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Let $f$ be a function. Then $f$ is holomorphic is equivalent to $f$ being analytic which in turn is also equivalent to satisfying the Cauchy-Riemann equations. All three concepts imply infinite differentiability.

My question is what do we need to add to infinite differentiability to recover analyticity/ holomorphicity/Cauchy Riemann equations?

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Seen this? –  J. M. Aug 7 '11 at 6:05
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...and very related. –  J. M. Aug 7 '11 at 6:06
    
Thanks a lot, didn't know about these examples! –  user786 Aug 7 '11 at 6:10
    
But those are real-analytic, right? So, do you extend them into complex-analytic functions then? –  gary Aug 7 '11 at 6:18
    
I assume you're talking about infinite real differentiability, as being even once complex differentiable entails complex analyticity. I doubt there are any conditions logically weaker than Cauchy-Riemann or equivalents that, when added together with real smoothness, imply holomorphicity, though analysis has a track record of surprising me. –  anon Aug 7 '11 at 6:36

1 Answer 1

There are a lot of equivalent characterizations of holomorphic functions, but the main part of the story is that for a given function $$ f: \mathbb{R}^2 \to \mathbb{R}^2 $$ all you need to know is that $f$ is continuous. Then some equivalent conditions are:

  1. f is complex differentiable (the differential $d f$ exists and is complex linear),

  2. f is locally integrable (edit: I meant to say that f has an antiderivative, this point and point 3 are usually called Morera's theorem),

  3. integrals along closed paths of f are zero,

  4. f is conformal,

  5. Cauchy's integral formula holds for f,

  6. f is represented by a power series around every point,

  7. the Cauchy-Riemann equations hold,

etc. For some of these conditions to make sense you need to know that $f$ is once differentiable as a real function, but none is related to f being smooth (infinitely often differentiable as a real function), so knowing that f is smooth is knowledge that is orthogonal to knowing that f is holomorphic.

Edit: Of course every holomorphic function is smooth. What I meant to say is that being smooth is not "closer to being holomorphic than simply being differentiable" in any sense. AFAIK, there is no statement like "a continuous function is holomorphic if it is smooth + some other condition".

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"Knowing that f is smooth is knowledge that is orthogonal to knowing that f is holomorphic." Well, if $f$ is holomorphic, then it is certainly smooth, no? –  Jesse Madnick Aug 8 '11 at 8:07
    
@Theo: that was probably an inept translation on my part of the German formulation of Morera's theorem, which I now reference explicitly. –  Tim van Beek Aug 8 '11 at 14:50
    
@Jesse Madnick: Sure, I meant to address the question if there is a criterion that one could add to "smooth" to get "holomorphic": AFAIK there isn't one that wouldn't also apply to a function that is only continuous or once real differentiable. –  Tim van Beek Aug 8 '11 at 14:53

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