Check continuity for complex function $F(z) = \frac{1}{2\pi i} \oint_\gamma \frac{f(\zeta)}{\zeta -z} d\zeta$ for $|z|<1$ First state the problem. 
Let $f$ be a continuous function on $\{z : |z|=1\}$. Define $\gamma=$ the unit circle traversed counterclockwise,
\begin{align}
  F(z)  =\left\{ \begin{array}{cc}
+ f(z), \quad \textrm{if} \quad |z|=1  \\
\frac{1}{2\pi i} \oint_\gamma \frac{f(\zeta)}{\zeta -z} d\zeta, \quad \textrm{if} \quad |z| <1
\end{array} \right.
\end{align}
Is $F$ continuous on $\bar{D}(0,1)$? 
The hint given by lecturer was think $f(z) =\bar{z}$, so i start 

For $|z|<1$, 
\begin{align}
\frac{1}{2\pi i} \oint_{\gamma} \frac{f(\zeta)}{\zeta-z} d \zeta 
= \frac{1}{2\pi i} \oint_{\gamma}\frac{1}{\zeta(\zeta-z)} d\zeta 
\end{align}
and since $\zeta=0$, $\zeta=z$ is inside the unit circle, by residue theorem, 
$\frac{1}{\zeta}|_{\zeta=z}$ and $\frac{1}{\zeta-z}|_{\zeta=0}$ gives vanishing integral, 
In order to show they are continuous the limit $|z|=1$, in this case i wonder the pole in on the contour, so i am confusing whether the poles on the contour can be candidated for poles or not. 
So my question here is what is the value of 
\begin{align}
\lim_{z\rightarrow 1} \frac{1}{2\pi i} \oint_{\gamma}\frac{1}{\zeta(\zeta-z)} d\zeta 
\end{align}
is it still vanish at $z=1$?
 A: 
The question as to the continuity of $F(z)$ has been addressed in another posted solution.  I thought it might be instructive to focus on the question of the integral $\oint_{|\zeta|=1}\frac{f(\zeta)}{z-\zeta}\,d\zeta$ when $|z|=1$

Let $I(z)$ be the integral given by 
$$\begin{align}
I(z)&=\frac{1}{2\pi i}\oint_{|\zeta|=1} \frac{f(\zeta)}{z-\zeta}\,d\zeta\\\\
\end{align}$$
for $|z|<1$.  In general, this integral fails to exist for $|z|=1$, even if $f$ is analytic, due to the singularity at $\zeta=z$ on the integration path.

However, we can evaluate the Cauchy Principal Value of $I(z)$ as defined by
$$\begin{align}
\text{P.V.}\left(\frac{1}{2\pi i}\oint_{|\zeta|=1}\frac{f(\zeta)}{z-\zeta}\,d\zeta\right)&=\lim_{\epsilon\to0^+}\left(\frac{1}{2\pi }\int_0^{\theta-\epsilon} \frac{e^{i\phi}f(e^{i\phi})}{e^{i\theta}-e^{i\phi}}\,d\phi+\frac{1}{2\pi }\int_{\theta+\epsilon}^{2\pi} \frac{e^{i\phi}f(e^{i\phi})}{e^{i\theta}-e^{i\phi}}\,d\phi\right)\tag1\end{align}$$
where $z=e^{i\theta}$, $0\le \theta<2\pi$.  


If $f(z)$ is analytic in and on the contour $|z|=1$, then Cauchy's Integral Theorem guarantees that the right-hand side of $(1)$ becomes
$$\begin{align}
\lim_{\epsilon\to0^+}\left(\frac{1}{2\pi }\int_0^{\theta-\epsilon} \frac{e^{i\phi}f(e^{i\phi})}{e^{i\theta}-e^{i\phi}}\,d\phi+\frac{1}{2\pi }\int_{\theta+\epsilon}^{2\pi} \frac{e^{i\phi}f(e^{i\phi})}{e^{i\theta}-e^{i\phi}}\,d\phi\right)&=\lim_{\epsilon \to 0^+}\left(\frac{1}{2\pi}\int_0^\pi  \frac{e^{i\psi}f(z+\epsilon e^{i\psi})}{e^{i\psi}}\,d\psi\right) \tag2\\\\&=\frac12 f(z)
\end{align}$$
In arriving at $(2)$ we deformed the circular contour $|\zeta|=1$ to exclude enclosing the pole at $z=e^{i\theta}$, $0\le \theta <2\pi$, with a semi-circular contour centered at $z$ with radius $\epsilon$.     

When $f(z)=\bar z$, we have

$$\begin{align}
\text{P.V.}\oint_{|\zeta|=1}\frac{f(\zeta)}{z-\zeta}\,d\zeta&=\lim_{\epsilon\to0^+}\left(\frac{1}{2\pi }\int_0^{\theta-\epsilon} \frac{1}{e^{i\theta}-e^{i\phi}}\,d\phi+\frac{1}{2\pi }\int_{\theta+\epsilon}^{2\pi} \frac{1}{e^{i\theta}-e^{i\phi}}\,d\phi\right)\\\\
&=\lim_{\epsilon\to 0^+}\left(\frac{1}{2\pi }\int_0^{\theta-\epsilon}\frac{e^{-i\phi}}{e^{i\theta}e^{-i\phi}-1}\,d\phi+\frac{1}{2\pi }\int_{\theta+\epsilon}^{2\pi}\frac{e^{-i\phi}}{e^{i\theta}e^{-i\phi}-1}\,d\phi\right)\\\\
&=\frac{ie^{-i\theta}}{2\pi } \lim_{\epsilon\to 0^+}\left(\log(e^{i\epsilon}-1)-\log(e^{i\theta}-1)+\log(e^{i\theta}-1)-\log(e^{-i\epsilon}-1)\right)\\\\
&=-\frac12 e^{-i\theta}\\\\
&=-\frac12 z
\end{align}$$

A: Your argument shows that for $f(z)=\bar{z}$ you have $F(z)=0$ for $|z|<1$.
Since $F(z)=\bar{z}$ for $|z|=1$ the function $F$ is manifestly not continuous.
The procedure is btw a way to extract the analytic part of a given funtion (works even for $L^2$). For $f(z)=\bar{z}$ there is no analytic part. If $f$ is analytic inside $D$ (and extends continuously to the boundary) then $F$ will indeed be continuous.
