The question is the following:
Let $f$ be a piecewise $\mathcal{C}^1(\mathbb{R})$, $2\pi$-periodic function, and let $a_{n}[f],b_{n}[f]$ be its real Fourier coefficients. Show that the series
\begin{align*}
a_{0}[f]+
\sum_{n=1}^{\infty} \rho(x, y)^n(a_{n}[f]\cos n\theta(x,
y)+b_n[f]\sin n\theta(x, y)),
\end{align*}
converges uniformly in any $\Omega_{\rho_0}=\{(x,y)\in\mathbb{R}^2:x^2+y^2<\rho_0^2\},\,\forall
\rho_0\in(0,1)$ to a function $u\in\mathcal{C}^{\infty}(\Omega_1)$, solution to
\begin{align*}
&u_{xx}+u_{yy}=0,\quad (x, y)\in\Omega_1
\end{align*}
which satisfies the boundary condition
\begin{align*}
&\lim_{\rho\to 1-}u(\rho\cos\theta, \rho\sin\theta)=\frac{f(\theta-)+f(\theta+)}{2},\,\forall \theta\in\mathbb{R}.
\end{align*}
I have indeed shown that the series converges uniformly in each $\Omega_{\rho_0}$ to a solution of Laplace's equation $u\in\mathcal{C}^{\infty}(\Omega_1)$. However, I'm struggling to show that the boundary condition is satisfied. I have also shown that
\begin{align*}
\lim_{m\to\infty}&\sum_{n=0}^{m}u_n(\cos\theta, \sin\theta)=\lim_{m\to\infty}\sum_{n=0}^{m}a_{n}[f]\cos(n\theta)+b_n[f]\sin(n\theta)=\\
&=\lim_{m\to\infty} T_m[f](\theta)=\frac{f(\theta-)+f(\theta+)}{2},\;\forall \theta\in\mathbb{R},
\end{align*}
but as the convergence is not uniform in $\Omega_1$, the infinite sumation and the limit cannot be exchanged. I have also tried to fix $\theta^*$ and show that
$$v(\rho)=u(\rho\cos\theta^*,\rho\sin\theta^*)$$
is continuous at $\rho=1$, but I cannot make it work. Do you have any ideas on how should I focus the proof or any references I could follow? I have seen some things based on the Poisson kernel, but I am not supposed to use complex numbers.
Best Answer
It seems like Abel's theorem for power series applies due to the pointwise convergence of $\sum_{n = 0}^{\infty} a_n[f]\cos(n\theta) + b_n[f]\sin(n\theta)$, meaning that you do actually have continuity of your function $v$ at $\rho = 1$.
And even if you were not supposed to know about that theorem, maybe you could adapt its proof to your case?