1

Claim: Let $D$ be a domain, let $C$ be a simple closed contour in $D$, f is analytic in $D/C$ and continuous in $D$, then $f$ is analytic in $D$.

I tried to show the contour integrals of $f$ in $D$ are zero and then by Morera's theorem, we can have the desired conclusion. For closed contour $\Gamma$ that does not cross with the given $C$, $\int_\Gamma f(z)dz=0$ by Cauchy-Goursat theorem, however, I don't quite have a clear picture on how to deal with the contours that does intersects with $C$. Any ideas on how I might continue my method or other ways to tackle the problem are much appreciated.

mathlearner
  • 45
  • What properties does this contour have? Is it closed, rectifiable or smooth? –  Apr 05 '20 at 12:41
  • The original statement in the question doesn't specify a property, but I guess let it be a simple closed contour. That may simplify the discussion. – mathlearner Apr 05 '20 at 13:02
  • It does not only simplify the discussion: without this assumption, by taking as contour something as pathological as the peano curve, the claim is false –  Apr 05 '20 at 13:06
  • I see. I guess the original question has a typo, since it is positioned as an application of Cauchy-Goursat theorem, I think we should add the property. I updated my post just now – mathlearner Apr 05 '20 at 13:11
  • It may be useful to note that Morera's theorem can be strengthened, requiring $\oint_\gamma f=0$ only for triangles –  Apr 05 '20 at 13:45

1 Answers1

3

Quick answer:

  1. if the contour is simply assumed to be a jordan curve (i.e. simple, closed and continuous), the claim is wrong
  2. if the contour is assumed to be rectifiable, the claim is true, but its proof is far from trivial

Explanation:

This claim, though it may seem naturally true, is actually quite complicated, without some kind of additional hypotesis other than the contour being a Jordan closed contour (and this is usually a ** of many textbooks in complex analysis). Let us state some results:

A compact set $K\subset \mathbb{C}$ is said to be removable if your statement holds, i.e. if, for every domain $D$ containing $K$, the set of functions analytic on $D$ is equal to the set of functions analytic on $D-K$ and continuous on $D$. This definition is strictly related to the concept of (continuous) analytic capacity , as you can see here and here. In particular, a theorem of Painlevé states that a rectifiable curve $\gamma$ is a removable set, while if the area of the curve is positive, the set is non-removable. For a proof of both statements, see "null sets for a class of analytic functions" by Zalcman.

As much as the concept of a curve with positive area is counterintuivite, this condition rules out the possibility of keeping the hypotesis on the contour as a jordan curve: in fact, there exists jordan curves and jordan arcs (called Osgood curves) with positive area (see here for more informations), and which are thus non removable.

I suppose the author of your exercise has tacitly assumed (or maybe has stated before) some properties of contours: for example, if the contour is "nice": for example, if we assume that every line intersect the curve in at most a finite number of points (this is true, for example, if the contour is an analytic curve), Morera's theorem (for triangles) is enough to show analyticity. Obviously such a requirement is very restrictive (even if it is geometrically quite intuitive): e.g., there exists $C^{\infty}$ curves which do not satisfy this requirement: $$\gamma(t)=\pmatrix{t\\ \sin\left(\frac{1}{t^2}\right)e^{-\frac{1}{t^2}}}$$

intersects the line $y=0$ in a countable set of points

Note: by area, I mean lebesgue measure on the plane

  • In the rectifiable case one can give an argument based on the fact that Cauchy (and residues etc) hold on rectifiable Jordan curves for analytic functions in the domain inside the curve that are continuous on the boundary and then one can prove the result locally by taking a small arc and completing it to Jordan rectifiable curves in the domains inside and outside the curve - one doesn't need Morera, just constructing two functions like that $2\pi if_j(z)=\int_{C_j}{\frac{f(\zeta)}{\zeta-z}}d\zeta$ and then $f_1+f_2$ is the required analytic continuation as the arc in cause gets canceled out – Conrad Apr 07 '20 at 00:14