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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 $z=1$?

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2 Answers 2

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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

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

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Commented Oct 5, 2016 at 23:28
  • $\begingroup$ very nice appoarch ! To prove the function was continuous couldn't you have constructed a bound for the upper half of your given contour then proved that, the limit of the given integral on the upper half plane vanishes ? $\endgroup$
    – Zophikel
    Commented Jul 7, 2018 at 21:16
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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.

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  • $\begingroup$ can you please explain why $F$ is not continuous when $F(z)=\bar{z}$ for $|z|=1$ $\endgroup$
    – HAC
    Commented Feb 6, 2018 at 16:40
  • $\begingroup$ $0=\lim_{z\rightarrow 1^-} F(z) \neq F(1)=1$ $\endgroup$
    – H. H. Rugh
    Commented Feb 7, 2018 at 11:45
  • $\begingroup$ how is this limit $0$? $\endgroup$
    – HAC
    Commented Feb 7, 2018 at 12:08
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    $\begingroup$ $F$ is identically zero for $|z|<1$ $\endgroup$
    – H. H. Rugh
    Commented Feb 7, 2018 at 17:39

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