# The Cauchy–Riemann equations and analyticity

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

1. $u_x$, $u_y$, $v_x$, $v_y$ exist almost everywhere in $D$.
2. $u$, $v$ satisfy the Cauchy–Riemann equations almost everywhere in $D$.
3. $f=f(x,y)$ is separately continuous (in $x$ and $y$) in $D$.
4. $f$ 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).

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just a doubt about assumption 2: what further assumption on $u_x$ $u_y$ $v_x$ $v_y$ are implicitly taken, in order that the CR equations make sense? – Pietro Majer Oct 6 '10 at 8:07
@ Pietro Majer: I'm not sure I follow. The derivatives are assumed to exist a.e. in $D$. I'm interested in locally integrable and separately continuous functions $u$, $v$ which satisfy the CR equations a.e. Am I missing something? – Andrey Rekalo Oct 6 '10 at 11:33
Just this: in order to have $u_{xx}$ classically defined at $(x_0,y_0)$ we need $u_x(x,y_0)$ to exist for all $x$ in a nbd of $x_0,$ and analogous condition for the other second order prtial derivatives. So I was wondering if these were implicitly assumed in 2 or there was a weaker sense. Also, a doubt about whether $u_{xy}=u_{yx},$ was assumed too, since that should not be ensured by Schwarz theorem, in this generality. Anyway, no matter, thanks to Nate's answer! – Pietro Majer Oct 6 '10 at 14:09

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.

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It's actually sufficient to require that $u,v$ solve the equation in distribution sense – Piero D'Ancona Oct 6 '10 at 13:21
Nice example! Thanks. – Andrey Rekalo Oct 6 '10 at 13:25
Your second comment is absolutely spot on, Nate. – Peter Luthy Oct 6 '10 at 18:55

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.

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Many thanks for the answer and links. – Andrey Rekalo Oct 6 '10 at 11:35