1
$\begingroup$

Following problem is an exercise in "Complex variables with application - S. Ponnusamy"

Suppose that $f(z)$ is continuous, but not necessarily analytic, on an contour $C$. Show that the function $$F(z)=\int_{C}\frac{f(\zeta)}{\zeta-z}d\zeta$$ is analytic at each $z$ not on $C$, with $$F'(z)=\int_{C}\frac{f(\zeta)}{(\zeta-s)^{2}}d\zeta.$$

Is it possible to apply the Morera's theorem showing this problem?

It seems related, but there is no progress.

My attempt is as follows:

For any $z\not\in C$, choose $|h|>0$ small enough so that the line segment from $z$ to $z+h$ don't across $C$. Then, we have $$\Bigg|\frac{F(z+h)-F(z)}{h}-F'(z)\Bigg|\le|h|\cdot\int_{C}\frac{|f(\zeta)|}{|\zeta-z-h|\cdot|\zeta-z|^2}|d\zeta|.$$ Now, take $|h|\to0$, we see that $F$ is analytic at each $z$ not on $C$.

Give some advice. Thank you!

$\endgroup$
  • 2
    $\begingroup$ Morera's theorem isn't applicable here. Following your work, notice that your last integral can be bounded by $2|h| \cdot \|f\|_{\infty} \operatorname{len}(C) / \operatorname{dist}(z, C)^3$ if you're careful about $h$. $\endgroup$ – user296602 Mar 3 '18 at 17:11
  • 1
    $\begingroup$ For Morera's theorem, you should try to integrate $F$ on any contour sufficiently small around $z$. Have you tried working out what that might be? $\endgroup$ – user357980 Mar 3 '18 at 17:11
  • $\begingroup$ @user296602 Thanks a lot, it is a helpful comment. $\endgroup$ – Primavera Mar 4 '18 at 7:37
  • $\begingroup$ @user357980 In fact, that's the way I wanted to do it. Because of the given conditions, it seemed natural to use the Morera's theorem. Unfortunately, I don't know what to do. Is it imitating the proof of Morera's theorem? How do I get started? $\endgroup$ – Primavera Mar 4 '18 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.