8
$\begingroup$

In Chapter 11 of Rudin's RCA, a harmonic function is defined to be a complex continuous function $u$ on a plane open set such that the Laplacian of $u$, i.e. the sum of its pure second-order partial derivatives $$u_{xx}+u_{yy}$$ is $0$. I was wondering if this was missing a condition, namely that $u_{xy}=u_{yx}$, because I was unable to show the following without this assumption.

For every harmonic function $u$ whose domain includes the image of a holomorphic function $f$ in a plane open set $\Omega$, the composite $u\circ f$ is harmonic in $\Omega$.

I attempted to show this by just brute calculation, and what remained was $u_{xy}-u_{yx}$ with some partials of $f$ multiplied on the outside.

Of course, Rudin later shows that harmonic functions have continuous partial derivatives of all orders because the real-valued ones are locally real parts of holomorphic functions, but I think his proof relies on a certain composite of functions being harmonic...

$\endgroup$
3
  • 1
    $\begingroup$ Theorem 11.9 proves that real-valued harmonic functions are locally the real part of holomorphic functions [and consequently that all harmonic functions are smooth, even real-analytic]. Well, those that are defined and continuous on the closed unit disk, but the general case is proved via totally straightforward translation and scaling. The proof of theorem 11.9 doesn't even require $u_{xy}$ or $u_{yx}$ to exist. After theorem 11.9, you're home and dry, all partial derivatives of a harmonic function exist and are continuous. $\endgroup$ Aug 24, 2016 at 21:20
  • $\begingroup$ @DanielFischer But isn't translation and scaling a composition? Sorry, this is my first time studying this material... $\endgroup$
    – user363464
    Aug 24, 2016 at 21:21
  • 1
    $\begingroup$ Yes, but with very simple functions, for which the matter is easy. Let $u$ be a real-valued harmonic function on the disk $D_r(z_0)$. Consider the function $v\colon D_2(0) \to \mathbb{R}$ given by $z \mapsto u(z_0 + r/2\cdot z)$. Note that $v_x(z) = \frac{r}{2} u_x(z_0 + r/2\cdot z)$, and similar for $v_{xx},\, v_y,\, v_{yy}$. $\endgroup$ Aug 24, 2016 at 21:27

1 Answer 1

2
$\begingroup$

You don't need to assume equality of the mixed partials in the definition. Here is a rough outline of one proof:

First, show the maximum principle for harmonic functions. Note that this implies that two functions which are continuous on a closed disk and harmonic in the disk's interior, and equal on the boundary, are equal in the interior as well.

Second, note that the Dirichlet problem on a disk with continuous boundary value has an explicit solution given by the separation of variables method in polar coordinates, and this solution is $C^\infty$.

Combine the above to show that any harmonic function is $C^\infty$ everywhere. In particular, mixed partials, being continuous, are equal.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .