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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.

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.

Steven Gubkin
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  • I would love to see an answer based on physical intuition: if $u, v$ satisfy the Cauchy-Riemann equations then $(u, -v)$ is a conservative and divergence-free vector field (such as the flow of an incompressible fluid). Is there a physical reason one expects such things to be $C^{\infty}$? – Qiaochu Yuan Apr 27 '11 at 18:56
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    The main problem, I think, is that it's not so easy to say what "twice differentiable" means geometrically. – Deane Yang Apr 27 '11 at 19:00
  • @Deane ya, I have been trying to draw pictures, but it is very hard. Drawing a pair of vectors at each point of R^2 to show what the derivative is there, being second differentiable should be about these pairs of vectors varying smoothly. I am still holding out hope that their might be a nice picture though. Maybe that a second differentiable function should look locally like a quadratic form. – Steven Gubkin Apr 27 '11 at 19:35
  • You can describe the property of a function being $C^\infty$ without actually using differentiation. It is equivalent to its finite differences being locally bounded (to each given order). In this case, what could be used as a bound for these finite differences other than a multiple of the maximum of $\Vert f\Vert$ in a region? We know this works, as the differences are equal to an integral over $f$ which, in turn, is bounded by a multiple of $\Vert f\Vert_\infty$. But, you are effectively asking to step straight to the bound without the in-between step involving the integral. – George Lowther Apr 27 '11 at 21:43
  • I think you could do this using summation rather than integration but, it would still essentially be the same proof. – George Lowther Apr 27 '11 at 21:43
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    Here's another non-answer. You might say that the most magical part of the following argument is the Fourier series.

    If $f(z)=f(re^{i\theta})$ is defined in $a\lt r\lt b$ then without using much smoothness (in the two real variable sense) you can write $f(z)=\Sigma_{n\in \mathbb Z}\ c_n(|z|)z^n$. Here $c_n(|z|)$, being the rotational average of $f(z)z^{-n}$, is differentiable in the complex sense if $f(z)$ is. Clearly a function of $|z|$ alone which is differentiable in the complex sense has derivative zero. Therefore $f(z)$ is the sum of a Laurent series.

     – Tom Goodwillie Apr 29 '11 at 12:23
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    I think your question has two distinct goals which are not necessarily compatible: to geometrically understand the unexpected regularity of complex differentiable functions and to establish that regularity without using integration. Given that so much of complex analysis is organized around the Cauchy integral formula, I think the right strategy for the first goal is to try to geometrically understand the role of integration. I don't completely understand the magic of complex analysis either, but I think it's because I don't completely understand the magic of Green's theorem. – Paul Siegel Mar 03 '12 at 12:17
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    @Paul - I feel like I have a pretty good geometric understanding of Green's theorem, and hence of the Cauchy integral formula. It is really just about chopping a region up into little tiny oriented rectangles, where your function essentially becomes linear. The rectangles are oriented so that edges of interior rectangles "cancel out". The picture is in Spivak's book. The part which is more troublesome to me is differentiation under the integral sign giving rise to infinite differentiability. – Steven Gubkin Mar 03 '12 at 16:39
  • Hi @StevenGubkin, Did you write any summary on undergraduate/graduate complex analysis ? What would you change/add in Visual Complex Analysis from Needham ? I'm asking because I saw on your profile that you are interested in improving undergraduate teaching, as I am. – Julien__ Nov 20 '15 at 13:57
  • @Julien__ I would certainly include a whole lot more pictures of phase plots, phase plots shaded by modulus, 3d graphs of modulus colored by phase, phase plots on the Riemann sphere, etc. Ideally these should be interactive. I will get there someday. Hopefully in the next year... – Steven Gubkin Nov 20 '15 at 17:33
  • Tools like this one should feature prominently: https://people.math.osu.edu/fowler.291/phase/ – Steven Gubkin Nov 20 '15 at 17:34
  • @StevenGubkin great webpage ! About phase portraits, do you know Visual Complex Analysis by Elias Wegert ? What's your email adress, mathoverflow might not be the best place to discuss this. I suppose it is something like name@university.edu ? – Julien__ Nov 21 '15 at 11:37
  • @Julien__ : lastname.1@osu.edu – Steven Gubkin Nov 21 '15 at 20:49

7 Answers7

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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–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.

Edit: After reading Gerald Edgar's answer, I should comment that I'm holding the Third Edition of Ahlfors.

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 $(z - z_0) f(z)$.

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.

Ryan Reich
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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 $\{0\}$.

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 ;)

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...

OK, certainly not straightforward technology but nice.

Stefan Waldmann
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    I agree - a very nice proof. It seems like it might unravel to the usual proof however: I think the integration is hidden in the definition of a distribution, and I am guessing Cauchy's integral formula is connected to 1/z being a fundamental solution. These are just my sneaking suspicions though. I should really break this argument down and see if it ends up being "the same" as the one I sketch in my question. – Steven Gubkin Apr 27 '11 at 18:52
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    Your suspicions appear correct to me. – Deane Yang Apr 27 '11 at 18:59
  • Dear Steven, dear Deane: Of course, one has to invest in some sense much more than the usual and almost elementary Cauchy-like proof. I agree with you completely. Also, you asked for something more geometric which this is certainly not... :( – Stefan Waldmann Apr 28 '11 at 12:22
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    No reason to frown dude! It is a nice proof, and I think looking from a more general and abstract perspective is also good. My first exposure to distributions was last year in a course which kind of culminated in the Malgrange-Ehrepries theorem (at east I thought of that as the culmination, after that it got a bit fuzzy for me). So I am glad to see your answer - it is nice to see an application of these ideas in a natural context. – Steven Gubkin Apr 28 '11 at 14:00
  • I would bet a pretty large amount of money that the not-too-hard-proof in distribution theory relies heavily on Fourier transforms or convolutions. A nice proof, though. – Johannes Ebert Apr 29 '11 at 21:08
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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): Weyl's Lemma, 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.

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 Cauchy–Kowalevski theorem but I don't see how this is related here.

j.c.
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Dirk
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  • Thanks for fixing the links, jc! Since this always happens to my links, it seems that I have a problem with the encoding... – Dirk Apr 29 '11 at 13:10
  • I assume you're just copying and pasting the URLs just like me? It might be worth starting a thread on tea.mathoverflow.net about it to see if this is a known bug. – j.c. Apr 29 '11 at 13:24
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    I sometimes have the same problem, so I usually delete the URL that is automatically placed at the bottom of the post by the javascript tool, and replace it by pasting again. – S. Carnahan Apr 29 '11 at 14:42
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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?

Gerald Edgar
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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 theory 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.

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...

leo
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godelian
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This appears to be the earliest "topological" proof:

A TOPOLOGICAL PROOF OF THE CONTINUITY OF THE DERIVATIVE OF A FUNCTION OF A COMPLEX VARIABLE BY ROBERT L. PLUNKETT 1958

http://www.ams.org/journals/bull/1959-65-01/S0002-9904-1959-10251-2/S0002-9904-1959-10251-2.pdf

G.T. Whyburn himself (whose research was foundational to the result) gives his own proof, allegedly inspired by Connell and Porcelli:

Developments in topological analysis, 1961 http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf

pereger
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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.

Adam Hughes
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  • Having never studied PDE's, I don't know whether it is more or less geometric to prove that harmonic functions are analytic than that holomorphic functions are. Are you sure this isn't just restating the problem? – Ryan Reich Apr 27 '11 at 18:38
  • I agree with Ryan. Spell out your proof that derivatives of Harmonic functions are Harmonic, and we might be onto something. – Steven Gubkin Apr 27 '11 at 18:48
  • Just that mixed partials commute, so apply the Laplacian to $u_x$, you get $u_{xxx}+u_{xyy}=}{\partial\over\partial x}(u_{xx}+u_{yy})={\partial\over\partial x}0=0$ – Adam Hughes Apr 27 '11 at 18:52
  • last line should be $u_{xxx}+u_{xyy}={\partial\over\partial x}(u_{xx}+u_{yy})={\partial\over\partial x}(0)=0 $ – Adam Hughes Apr 27 '11 at 18:54
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    Doesn't that assume that the derivatives in question exist? – Steven Gubkin Apr 27 '11 at 18:55
  • Indeed it does, but I remember something about harmonic functions being smooth just because they satisfy the Laplace equation as a general fact not restricted to two dimensions, so it wouldn't be from the complex analysis proof. – Adam Hughes Apr 27 '11 at 20:02
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    The derivatives do exist in the sense of distributions and you can show that any harmonic function ($\Delta f=0$ in the sense of distributions) is infinitely differentiable. This works on any Riemann manifold, so it is not a complex analysis proof. It is also very useful. For example you can construct harmonic functions by minimising the Dirichlet integral $\int\Vert\nabla f\Vert^2dV$ and it is not obvious that the result is even once comtinuously differentiable. Luckily, you are guaranteed that it is $C^\infty$. However, even the definition of distributions involves integration. – George Lowther Apr 27 '11 at 20:17
  • So, it seems that integration is hard to a avoid. It is just too basic a concept to do without. – George Lowther Apr 27 '11 at 20:18
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    Asking one of my analysis friends, the proof of harmonic functions being smooth uses a similar argument - you use the Poisson kernel to represent the harmonic function as an integral around the boundary of a region, and then differentiate under the integral sign. – Steven Gubkin Apr 27 '11 at 20:19
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    @George - Maybe avoiding integration is too much to ask. I just can't help feeling like I have been tricked with this proof - it seems to buy to much for what it brings to the table. I don't know, it is more about my personal psychology than it is about mathematics. In talking to a number of my fellow grad students, I find that I am not alone in feeling that there is something magical about the proof. My goal when learning math is always to kill all of the magic, and make everything seem totally reasonable. – Steven Gubkin Apr 27 '11 at 20:23
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    @Steven: I think I could concoct a proof using summation rather than integration. That is, look at the function on a discrete grid and bound it's finite differences (to all orders) in terms of a sum of f over grid points. This would still be morally similar to the integration proof, while avoiding any actual integration. – George Lowther Apr 27 '11 at 20:31
  • ...and it would still be missing the geometric motivation. – George Lowther Apr 27 '11 at 20:38
  • Haha, I think the "magic" is just what makes it so nice. Classical complex analysis is a beautiful subject and has a lot to bring to the table in the ways of powerful tools. I'm certain you'll never find success at removing all the magic from things though, things are far too complicated for that. – Adam Hughes Apr 27 '11 at 21:17
  • And I should add that I think that perhaps the so-called "magic" is less mind-boggling as you get used to it. Just because you don't see some masterful underlying reason which trivializes things now, doesn't mean you won't later on. And if you don't, then I don't think it's that much of a loss, you just move on. – Adam Hughes Apr 27 '11 at 21:21
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    @Steven: btw the simpler reason why harmonic functions are smooth, as I seee it, is the mean value property, that one can express as invariance under convolution with characteristic functions of balls. Convoultion regularizes, and that's it. Details here http://en.wikipedia.org/wiki/Harmonic_function#The_mean_value_property – Pietro Majer May 19 '11 at 04:12