This question is from Brown and Churchill's Complex Variables and Applications, 8ed., Section 52, Question 6.

Let $f(s)$ denote a continuous function taken along a simple contour, $C$ enclosing a region $D$, prove that the function:

$$ g(z) = \frac{1}{2 \pi\mathrm{i}}\int \frac{f(s)}{s-z} ds$$

is analytic $\forall z \in D$ and that

$$ g'(z) = \frac{1}{2 \pi\mathrm{i}}\int \frac{f(s)}{(s-z)^2} ds$$

I know to prove Cauchy's Integral formula I'm to assume that some ball with radius $R$ creates a region $D$ and $f(s)$ is holomorphic on $D$, except at the point $z$. I can create an open ball with radius $r$ centered at $z$. Then using a limit, I reduce the radius $r$ until it converges on the singular point $z$ and show that this limit (or residue?) is zero. i.e. Show:

$$\lim_{r\to 0}\, \left\lvert \int_{C_R} \frac{f(s)}{s-z}\ ds - 2\pi\mathrm{i} f(z) \right\rvert = 0. $$

The problem is that in the above proof I've assumed that $f(s)$ is holomorphic, therefore analytic everywhere therefore continuous everywhere. If I'm only allowed to assume that $f(s)$ is continuous how can I show that $f(s)$ is also analytic? Does this have something to do with existence of a derivative in an open ball around the singularity? How can I make the jump between continuity and analyticity?


In the statement of the problem you’ve referred to, Brown and Churchill give a big hint by recommending that you follow “… a procedure used in Sec. 51 …”. In fact, this exercise appears to be little more than a nudge from the authors for the reader to follow and understand the text’s Section 51 extension of Cauchy's integral formula. Note that in their derivation, the only place where the assumption that $f$ is analytic is used is in the application of the Cauchy integral formula. But the Cauchy integral formula is not required in the specific problem that’s been presented in Section 52’s problem #6, and a quick review of the author's Section 51 demonstration shows that only continuity for $f$ is required to show that the function $g$ is analytic at all points in $D$ and that $g'$ has the indicated form.

For anyone interested in seeing the Brown-Churchill argument, here (as of April 2016) is a link the the full-text 8th edition. The exercise is on page 171 and the Section 51 demonstration is on pages 166-167. A slightly different argument is given in Ahlfors' classic "Complex Analysis" in section 2.3 (third edition).

  • $\begingroup$ $g(z) = \int_C \frac{f(s)}{s-z}ds$ is obviously holomorphic, since $$\frac{g(z+h)-g(z)}{h} = \int_C f(s) \frac{\frac{1}{s-z-h}-\frac{1}{s-z}}{h}ds =\int_C f(s)(\frac{-1}{(s-z)^2}+\mathcal{O}(h))ds$$ $\endgroup$ – reuns Oct 22 '16 at 3:05
  • $\begingroup$ To clarify the Ahlfors reference: Third Edition, Chapter 4, section 2.3, lemma 3, page 121. $\endgroup$ – Tom Collinge Jul 16 '18 at 12:40

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