Let $F:[-π,π]→[0,∞]$ be $2π$_ periodic, integrable and "continuous" function. For $-π≤θ≤π$ and $0≤r<1$. Show that $$ P_r*F(θ)=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\dfrac{1-r^2 }{1-2r \cos (θ-t)+r^2}F(t) dt$$ can be continuously extended to the unit circle, as mapping again to$ [0,∞]$?
I know that $ P_r*F(θ)→ F(\theta)$ as $r→ 1$, when $F$ is continous on the unit circle, but I'm not sure how to prove this 'continuous extension' when $F$ takes $∞$, If any one can help!
Thanks.
The proof for the usual continuous (finite) case is local, so in other words, showing that $P_r*F(e^{i\theta})→ F(e^{i\theta})$ depends only on a small neighborhood of $e^{i\theta}$ on the circle (and moreover one can show "full" continuity in the sense that if $u(re^{i\theta})=P_r*F(e^{i\theta})$ we actually have $u(z) \to F(e^{i\theta}), z \to e^{i\theta}, |z| \le 1$.
In particular, for finite points of continuity, $\theta$ nothing changes as long as we can show that away from $e^{i\theta}$ the Poisson integral converges to zero with $r$. Fixing $e^{i\theta}=1$ for notational convenience, this would mean to show that
$\dfrac{1}{2\pi}\int_{\delta}^{2\pi-\delta}\dfrac{1-r^2 }{1-2r \cos (t)+r^2}F(t) dt \to 0, r \to 1, \delta >0$ arbitrary; in the usual case this follows immediately from the Poisson Kernel properties and the boundness of $F$, while here we use that $F \ge 0$ integrable so we can apply the Monotone convergence theorem since $g(r,t)=\dfrac{1-r^2 }{1-2r \cos (t)+r^2}F(t)$ is decreasing in $r$ to zero (at least where $F(t)$ is finite hence ae) for $r \ge |\cos \delta|$ - as the derivative in $r$ is clearly negative for $r \ge |\cos t|$ and $|\cos t| \le \cos \delta$
Hence the only thing you need to do is to prove the result at points $\theta$ where $F$ is infinite and assume for notational convenience $e^{i\theta}=1$ - so we need to prove that $u(z) \to \infty, z \to 1$; but for each $N$ there there is a neighborhood of $1$ where $F(e^{i\theta}) \ge N, |\theta| \le 2\alpha_N$ for some small $\alpha_N >0$ so
$u(re^{i\theta})=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\dfrac{1-r^2 }{1-2r \cos (θ-t)+r^2}F(t) dt=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\dfrac{1-r^2 }{1-2r \cos t+r^2}F(t+\theta) dt \ge$
$\ge \dfrac{1}{2\pi}\int_{-\alpha_N}^{\alpha_N}\dfrac{1-r^2 }{1-2r \cos t+r^2}F(t+\theta) dt \ge \dfrac{N}{2\pi} \int_{-\alpha_N}^{\alpha_N}\dfrac{1-r^2 }{1-2r \cos t+r^2}dt$, for $|\theta| \le \alpha_N$
But now as before $\dfrac{1}{2\pi}(\int_{-\pi}^{-\alpha_N}+\int_{\alpha_N}^{\pi})\dfrac{1-r^2 }{1-2r \cos (t)+r^2} dt \to 0, r \to 1$ so one can pick $r_N$ for which the integral above is at most $1/N, r \ge r_N$ and putting things together we get:
$u(re^{i\theta}) \ge \dfrac{N}{2\pi} \int_{-\alpha_N}^{\alpha_N}\dfrac{1-r^2 }{1-2r \cos t+r^2}dt \ge \dfrac{N}{2\pi} (\int_{-\pi}^{\pi}\dfrac{1-r^2 }{1-2r \cos t+r^2}dt-1/N)=\dfrac{N-1}{2\pi}$ for $|\theta| \le \alpha_N, r \ge r_N$ hence indeed $u(z) \to \infty, z \to 1$ and we are done!