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When I studied complex analysis, I could never understand how once-differentiable complex functions could be possibly be infinitely differentiable. After all, this doesn't hold for functions from R2 to R2. Can anyone explain what is different about complex numbers?

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"After all, this doesn't hold for functions from R2 to R2" ??? is there an easy counterexample? this is new to me, but i'm just an amateur. –  Jason S Jul 24 '10 at 12:54
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or f(x) = x^q where q is a noninteger bigger than 1. –  Eric O. Korman Jul 24 '10 at 14:11
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This is new to me too, perhaps a way to gain intuition is too think how every complex number can be put in exponential form, re^xi. It might come from the fact that e^x is infinitely differentiable. –  Jonathan Fischoff Jul 24 '10 at 18:34
    
Look at Needham's Visual Complex Analysis for an excellent intuitive text. –  lhf Apr 15 '11 at 18:51
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Thanks, @AndresCaicedo. Let's fix that. Take your favorite non-differentiable continuous function and integrate it, and that will be once differentiable but not twice differentiable. To make things concrete, start with the function which is 0 for negative x, and x for positive x. An integral of this is the function which is 0 for negative x and $x^2/2$ for positive x. –  Noah Snyder Jan 24 at 4:09
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6 Answers 6

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As Akhil mentions, the keyword is elliptic regularity. Since I don't know anything about this, let me just say some low-level things and maybe they'll make sense to you.

A differentiable function $f : \mathbb{R} \to \mathbb{R}$ can be thought of as a function which behaves locally like a linear function $f(x) = ax + b$. So, very roughly, it is a collection of tiny vectors which fit together. These tiny vectors can, however, fit together in a very erratic manner. That's because since you only have to fit one vector to the two vectors that are its neighbors, there is a lot of room for bad behavior.

A differentiable function $f : \mathbb{C} \to \mathbb{C}$ has to satisfy a much more stringent requirement: locally, it has to behave like a linear function $f(z) = az + b$ where $z, a, b$ are complex, which is a rotation (and scale, and translation). So, very roughly, it is a collection of tiny rotations which fit together. Now one rotation has a continuum of neighbors to worry about, and it becomes much harder for erratic behavior to persist.

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I really like this answer, but maybe that is because I don't much know much about complex analysis, just a lot about euclidean transformations, and how they relate to complex numbers. +1 –  Jonathan Fischoff Jul 27 '10 at 21:55
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The existence of a complex derivative means that locally a function can only rotate and expand. That is, in the limit, disks are mapped to disks. This rigidity is what makes a complex differentiable function infinitely differentiable, and even more, analytic.

For a complex derivative to exist, it has to exist and have the same value for all ways the "h" term can go to zero in (f(z+h) - f(z))/h. In particular, h could approach 0 along any radial path, and that's why an analytic function must map disks to disks in the limits.

By contrast, an infinitely differentiable function of two real variables could map a disk to an ellipse, stretching more in one direction than another. An analytic function can't do that.

A smooth function of two variables could also flip a disk over, such as f(x, y) = (x, -y). An analytic function can't do that either. That's why complex conjugation is not an analytic function.

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Ok, so you say a necessary condition for a function to be analytic is to preserve disks in some sense. Is this also sufficient. Moreover, does this work for complex Banach Spaces as well? –  Freeze_S Jan 24 at 0:44
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The proofs I have seen derive this as a corollary of Cauchy's integral formula. Look at the difference quotient as an integral, play around with it, and you get that it converges to what you'd get if you differentiated under the integral sign.

Note that since harmonic functions also satisfy a similar integral equation, they are also infinitely differentiable in the same way (this also follows since they are real and imaginary parts of holomorphic functions).

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The fact that harmonic functions are infinitely differentiable is a special case of elliptic regularity: solutions to elliptic equations $P(u)=0$ are $C^\infty$. –  Akhil Mathew Jul 24 '10 at 13:31
    
I've seen the formal proof, I just can't "get it" intuitively –  Casebash Jul 24 '10 at 19:40
    
Holomorphic case too: The Cauchy-Riemann equations are elliptic. –  timur Dec 24 '10 at 6:23
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When one uses the complex plane to represent the set of complex numbers ${\bf C}$,

$z=x+iy$

looks so similar to the point $(x,y)$ in ${\bf R}^2$.

However, there is a difference between them which is not that obvious. The linear transformation in ${\bf R}^2$, can be represented by a $2\times 2$ matrix as long as one chooses a basis in ${\bf R}^2$, and conversely, any $2\times 2$ matrix can define a linear transformation by using the matrix multiplication $A(x,y)^{T}$.

On the other hand, the linear transformation on $\bf C$ is little different. Let $f:{\bf C}\to{\bf C}$ where $f(z)=pz$, $p \in{\bf C}$. If one writes $p=a+ib$ and $z=x+iy$, this transformation can be written as

$$ \begin{bmatrix} x\\ y \end{bmatrix}\to \begin{bmatrix} a &-b\\ a &b \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} $$

when one sees it as in the complex plane. Hence, not all the matrices can define a linear transformation $f:\bf C\to C$.


The derivative, which can be regarded as a "linear transformation", is also different for $f:{\bf R}^2\to {\bf R}^2$ and $f:\bf C\to C$. In the real case

$$ f(\begin{bmatrix} x\\ y \end{bmatrix})= \begin{bmatrix} f_1(x,y)\\ f_2(x,y) \end{bmatrix} $$

$f_1$ and $f_2$ are "independent" for the sake of $f$ being differentiable.

While in the complex case $f_1$ and $f_2$ have to satisfy the Cauchy-Riemann equations.

Edit: The relationship between $f:{\bf R}^2\to{\bf R}^2$ and $f:{\bf C}\to{\bf C}$ is also discussed here.

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Interesting, although the derivative is not a linear transformation from R2->R2, but from f(R2)->f(R2) –  Casebash Apr 15 '11 at 23:14
    
By definition, the derivative of $f:E\subset{\bf R}^2\to{\bf R}^2$ is indeed a linear transformation $Df:{\bf R}^2\to{\bf R}^2$. And more generally, the derivative of $f:E\subset{\bf R}^m\to{\bf R}^n$ is a linear transformation $Df:{\bf R}^m\to{\bf R}^n$, where $E$ is an open set. See Terence Tao's Analysis II, Chapter 17. –  Jack Apr 15 '11 at 23:56
    
Correct, but the derivative of a constant is 0, so the linearity is trivial. It only becomes interesting when we consider the derivative as a transformation from functions to functions –  Casebash Apr 16 '11 at 0:01
    
For details on the relation between $f:{\bf R}^2\to{\bf R}^2$ and $f:{\bf C}\to{\bf C}$ and their differentiability, see math.stackexchange.com/questions/20986/complex-differentiabilty/… –  wildildildlife Apr 16 '11 at 0:04
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On an intuition level:

  • Complex numbers are somewhat equivalent to $\mathbb R^2$ (homeomorphic to a $2$-dimensional real space), on the other hand they are also a $1$-dimensional complex space. This gives them an extra structure with consequences like this theorem.

  • Another view on this is that differentiable complex functions must also satisfy the Cauchy-Riemann equations and this extra hypothesis is what makes them holomorphic.

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It isn't true. Take any singly but not doubly differentiable function f, with the property that f(x)=f(-x). Then obtain F from f by rotating f around the origin: F(r e^it)=f(r). This function is then singly but not doubly differentiable over the whole complex plane.

I think you are confusing differentiable with holomorphic.

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What is your definition of complex differentiable, if not holomorphic? –  Larry Wang Jul 24 '10 at 20:41
    
@Kaestur: Has partial derivatives in the neighbourhood, but, unlike Wolfram, need not satisfy the Cauchy-Riemann equations. –  Charles Stewart Jul 25 '10 at 16:41
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I've never seen the term "complex differentiable" used in that manner. This seems like a very nonstandard use of terminology. –  Zach Conn Jul 27 '10 at 22:10
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@Zach: Neither the question nor I used the term "complex differentiable". The question asks about "once-differentiable complex functions", which I took to be as I said. –  Charles Stewart Jul 28 '10 at 20:35
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