Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Possible Duplicate:
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.

share|cite|improve this question

marked as duplicate by Deane Yang, Andrés 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
I'm taking an undergrad complex analysis class, and I felt that this result (that holomorphic functions are infinitely differentiable) was surprising. I was curious whether it could be proved without complex variable theory. – Bill Sep 28 '11 at 18:20
As the answers below, there are ways to do this without complex analysis, but they are not easier to learn. The usual proof, using the Cauchy integral formula, is both miraculous and elegant. I suggest you try to live with that for now. – Deane Yang Sep 28 '11 at 18:50
@Benoit: can't someone just be curious? – euklid345 Sep 28 '11 at 19:00

3 Answers 3

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).

share|cite|improve this answer

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".

share|cite|improve this answer
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
Or just use "weakly harmonic". Proof given by Christopher Wong still works as is. – Deane Yang Sep 28 '11 at 18:35
Yeah, using weak notions works. I've amended my solution to make that idea more concrete (since the OP said he's an undergrad). – Christopher A. Wong Sep 28 '11 at 18:45
But what do you mean by the antiderivative of a function of two variables? – Deane Yang Sep 28 '11 at 18:47

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?

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.