$ s.t. $$|\delta|\leqq \delta_1\Rightarrow \underset{1-\delta\leqq r<1}{\sup_{|a-b|\leqq \delta}}|u(r\cos a,r\sin a)-f(b)| <\epsilon.$$
And from the continuity of $f$, there is $\delta_2>0$ s.t. $$|\eta-\xi|<\delta_2 \Rightarrow |f(\eta)-f(\xi)|<\epsilon.$$
Let $\delta_3:=\min\{\delta_1, \delta_2\}$, and let $(x,y)\in \overline D$ satisfy $|(x,y)-(c,d)|<\delta_3.$ Denote $x=r\cos a, y=r\sin a.$
Then, if I could show $|v(x,y)-v(c,d)|<\epsilon$, the proof will finish.
If $(x,y)\in \partial D$, then
$|v(x,y)-v(c,d)|=|f(a)-f(\xi)|$, so $|a-\xi|<\delta_3$ is desired, but this doesn't seem to work because $|a-\xi|<|(x,y)-(c,d)|$ doesn't hold. (Actually, the opposite inequality holds.)
If $(x,y)\in D$, I want to do
\begin{align}
|v(x,y)-v(c,d)|&=|u(x,y)-f(\xi)|\\
&\leqq\underset{1-\delta_3\leqq r<1}{\sup_{|a-b|\leqq \delta_3}}|u(r\cos a,r\sin a)-f(b)|
\end{align},
so I have to check $|a-\xi|\leqq\delta_3$ and $1-\delta_3\leqq r<1$
I get $1-\delta_3\leqq r<1$ since $r<1$ follows from $(x,y)\in D$ and $1-\delta_3\leqq 1-|(x,y)-(c,d)|\leqq r.$
Thus, in both cases $(x,y)\in \partial D$ and $(x,y)\in D$, $|a-\xi|<\delta_3$ is desired.
I'd like you to share the idea for showing $|a-\xi|<\delta_3$. Another proof for the continuity of $v$ is also welcomed.
 A: The two hypothesis (on $u$ and $f$) simply mean that for any $p\in\partial D$,
$$\lim_{m\in D,\;m\to p}v(m)=v(p)\quad\text{and}\quad\lim_{m\in\partial D,\;m\to p}v(m)=v(p),$$which is equivalent to
$$\lim_{m\in \overline D,\;m\to p}v(m)=v(p)$$
and this ends the proof.
If you prefer a more "pedestrian" proof, for any $\xi\in\mathbb R/(2\pi\mathbb Z)$, of the continuity at $(\cos\xi,\sin\xi)\in\partial D$, let $\epsilon>0$.
Choose $\delta_1\in(0,\pi]$ such that
$$\underset{1-\delta_1\le r<1}{\sup_{|a-\xi|\le\delta_1}}|u(r\cos a,r\sin a)-f(\xi)|<\epsilon$$
and $\delta_2\in(0,\pi]$ such that
$$|a-\xi|\le\delta_2\Rightarrow|f(a)-f(\xi)|<\epsilon.$$
Let $\|\cdot\|$ denote the Euclidean distance and let $$\delta_3:=\min(\sqrt2,2\sin(\delta_1/2),2\sin(\delta_2/2))\in(0,\delta_1].$$
Then, for all $a\in\mathbb R/(2\pi\mathbb Z)$, we have both
$$\forall r\in[0,1)\quad\|(r\cos a,r\sin a)-(\cos\xi,\sin\xi)\|\le\delta_3\Rightarrow r\ge1-\delta_1\text{ and }|a-\xi|\le\delta_1$$
and
$$\|(\cos a,\sin a)-(\cos\xi,\sin\xi)\|\le\delta_3\Rightarrow|a-\xi|\le\delta_2,$$
the inequalities $|a-\xi|\le\delta_i$ being due to the fact that $|a-\xi|\le\frac\pi2$ and $\sin|(a-\xi)/2|\le\delta_3/2\le\sin(\delta_i/2).$
Thus, $\forall(x,y)\in\overline D$,
$$\|(x,y)-(\cos\xi,\sin\xi)\|\le\delta_3\Rightarrow|v(x,y)-f(\xi)|<\epsilon.$$
