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Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions?

Is there a way to show that holomorphic functions, thought as real functions, are $C^\infty$ , without using any Complex Analysis?

That is,

Let $f:A\rightarrow R^2$ where $A \subseteq R^2$ is open, f is differentiable on A, and satisfies the Cauchy Riemann Equations. Show that f is $C^\infty$ on $A$, without using any Complex Analysis definiton, theorem, or tool.

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    $\begingroup$ Cauchy's integral formula is basically Green's Theorem and that's where it starts. So in a sense, the proof is very much not a complex analysis proof. $\endgroup$
    – Ryan Budney
    Commented Sep 28, 2011 at 17:36
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    $\begingroup$ Your question lacks motivation. Why are you interested in doing complex analysis without doing complex analysis? $\endgroup$
    – Benoît Kloeckner
    Commented Sep 28, 2011 at 17:56
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    $\begingroup$ @Benoit: can't someone just be curious? $\endgroup$
    – euklid345
    Commented Sep 28, 2011 at 19:00
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    $\begingroup$ It's a good question, but maybe more for math.stackexchange.com than here? It's not really a research level question. $\endgroup$
    – Deane Yang
    Commented Sep 28, 2011 at 19:02
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    $\begingroup$ But if you show that holomorphic functions are infinitely differentiable, doesn't that constitute complex analysis regardless of the method? I'm inclined to construe the question as meaning without power series. $\endgroup$
    – Michael Hardy
    Commented Sep 28, 2011 at 22:30

3 Answers 3

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It can be instructive to consider the Cauchy-Riemann equations as an elliptic system of partial differential equations. For elliptic systems, there is a $C^\infty$-regularity result similar to the one known for a single equation (see L. H\"ormander, Linear partial differential operators, Springer, 1963).

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Edit: Avoided by $C^2$ by using integrals.

Edit 2: I believe that the following manipulations of $U,V$ work out right...

If $f$ satisfies the Cauchy-Riemann equations, then write $f(x,y) = (u(x,y), v(x,y))$. Now, let $U,V$ be the antiderivatives of $u,v$ with respect to $x$. Since $u,v \in C^1$, we claim that $U,V \in C^2$ and are harmonic. Using uniform convergence to exchange integral and derivative,

\begin{equation} U_{xx} = D_x^2 \int u \ dx = D_x \int u_x \ dx = u_x \end{equation} \begin{equation} U_{yy} = D_y^2 \int u \ dx = D_y \int u_y \ dx = D_y \int -v_x \ dx = -v_y = -u_x \end{equation}

The computations for $V$ should be similar. Hence $U,V$ are harmonic.

A harmonic function is defined by satisfying the Laplace equation. We can prove that every real harmonic function is smooth by showing that the convolution of any harmonic function by a mollifier is in fact equal to the same harmonic function; this proves that the harmonic function is smooth. If you want a reference, see, for example, L. Craig Evans' textbook "Partial Differential Equations".

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  • $\begingroup$ But to get that the real and imaginary parts are harmonic you need them to be $C^2$, and that wasn't assumed. $\endgroup$
    – Bill
    Commented Sep 28, 2011 at 18:01
  • $\begingroup$ One could always use the Dirac operator. $\endgroup$
    – Paul Siegel
    Commented Sep 28, 2011 at 18:14
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    $\begingroup$ No. Sorry but that doesn't work. If you integrate a $C^1$ function only in the $x$ variable, the result is not necessarily $C^2$, because you can't necessarily differentiate twice in the $y$ variable. You really need to define the notion of a weak derivative and define "weakly harmonic" in terms of weak derivatives. $\endgroup$
    – Deane Yang
    Commented Sep 28, 2011 at 19:04
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    $\begingroup$ Maybe I'm tired. If $U$ is the antiderivative of $u$ with respect to $x$, is not true that $U_{yy} = D_y \int u_y \, dx = D_y \int -v_x \, dx = - v_y$? $\endgroup$
    – Christopher A. Wong
    Commented Sep 28, 2011 at 19:09
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    $\begingroup$ Ah, maybe you're right about that. But maybe you should put the details in your answer? $\endgroup$
    – Deane Yang
    Commented Sep 28, 2011 at 19:14
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This already got asked, and I answered it: Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions?

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  • $\begingroup$ You are right! I'm sorry. Should I ask the question to be closed as duplicate? $\endgroup$
    – Bill
    Commented Sep 28, 2011 at 19:44

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