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As we all know, all the basic properties of holomorphic functions (i.e. functions which are differentiable in the complex sense) can be deduced from Cauchy's formula. Moreover, Cauchy's formula itself can be viewed as a rather simple consequence of the Green-Riemann formula, provided that the holomorphic function you have at hand is assumed to have a continuous derivative.

Of course, Cauchy's formula holds without assuming continuity of the derivative, and it yields continuity of the derivative and much more since it implies power series expansion. But Cauchy's formula (or, if you prefer, Cauchy's theorem) without assuming continuity of the derivative is a rather subtle thing, and this gives a rather "indirect" proof of the fact that holomorphic functions are in fact $\mathcal C^1$.

So my question is the following: does anybody know a direct proof of the fact that if a function $f$ defined on an open subset of $\mathbb C$ is differentiable in the complex sense, then its derivative $f'$ is continuous?

I'm pretty sure I am not the first one and will not be the last one to ask this question, at least for himself (or herself). So please feel free to close it if it has indeed been asked previously on this site.

Edit. Perhaps I should say a few more words about what I mean by a "direct proof". Anything that relies in one way or another to Cauchy's formula or Cauchy's theorem is not considered as a direct argument. A direct proof should somehow establish "from scratch", or "from very basic principles" that holomorphic implies $\mathcal C^1$.

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  • $\begingroup$ $f$ is an analytic function on the open subset. If a complex function is analytic on the open subset, it is infinitely differentiable on the subset. Thus, $f^{'}$ is differentiable, and thus continuous. $\endgroup$
    – user122283
    Feb 20, 2014 at 23:48
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    $\begingroup$ @SanathDevalapurkar Thanks for the comment; but this does not answer the question at all. $\endgroup$
    – Etienne
    Feb 20, 2014 at 23:51
  • $\begingroup$ @Etienne I am not sure if I understand your question correctly. If one considers $f: \mathbb C \to \mathbb C$ as a function from $f: \mathbb R^2 \to \mathbb R^2$ then if $f$ is (complex) differentiable at a point $x$ its derivative at $x$ is given by the Jacobian of $f$. But in finite dimension every linear map is continuous which would yield what I understand you are asking. But I think I misunderstand your question or I am probably making a fundamental mistake in what I wrote. $\endgroup$ Feb 21, 2014 at 16:29
  • $\begingroup$ @MattN. For every fixed $x$, the derivative at $x$ is of course a continuous linear map; but the map $x\mapsto f'(x)$ has no a priori reason for being continuous. $\endgroup$
    – Etienne
    Feb 21, 2014 at 17:36
  • $\begingroup$ @Etienne Thank you for your comment. Indeed a seemingly not very thought out comment of mine, I'm sorry about that. Especially if one thinks about real functions where the linear map that is the derivative becomes the tangent! $\endgroup$ Feb 21, 2014 at 19:54

3 Answers 3

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There is an efficient and straightforward proof of the Cauchy-Goursat theorem, by a `lion-hunting' argument based only on differentiability at every point in $\Omega$, available at this link.

In an honors undergraduate analysis class that has dealt with path integrals, one can use it to provide an account of the Cauchy integral formula, analyticity and the classification of zeros and poles in only a couple of weeks.

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  • $\begingroup$ Thanks for the reference. The proof is indeed nice; but this does not answer my question. The point is that if you use something like Cauchy's formula to prove that holomorphic implies $\mathcal C^1$, this is not a "direct" proof. $\endgroup$
    – Etienne
    Feb 21, 2014 at 15:17
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    $\begingroup$ I appreciate that I didn't answer the question you asked, and I should have mentioned it. A point I'd like to stress, though, is that proving the Cauchy integral formula without continuity of the derivative does not need to seem nearly as subtle and complicated as the roundabout approach taken in standard books suggests. In particular, it does not require a 2D integration theory, which arguably makes this approach considerably less complicated than one invoking Green's theorem. $\endgroup$
    – Bob Pego
    Feb 28, 2014 at 19:49
  • $\begingroup$ I agree that the Green-Riemann approach is perhaps not the simplest one. still, I think it would be really nice to have a direct proof that holomorphic implies $\mathcal C^1$; perhaps one which wouldn't use line integrals at all... $\endgroup$
    – Etienne
    Mar 2, 2014 at 13:53
  • $\begingroup$ @Etienne an underappreciated difficulty is that Green-Riemann is typically proved (in undergrad classes) for domains that are the inside of a rectifiable Jordan curve. I agree that it's good for building intuition though. The proof given here is really elementary. $\endgroup$ Jun 24, 2021 at 6:59
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Does this link help you? Is is entitled: "A TOPOLOGICAL PROOF OF THE CONTINUITY OF THE DERIVATIVE OF A FUNCTION OF A COMPLEX VARIABLE": https://projecteuclid.org/download/pdf_1/euclid.bams/1183522992

In the demonstration no use is made of complex integration, only results from differential topology and Rouche's theorem to arrive to: If f'(z) exists for all z in an open connected set E of the complex plane, then f'(z) is continuous in E.

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  • $\begingroup$ This looks fine, thanks! $\endgroup$
    – Etienne
    Mar 2, 2017 at 20:50
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This does seem to require the Cauchy-Goursat theorem. And, really, there is some strong intuition in this. If you have an open region $\Omega$ where $f$ is differentiable, then $$ \mu(R)=\oint_{R}f(z)dz $$ defines a finitely-additive set function on solid rectangles $R$ contained in $\Omega$ with the property that $$ \lim_{|S|\rightarrow 0}\frac{\mu(S)}{|S|}=0 $$ for sequences of squares $S$ which contain a common point $z$ (Here $S$ denotes the usual area measure of $S$.) Intuitively, there's just nothing such a complex measure could be but 0 when the derivative of $\mu$ with respect to usual area measure is 0 everywhere. Everything after that is contour integral magic a la Cauchy. I don't think you can get away from the measure-theoretic aspect of this.

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