Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:56:36Z http://mathoverflow.net/feeds/question/63205 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? Steven Gubkin 2011-04-27T18:13:51Z 2012-11-01T23:20:04Z <p>The title pretty much says it all. I am revisiting complex analysis for the first time since I "learned" some as an undergraduate. I am trying to wrap my head around why it should be the case that a function which is differentiable once should be differentiable twice. I know a proof (Use Cauchy integral formula and differentiate under the integral sign), but that proof doesn't do a whole lot to explain the magic.</p> <p>Say you had never heard of complex numbers before, and someone told you that you had a function $f: \mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y) = (u(x,y), v(x,y))$, which locally looks like a rotation/expansion (i.e. it satisfies the Cauchy-Riemann equations everywhere). Then why on Earth should this function be $C^\infty$? I would love to see a picture, or just a proof that makes this feel less like a magic trick. I hoped that "Visual Complex Analysis" would help me out here, but this seems to be the one theorem in the book which is not given a geometric motivation.</p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/63209#63209 Answer by Adam Hughes for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? Adam Hughes 2011-04-27T18:23:46Z 2011-04-27T18:23:46Z <p>Well it seems to me the Cauchy-Riemann equations say quite a lot. After all once you know the real and imaginary parts are harmonic, then you know their derivatives are as well, and $$f'(z)=u_x(z)+iv_x(z)$$ so you can do it again. I don't think this is boot-strapping it, but if anyone greater on the complex food-chain cares to disagree, I'm likely retreat quickly at their word.</p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/63210#63210 Answer by Stefan Waldmann for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? Stefan Waldmann 2011-04-27T18:24:51Z 2011-04-27T18:24:51Z <p>well, this is probably much more magic than the usual proof. Nevertheless, I like it a lot: coming for linear PDE one encounters the notion of a hypoelliptic differential operator (say with constant coefficients). Then in the theory of distributions it is not too hard to show that hypoellipticity is equivalent to have a fundamental solution with singular support being just ${0}$.</p> <p>For the $\overline{\partial}$ operator it is then easy to see that $\frac{1}{z}$ is a fundamental solution. In some sense here you only have to differentiate once ;)</p> <p>Now why is this nice. Well, having a hypoelliptic diffop $D$, the solutions to the homogeneous equation $Du = 0$ in the very very weak sense of distributions will yield smooth solutions right away. In this sense, you get along with even much less differentiabily...</p> <p>OK, certainly not straightforward technology but nice.</p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/63395#63395 Answer by Dirk for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? Dirk 2011-04-29T10:27:14Z 2011-04-29T13:00:07Z <p>I asked a colleague who works in complex analysis and he also does not know any "integration free" proof or argument. However, he pointed me to a version free of complex numbers but equivalently astonishing (at least for me): <a href="http://en.wikipedia.org/wiki/Weyl%27s_lemma_%28Laplace_equation%29" rel="nofollow">Weyl's Lemma</a>, stating that any function $u\in L^1_\text{loc}$ satisfying $$\int u\Delta \phi = 0$$ for any test function $\phi$ is $C^\infty$. However, the proof also uses integration in the form of convolutions and hence, integrals are not at all avoided.</p> <p>Another comment: Many theorems about (unsespected) smoothness of solutions of partial differential equation use some integral formula to deduce higher smoothness. One exception is the <a href="http://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem" rel="nofollow">Cauchy–Kowalevski theorem</a> but I don't see how this is related here.</p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/63406#63406 Answer by Gerald Edgar for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? Gerald Edgar 2011-04-29T12:53:58Z 2011-04-29T12:53:58Z <p>When I took complex analysis from Ahlfors in 1970 or so (using his textbook), when we came to the section on this, he made a remark in class. For a long time it was thought this could not be proved without integration. Today proofs without integration are known. But they are much more difficult than the proof in the textbook with integration. He did not give any more explanation. Maybe (?) if there is a more recent edition of his book he added the remark there?</p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/63408#63408 Answer by Ryan Reich for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? Ryan Reich 2011-04-29T13:04:46Z 2011-04-29T13:43:33Z <p>I recalled a comment of Ahlfors at the beginning of Chapter 4 of his book "Complex Analysis" which asserted the existence of such a proof, and it led me to the paper "A proof of the power series expansion without Cauchy's formula" by E. H. Connell and P. Porcelli (Bull. Amer. Math. Soc. 67 (1961), 177-181), where they give just such a proof based on a topological theorem from G. T. Whyburn, "Topological analysis" stating that any holomorphic function is an open mapping. I have not checked the latter source to see that it is independent of integration, but it likely is; I can't get my hands on it at the moment since it is in a book rather than an article. In fact, the Connell&ndash;Porcelli article is actually an announcement of results and so omits some details. There is an article just by Connell, "On properties of analytic functions" (Duke Math. J. 28 1961 73–81) which allegedly gives the details and I haven't looked up.</p> <p><strong>Edit:</strong> After reading Gerald Edgar's answer, I should comment that I'm holding the Third Edition of Ahlfors.</p> <p>A note on the proof. As you might expect, the topological content is that the complex plane has a basis of open balls such that even after removing finitely many points, they are connected. This implies that a function which is continuous on an open set and open away from a point is again open (contrast that with the absolute value function on the real line), and this trivially implies the maximum modulus principle if you know that holomorphic functions are open maps. From there you can just do elementary epsilon-delta reasoning (applied to the "difference quotient function" $(f(z) - f(z_0))/(z - z_0)$) to get that a continuous function which is holomorphic away from a point is also holomorphic at the point. You can replace "continuous" by "bounded" by doing this with $(z - z_0) f(z)$.</p> <p>Alas, it is at this point that the summary paper starts summarizing. However, they do go on to prove (completely) the existence of a power series development, which is stronger than mere twice-differentiability.</p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/63422#63422 Answer by godelian for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? godelian 2011-04-29T14:51:31Z 2011-04-29T14:51:31Z <p>There is another approach to complex analysis that was initiated already by Lagrange but whose main development is due to Weierstrass. While the main application of Cauchy's thery is to prove that analytic functions have power series expansions, for Weierstrass the power series plays a fundamental rôle. It is possible to define an analytic function as a function that admits a convergent power series expansion. Then one sees that there is uniform convergence in disks and the next step is to prove that such a power series has all orders derivatives.</p> <p>There is a simple integration-free proof that a power series can be derived term by term inside its radius of convergence, which already gives you that it is infinitely derivable and the power series for the derivatives have the same radius of convergence. This makes Weierstrass approach possible. Elementary functions like exponential and trigonometric functions can be defined by means of corresponding power series and then one shows its usual properties. Consequences of Cauchy's integral formula like Liouville's theorem or Cauchy's inequality can be recovered in this context without integration by means of Parseval's identity (which only involves real integration). And so on... </p> http://mathoverflow.net/questions/63205/is-there-an-integration-free-proof-or-heuristic-that-once-differentiable-implie/90104#90104 Answer by pereger for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? pereger 2012-03-03T03:33:14Z 2012-03-03T04:01:57Z <p>This appears to be the earliest "topological" proof:</p> <p>A TOPOLOGICAL PROOF OF THE CONTINUITY OF THE DERIVATIVE OF A FUNCTION OF A COMPLEX VARIABLE BY ROBERT L. PLUNKETT 1958</p> <p><a href="http://www.ams.org/journals/bull/1959-65-01/S0002-9904-1959-10251-2/S0002-9904-1959-10251-2.pdf" rel="nofollow">http://www.ams.org/journals/bull/1959-65-01/S0002-9904-1959-10251-2/S0002-9904-1959-10251-2.pdf</a></p> <p>G.T. Whyburn himself (whose research was foundational to the result) gives his own proof, allegedly inspired by Connell and Porcelli:</p> <p>Developments in topological analysis, 1961 <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf" rel="nofollow">http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf</a></p>