# Is $F$ continuous on the closed unit disk $D(0, 1)$?

Let $f$ be a continuous function on the unit circle $\partial D=\{z : |z| = 1\}$. Define $$F(z) = \begin{cases} f(z), & \text{if |z|=1,} \\ \displaystyle \frac{1}{2\pi i}\int _{\partial D} \frac{f(w)}{w-z} dw, & \text{if |z| \lt 1.} \end{cases}$$ Is $F$ continuous on the closed unit disk $\overline{D}(0, 1)=\{z : |z| \leq 1\}$?

My take: well I suppose since $f(z)$ is continuous according to the assumption, that makes $F(z)$ continuous too but how do you justify that? Any hints?

• This is not as elementary as it sounds. See Theorem 11.8 of third edition of Rudin's RCA. – Kavi Rama Murthy Feb 14 '18 at 7:58
• I'm sorry, is there a link to that source? I couldn't find it online – ISuckAtMathPleaseHELPME Feb 14 '18 at 8:00
• Sorry. I mistook F to be the Poisson integral of f. The correct answer is F is not necessarily continuous. – Kavi Rama Murthy Feb 14 '18 at 10:08

No, $F$ is not necessarily continuous in $\{z : |z| \leq 1\}$.
Take for example $f(z)=\overline{z}$ which is continuous in $\mathbb{C}$ (but not holomorphic). Then for $r\in (0,1]$, $$\frac{1}{2\pi i}\int _{\partial D} \frac{f(w)}{w-r} dw= \frac{1}{2\pi i}\int _{0}^{2\pi} \frac{e^{-it}e^{it}i}{e^{it}-r} dt =\frac{1}{2\pi}\int _{0}^{2\pi} \frac{dt}{e^{it}-r}=-\frac{i\left[\ln(1-re^{-it})\right]_0^{2\pi}}{2\pi r}=0.$$ Hence $$\lim_{r\to 1^-}F(r)=\lim_{r\to 1^-}\frac{1}{2\pi i}\int _{\partial D} \frac{f(w)}{w-r} dw=0\not=1=f(1)=F(1).$$ and we may conclude that $F$ is not continuous at $1$.
Suppose that $f(z)=\dfrac1z$. Then, when $|z|<1$ and $z\neq0$,\begin{align}F(z)&=\int_{\partial D}\frac1{w(w-z)}\,\mathrm dw\\&=\frac1z\int_{\partial D}\frac1{w-z}-\frac1w\,\mathrm dw\\&=\frac1z\left(\int_{\partial D}\frac1{w-z}\,\mathrm dw-\int_{\partial D}\frac1w\,\mathrm dw\right)\\&=0,\end{align}since each of the integrals is $2\pi i$ times the winding number of the loop $t\mapsto e^{it}$ ($t\in[0,2\pi]$) with respect to z and to $0$ respectively, both of which are equal to $1$ (a conclusion that you can also reach using Cauchy's integral theorem).
So, $F(z)=0$ when $|z|<1$. But $F(1)=1$. So…