9

This might be very silly to ask, but somehow this sequence of results are leading me to this wrong result. I am dealing with complex analysis and the mistake I am making might be because I am using some results from real analysis.

If a function, $f(z)$, is continuous in simply connected domain, then it will be Riemann integrable and hence its antiderivitive, $F(z)$, will exist and moreover the antiderivative will be differentiable in the domain.

This implies that $F(Z)$ is analytic since it is differentiable in the neighborhood of all points.

Which also means that it is infinitely times differentiable.

And hence even $f(z)$ is infinitely times differentiable and hence, $f(z)$ is also analytic.

Peter Mortensen
  • 670
Shubham Ugare
  • 703

3 Answers3

12

Your second point is where it breaks. The antiderivative exists iif the integral from one point to another is independent from the path that is taken which is not guaranteed by continuity.

curiosity
  • 566
  • 1
    Are not all continuous functions Riemann integrable? – Shubham Ugare Aug 29 '15 at 06:02
  • 2
    Yes on a compact for instance. But to define the antiderivative you have to choose a point, lets say (0,0) , an then define F(x,y) as the integral from you get from going from (0,0) to (x,y) and to do that you have to choose a path to integrate on. The problem is that you have no guarantee that the "F(x,y)" you just calculated doesn't depend on the path you arbitrary chose. For example the result may be different if you go (0,0)-->(0,y)-->(x,y) or if you go (0,0)-->(x,0)-->(x,y) in straight lines. – curiosity Aug 29 '15 at 06:07
  • 2
    Part of the problem for the OP is the misleading name of "integral" that is often given to the antiderivative. – Martin Argerami Aug 29 '15 at 11:07
4

You are mixing up complex differentiability and real differentiability.

Thomas
  • 22,361
0

Take $f$ defined on the real line by $f(x) = e^{-1/x}$ for $x>0$ and $f(x)=0$ if $x\leq 0$. The function $f$ is continuous, and is even infinitely differentiable, on the real line, but is not analytic. Indeed, you prove that $f^{(n)}(0)=0$ for all $n$, which shows that $f$ that cannot be analytic on any interval containt $0$ in its interior, as it would then be (by analytic continuation) equal to the zero function.

Olórin
  • 12,040
  • 1
    That doesn't really answer the question. OP is aware that the result does not hold, (s)he asks to point a flaw in the argument. – Marcin Łoś Aug 29 '15 at 08:48