The Cauchy-Riemann equations and analyticity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:13:57Z http://mathoverflow.net/feeds/question/41212 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41212/the-cauchy-riemann-equations-and-analyticity The Cauchy-Riemann equations and analyticity Andrey Rekalo 2010-10-05T21:54:05Z 2010-10-06T12:58:10Z <p>I would be interested to learn if the following generalization of the classical <a href="http://en.wikipedia.org/wiki/Looman%25E2%2580%2593Menchoff_theorem" rel="nofollow">Looman-Menchoff theorem</a> is true. </p> <hr> <p>Assume that the function $f=u+iv$, defined on a domain $D\subset\mathbb{C}$, is such that </p> <ol> <li>$u_x$, $u_y$, $v_x$, $v_y$ exist almost everywhere in $D$.</li> <li>$u$, $v$ satisfy the Cauchy-Riemann equations almost everywhere in $D$.</li> <li>$f=f(x,y)$ is separately continuous (in $x$ and $y$) in $D$.</li> <li>$f$ is locally integrable.</li> </ol> <blockquote> <p><strong>Question:</strong> Does it follow that $f$ is analytic <em>everywhere</em> in $D$?</p> </blockquote> <hr> <p><strong>Remark 1.</strong> Condition 3 is essential (take $f=1/z$).</p> <p><strong>Remark 2.</strong> G. Sindalovskiĭ proved analyticity of $f$ under conditions 2-4 when the partial derivatives exist everywhere in $D$, <em>except on a countable union of closed sets of finite linear Hausdorff measure</em> (<a href="http://iopscience.iop.org/0025-5734/56/2/A06" rel="nofollow">link</a>).</p> http://mathoverflow.net/questions/41212/the-cauchy-riemann-equations-and-analyticity/41244#41244 Answer by Piero D'Ancona for The Cauchy-Riemann equations and analyticity Piero D'Ancona 2010-10-06T09:08:37Z 2010-10-06T09:08:37Z <p>See <a href="http://mathoverflow.net/questions/34763/continuous-holomorphic-on-a-dense-open-holomorphic/34955#34955" rel="nofollow">this</a> 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.</p> <p>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 <a href="http://www.springerlink.com/content/a375r163l1x65153/" rel="nofollow">here</a> for a different proof of his result.</p> http://mathoverflow.net/questions/41212/the-cauchy-riemann-equations-and-analyticity/41268#41268 Answer by Nate Eldredge for The Cauchy-Riemann equations and analyticity Nate Eldredge 2010-10-06T12:58:10Z 2010-10-06T12:58:10Z <p>No.</p> <p>Let $c$ be the <a href="http://en.wikipedia.org/wiki/Cantor_function" rel="nofollow">Cantor function</a> 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.</p> <p>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.</p>