Show that holomorphic functions are infinitely differentiable without complex analysis [duplicate]

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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marked as duplicate by Deane Yang, Andrés E. Caicedo, S. Carnahan♦Sep 29 '11 at 16:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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. – Ryan Budney Sep 28 '11 at 17:36
Your question lacks motivation. Why are you interested in doing complex analysis without doing complex analysis? – Benoît Kloeckner Sep 28 '11 at 17:56
@Benoit: can't someone just be curious? – euklid345 Sep 28 '11 at 19:00
It's a good question, but maybe more for math.stackexchange.com than here? It's not really a research level question. – Deane Yang Sep 28 '11 at 19:02
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. – Michael Hardy Sep 28 '11 at 22:30

3 Answers

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,

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

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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But to get that the real and imaginary parts are harmonic you need them to be $C^2$, and that wasn't assumed. – Bill Sep 28 '11 at 18:01
One could always use the Dirac operator. – Paul Siegel Sep 28 '11 at 18:14
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. – Deane Yang Sep 28 '11 at 19:04
Maybe I'm tired. If $U$ is the antiderivative of $u$ with respect to $x$, is not true that $U_{yy} = D_y \int u_y \, dx = D_y \int -v_x \, dx = - v_y$? – Christopher A. Wong Sep 28 '11 at 19:09
Ah, maybe you're right about that. But maybe you should put the details in your answer? – Deane Yang Sep 28 '11 at 19:14

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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You are right! I'm sorry. Should I ask the question to be closed as duplicate? – Bill Sep 28 '11 at 19:44