Approximately holomorphic functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:53:01Z http://mathoverflow.net/feeds/question/19772 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19772/approximately-holomorphic-functions Approximately holomorphic functions Johan 2010-03-29T21:23:32Z 2010-04-01T16:44:49Z <p>In real analysis one can define something known as the approximative derivative of a function. <a href="http://eom.springer.de/a/a012850.htm" rel="nofollow">See here eg</a> Roughly speaking one asks that the limit of the difference quotient exists as long as h goes to zero while only taking values in some subset that is sufficiently dense. </p> <p>Does anyone know if this concept has been studied for complex-valued valued functions of a complex variable? The basic definition should go through without problem so it should make sense to speak of an approximately holomorphic function as one that has an approximate complex derivative at every point of some open set. It would be interesting how much of the classical complex analysis one could generalize. Even knowing whether there exists functions that are approximately holomorphic but not holomorphic in the normal sense would be interesting. </p> http://mathoverflow.net/questions/19772/approximately-holomorphic-functions/19782#19782 Answer by Harald Hanche-Olsen for Approximately holomorphic functions Harald Hanche-Olsen 2010-03-29T23:23:25Z 2010-03-29T23:23:25Z <p>One interesting class is that of <a href="http://en.wikipedia.org/wiki/Quasiconformal_mapping" rel="nofollow">quasiconformal mappings</a>.</p> http://mathoverflow.net/questions/19772/approximately-holomorphic-functions/19784#19784 Answer by Andy Putman for Approximately holomorphic functions Andy Putman 2010-03-30T00:04:59Z 2010-03-30T00:46:54Z <p>I've got some thoughts, but they should be treated somewhat as speculation. Harald above brought up the notion of quasiconformality. It would not surprise me if the "approximately holomorphic" functions you described were quasiconformal -- the quasiconformality condition is a very soft condition. I would look at the discussions of it in Hubbard's book on Teichmuller theory and Ahlfors's book on quasiconformal mappings and see if you can prove that it holds.</p> <p>If it does hold, then I'm pretty certain that that the quasiconformality constant would be $1$. If it is, then you are in luck -- a famous theorem of Weyl says that quasiconformal mappings with quasiconformality constant $1$ are actually holomorphic (well, maybe you wouldn't call it luck, as you then wouldn't actually have a generalization).</p> <p>This sequence of speculations fits into another important intuition about holomorphic functions, namely that they end up being more differentiable than you might guess a priori. For instance, if $f$ has partial derivatives with respect to $x$ and $y$ and the Cauchy-Riemann conditions hold, then $f$ is automatically infinitely differentiable. You don't even have to assume that the partial derivatives of $f$ are continuous or that $f$ is actually differentiable. The theorem of Weyl I mentioned above is also in this vein, as it says that you can even assume that the partial derivatives of $f$ only exist in the weak sense.</p> http://mathoverflow.net/questions/19772/approximately-holomorphic-functions/20084#20084 Answer by 002 for Approximately holomorphic functions 002 2010-04-01T16:44:49Z 2010-04-01T16:44:49Z <p>Men'shov proved in 1936 that if $f\colon D\to\mathbb C$ is continuous and approximately differentiable outside of a countable set, then it is holomorphic in $D$ <a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=sm&amp;paperid=5370&amp;option_lang=rus" rel="nofollow">1</a> (Russian original with French summary). In the same paper he gives an example, attributed to Lusin, which shows that continuity cannot be dropped entirely even if the function is approximately differentiable at every point. Let $\varphi\colon\mathbb C\to\mathbb C$ be an entire function that tends to $0$ as $z\to\infty$ within any sector <code>$\{|\arg z|&lt;\pi-\epsilon\}$</code> (approximation theory can be used to create such examples). Then $f(z)=z\varphi(1/z)$ has an approximate derivative everywhere in $\mathbb C$, but of course it is not differentiable or even continuous at $0$. </p> <p>Men'shov asked if continuity can be replaced with boundedness. This was answered affirmatively by Telyakovskii 50 years later <a href="http://www.ams.org/mathscinet-getitem?mr=837136" rel="nofollow">2</a>. In fact, boundedness can be weakened to logarithmic integrability. </p> <p>Brodovich <a href="http://www.ams.org/mathscinet-getitem?mr=1241165" rel="nofollow">3</a> proved that an injective function with an approximate derivative at every point is holomorphic. (MR review omits the injectivity assumption). A survey of results in this area was written by Dolzhenko <a href="http://www.ams.org/mathscinet-getitem?mr=1211822" rel="nofollow">4</a>, but it predates the work of Brodovich. </p>