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

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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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 2011 at 17:36
Your question lacks motivation. Why are you interested in doing complex analysis without doing complex analysis? – BenoĆ®t Kloeckner Sep 28 2011 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. – Guillermo Tochi Sep 28 2011 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 2011 at 18:50
@Benoit: can't someone just be curious? – euklid345 Sep 28 2011 at 19:00

## closed as exact duplicate by Deane Yang, Andres Caicedo, S. Carnahan♦Sep 29 2011 at 16:16

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. – Guillermo Tochi Sep 28 2011 at 18:01 One could always use the Dirac operator. – Paul Siegel Sep 28 2011 at 18:14 Or just use "weakly harmonic". Proof given by Christopher Wong still works as is. – Deane Yang Sep 28 2011 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 2011 at 18:45 But what do you mean by the antiderivative of a function of two variables? – Deane Yang Sep 28 2011 at 18:47