Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
    
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
1  
@ 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
add comment

2 Answers

up vote 14 down vote accepted

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.

share|improve this answer
3  
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
add comment

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.

share|improve this answer
    
Many thanks for the answer and links. –  Andrey Rekalo Oct 6 '10 at 11:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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