Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For homework, I had to answer the question

Construct a polynomial function of $x$ and $y$ which is differentiable on the parabola $y=x^2$ but at no other point on the complex plane. Then write the function in terms of $z=x+i y$.

(This is in Palka, $\S 3.2 \# 6.12$.)

Using Cauchy Riemann I get $$u(x,y)=\frac{x^3}{3}+\frac{y^2}{2}$$ $$v(x,y)=-\frac{x^3}{3}+\frac{y^2}{2}$$ is differentiable if and only if $y=x^2$. This answers the question, but now I have other things I want to know.

  1. I think this solution is unique (up to adding/multiplying by constants), but I am not sure. If so, does this fail to be unique if we drop the assumption that $u$ and $v$ should be polynomial in $x$ and $y$? Of course it is not unique. If $g=u'+i v'$ is holomorphic everywhere then letting $f+g=\overline{u}+i\overline{v}$, the Cauchy Riemann equations for $f+g$ are $$\left\{\begin{array}{l}\overline{u}_x=u_x+u_x^\prime=v_y+v_y^\prime=\overline{v}_y^\prime \\ \overline{u}_y=u_y+u_y^\prime=-v_x-v_x^\prime=\overline{v}_x^\prime\end{array}\right.$$ but since $u_x^\prime=v_y^\prime$ and $u_y^\prime=-v_x^\prime$ everywhere, this reduces to the condition that $u$ is differentiable which is only true iff $y=x^2$ by assumption. So $f+g$ is differentiable only on $y=x^2$ for any holomorphic $g$. So, my only remaining question is number 2:

  2. Suppose I were given some other set $\mathcal{S}\subset \mathbb{C}$ other than $y=x^2$. How would I go about finding a function differentiable only on $\mathcal{S}$? With a nice algebraic variety like $y=x^2$ we can just do the integration trick to make Cauchy Riemann always work there, but what if the set has a less nice closed form, like $(y=x)\vee (y=x^2)$ or the Cantor set? Is it always possible? If not, is there an 'only if' condition characterizing for which $\mathcal{S}$ we can find a function differentiable only on $\mathcal{S}$?

Apologies if these questions are basic. I am not very good at analysis yet.

share|improve this question
2  
For any solution $f$ and any entire (polynomial) function $g$, the function $f+g$ is a solution. So solutions aren't unique. –  jspecter Jan 22 '13 at 1:47
    
@jspecter Oh right. obviously. –  Samuel Handwich Jan 22 '13 at 2:06
    
@jspecter So the solutions are an affine space over the vector space of entire functions? The translation by $f$? –  Samuel Handwich Jan 22 '13 at 2:07
    
For any solution $f$, $\lambda f$ is also a solution for any constant $\lambda\neq 0$. –  Jonas Meyer Jan 22 '13 at 2:15
    
@Jonas Yes, hence "up to adding and multiplying by constants." Also $\lambda f + c$. –  Samuel Handwich Jan 22 '13 at 2:17
show 1 more comment

1 Answer

up vote 2 down vote accepted

Let $S$ be a closed subset of $\mathbb C$; we want $f:\mathbb C\to\mathbb C$ that is differentiable on $S$ only. If we don't require $f$ to be continuous, there is a cheap trick: let $f(z)=\operatorname{dist}(z,S)^2$ if $\operatorname{Re} z$ is rational and $f(z)=0$ otherwise. This function satisfies $f'(z)=0$ for all $z\in S$ and is not continuous (hence not differentiable) anywhere else.

But the question is more interesting if we require $f$ to be nice in the real variable sense. Let $S$ be a closed subset of $\mathbb C$. It is well-known that there is a $C^\infty$ function $g:\mathbb C\to [0,\infty)$ such that $S=g^{-1}(0)$. Being real-valued, $g$ is complex-differentiable precisely on the set $\{\nabla g=0\}$. By construction this set contains $S$. It may contain other points, though, and I don't see an easy way to get rid of them.

From here I give two constructions.


Version 1. The aforementioned function $g$ can be taken to be bounded (either it is already bounded by construction, or we change it to $g/(1+g)$). Multiply it by a smooth decaying weight such as $e^{-|z|^2}$. Let $h$ be this product. The Cauchy transform of $h$ is $$ f(z)= \frac{1}{\pi}\int_{\mathbb C}\frac{h(\zeta)}{z-\zeta} \,dA(\zeta) $$ where $dA$ indicates an integral with respect to area. Since this is a convolution with integrable kernel, $f$ is as smooth as $h$, that is infinitely smooth. And since the kernel of the Cauchy transform is the fundamental solution of the $\bar \partial$ operator, we have $$ \frac{\partial f}{\partial \bar z} = h $$ which means $f$ is complex differentiable precisely on the set $S$.

See Chapter 4 of this book for details of the Cauchy transform.


Version 2 uses heavier machinery of the Beltrami equation, see Chapter 5 of the same book. The equation $$f_{\bar z}=\frac{g}{2+g}f_z \tag{1}$$ is uniformly elliptic, and therefore its solutions are as smooth as the coefficient allows: namely, $C^\infty$ smooth. Moreover, (1) admits a solution $f$ that is a diffeomorphism of $\mathbb C$. For such a solution $f_{z}\ne 0$ everywhere, and therefore the set $\{z:f_{\bar z}= 0\}$ coincides with $\{z:g(z)=0\}$, which is $S$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.