The Cauchy-Riemann equations and analyticity - MathOverflow

The Cauchy-Riemann equations and analyticity

2010-10-05T21:54:05Z

I would be interested to learn if the following generalization of the classical Looman-Menchoff theorem is true.

Assume that the function $f=u+iv$, defined on a domain $D\subset\mathbb{C}$, is such that

$u_x$, $u_y$, $v_x$, $v_y$ exist almost everywhere in $D$.
$u$, $v$ satisfy the Cauchy-Riemann equations almost everywhere in $D$.
$f=f(x,y)$ is separately continuous (in $x$ and $y$) in $D$.
$f$ is locally integrable.

Question: Does it follow that $f$ is analytic everywhere in $D$?

Remark 1. Condition 3 is essential (take $f=1/z$).

Remark 2. G. Sindalovskiĭ proved analyticity of $f$ under conditions 2-4 when the partial derivatives exist everywhere in $D$, except on a countable union of closed sets of finite linear Hausdorff measure (link).

Answer by Piero D'Ancona for The Cauchy-Riemann equations and analyticity

2010-10-06T09:08:37Z

See this related question. Denjoy proved that there exist a continuous function $f$ on the unit square and a continuous curve $\gamma$, which is the graph of a continuous function, such that $f$ is holomorphic on the square minus $\gamma$ but not on the whole square. Thus $f$ satisfies the CR equations almost everywhere, and actually on the whole square minus the support of $\gamma$, but not on the whole square.

Thus the general answer to your question seems to be a solid no. Using the postive parts of Denjoy's result, one can imagine to answer in the affirmative if the set where CR fails is a countable union of curves with sufficiently nice behaviour, but it seems difficult to do better than Sindalovskii. See also here for a different proof of his result.

Answer by Nate Eldredge for The Cauchy-Riemann equations and analyticity

2010-10-06T12:58:10Z

No.

Let $c$ be the Cantor function on $[0,1]$, so that $c$ is continuous, $c' = 0$ a.e., but $c$ is not constant. Then take $u(x+iy) = v(x+iy)=c(x)c(y)$. We have $u_x=u_y=v_x=v_y=0$ a.e. so the Cauchy-Riemann equations are trivially satisfied, and $f(z)=u(z)+iv(z)$ is bounded and continuous on the unit square, but certainly not analytic.

Almost everywhere differentiability is almost never the right condition for solutions to a PDE. A better condition would be to have $u,v$ in some Sobolev space.