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Here's what I'm wondering.

prove that : $f(a) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - a}dz, C(\theta) = e^{i\theta}(0\le\theta\le 2 \pi), |a| = 1, a \in \mathbb{C}$ where $f(z)$ is continuous on $C$

I know that this statement is true when $a$ is strictly inside the unit disk but I'm lost how to find out the case when it's on the boundary. I posted original problem but didn't get the answer(here) and it's the second part of it. I figured out to this point according to this post, and I used the same method(to show uniform convergence of the integrand) to prove that second part of original problem reduces to the problem I stated. Can you give me some help regarding this?

  • Well, you also need some assumptions on the function...don't you? – DonAntonio May 07 '20 at 21:08
  • And after the above, check the lemma and in particular its corollary in the most upvoted answer here: https://math.stackexchange.com/questions/83828/definite-integral-calculation-with-poles-at-0-and-pm-i-sqrt3/184874#184874 This is what you need. – DonAntonio May 07 '20 at 21:09
  • @DonAntonio right.. I forgot the assumptions I'll add those –  May 07 '20 at 21:17
  • @DonAntonio And I'll check the lemma out! –  May 07 '20 at 21:19
  • When $a$ is on the boundary the integral is, in general, not convergent. Do you mean to take the P.V. of the integral? –  May 07 '20 at 21:29
  • @Caffeine what is P.V.? –  May 07 '20 at 21:32
  • https://en.wikipedia.org/wiki/Cauchy_principal_value?wprov=sfla1 –  May 07 '20 at 21:32
  • Well, I haven't heard of it but I'm sure that I don't mean P.V. but I mean the actual integral. The original problem goes like : How do I prove that $g(z) = \oint_C \frac{f(t)} {t - z} dt$ ($|z| \ne 1$, $C = e^ {I\theta}, 0 \le \theta \le 2\pi$) is analytic with $f$ continuous on unit circle? And that $\lim_ {z\to a} g(z) = f(a),|a| = 1$, limit taken from inside the unit disc? –  May 07 '20 at 21:39
  • @Caffeine And here, I showed the first part and reduced the second part which is computing the limit, and by uniform convergence of the integrand this became the problem I wrote above: prove or disprove : $f(a) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - a}dz, C(\theta) = e^{i\theta}(0\le\theta\le 2 \pi), |a| = 1, a \in \mathbb{C}$ where $f(z)$ is continuous on $C$ –  May 07 '20 at 21:41

1 Answers1

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Take $f(z)=1; a=1$. The integral is

$$\int_0^{2\pi}\frac{ie^{i\vartheta}}{e^{i\vartheta}-1}d\vartheta=i\left(\int_0^{2\pi}1+\frac{1}{e^{i\vartheta}-1}d\vartheta\right)=\\ =2\pi i+i\int_0^{2\pi}\frac{1}{e^{i\vartheta}-1}d\vartheta$$

The remaining integral is not convergent (as it is asymptotic to $1/\vartheta$ as $\vartheta\to 0$; $\vartheta\to 2\pi$)

As for your original problem, a hint: The statement is true for constant functions. Thus, to prove that it is true for $f$, it suffices to prove it for $f-f(a)$. How does this simplify the problem?

  • You're saying that the original statement is true; Do you mean the original limit and the one I wrote in the main question are not equivalent? In other words I made some mistake? –  May 07 '20 at 21:49
  • @pratjohn the original statement is true, while this one is not. In particular, I think you made a mistake in assuming the convergence was uniform –  May 07 '20 at 21:51
  • Got it. I'll try to check where it got wrong. Thanks for you help though. I'll try to solve the original one with your suggestion. –  May 07 '20 at 21:52
  • And also, can you give a little bit more reason why your example diverges? because I think being asymtotic at some point is not quite sufficient for the integral to diverge. Maybe some inequalities or so that can prove divergence? –  May 07 '20 at 21:54
  • I tried to prove that using your advice and I suppose you are suggesting to prove like: $\lim_{z\to a}g(z)=f(a),|a|=1$ here, by cauchy intergral foumula, for each z inside the unit disk, the integral $g(z) = \frac{1}{2\pi i}\oint_C\frac{f(t)}{t-z}dt=f(z)$ but actually $f$ is defined only on the boundary with condition that it is continuous. Is there something more I could do from now? (I messed up with definition of $g(z)$ in above comment. it should have factor at front. –  May 07 '20 at 22:23