3

The following is a theorem from baby Rudin:

Suppose $f:E\to {\bf R}^m$ be a differentiable function, where $E\subset {\bf R}^n$ is convex and open, and there is $M>0$ such that $$ \|f'(x)\|<M $$ for every $x\in E$. Then $$ |f(b)-f(a)|\leq M \ |b-a| $$ for all $a,b\in E$.

Here is my question:

Can one replace "convex" with "connected" in this theorem?

When $n=1$, this is obviously still true. But when $n\geq 2$, the method in the proof, which uses convexity to reduce the dimension to $1$, would fail if one has connectedness instead of convexity. But I don't have a counterexample.

  • 1
    Have you checked "domains with cusps"? Usually, one finds counter examples on such domains. I think a counter examples is provided in the book of Gilbarg & Trudinger about elliptic PDE – Quickbeam2k1 Sep 22 '14 at 19:05
  • @Quickbeam2k1: Thanks for your comment. Would you elaborate a little bit more? –  Sep 24 '14 at 23:24
  • Sorry, the example in Gilbarg trudinger only works for showing that $C^1$ functions are not Hölder regular for Hölder exponents less than $1$. – Quickbeam2k1 Sep 25 '14 at 06:01

1 Answers1

6

Take an annulus with a radius removed, say

$$A = \{ z : 1 < \lvert z \rvert < 2\} \cap (\mathbb{C}\setminus [0,+\infty)).$$

As a function, take for example the argument (with values in $(0,2\pi)$). That is differentiable, the derivative is bounded, but since the argument jumps at the positive real axis, and we can come arbitrarily close to it from above and from below, there is no $M < +\infty$ with

$$\lvert \arg z - \arg w\rvert \leqslant M\cdot \lvert z-w\rvert$$

for all $z,w\in A$.

Domains where the boundedness of the derivative implies Lipschitz continuity cannot be too far from convex.

A bound on the derivative, $\lVert f'(x)\rVert \leqslant M$ implies

$$\lvert f(x) - f(y)\rvert \leqslant M \cdot \inf \left\{L(\gamma) : \gamma \text{ is a piecewise differentiable path from } x \text{ to } y \text{ in } E\right\},$$

as follows from

$$f(y) - f(x) = \int_0^1 f'(\gamma(t))\cdot \gamma'(t)\,dt$$

if $\gamma \colon [0,1] \to E$ is such a path. If the quotient

$$\frac{\inf \left\{L(\gamma) : \gamma \text{ is a piecewise differentiable path from } x \text{ to } y \text{ in } E\right\}}{\lvert y-x\rvert}$$

is bounded for all $x,y\in E$, then the boundedness of the derivative implies Lipschitz continuity.

Daniel Fischer
  • 211,575
  • Could "too far from convex" be replaced with "nonconvex"? – Jonas Meyer Sep 22 '14 at 19:13
  • No, If you take a rectangle, and bend it only slightly, the boundedness of the derivative still implies Lipschitz continuity. I guess I should add a little to the answer. – Daniel Fischer Sep 22 '14 at 19:15
  • Thanks! Sorry, that isn't even what I meant to ask, and my question was based on a misreading of your answer. But nonetheless, thanks for the additional info. – Jonas Meyer Sep 22 '14 at 19:30
  • Thank you for your answer! I'm assuming that one needs to convert the example in the first paragraph into the $f:A\to\Bbb{R}^2$ version? Then the complex derivative is different from the real one $Df(x,y)$ which is a matrix? –  Sep 24 '14 at 19:33
  • @Jack I took the argument to have a real-valued function. Since it is non-constant and real-valued, it is not complex differentiable and you only have the real differential. Taking the logarithm is probably not more difficult, then you immediately have the complex derivative, and translating from the complex derivative to the real is very straightforward, if $f'(z) = a+ib$, then the real derivative is given by the matrix $$\begin{pmatrix} a & -b\ b & a \end{pmatrix}.$$ – Daniel Fischer Sep 25 '14 at 14:24
  • According to the description here (http://en.wikipedia.org/wiki/Argument_(complex_analysis)#Computation), I don't quite understand how to you convert $f(z)$ to $f(x,y)$ so that it is differentiable with bounded derivative. –  Sep 26 '14 at 02:27
  • @Jack You identify $\mathbb{C}$ with $\mathbb{R}^2$ via $x+iy \leftrightarrow (x,y)$. So if you have a complex form $f_{\mathbb{C}}(z)$, it becomes $f_{\mathbb{R}}(\operatorname{Re} z, \operatorname{Im} z)$. For the argument (which you unfortunately can't give a uniform description in the real coordinates) you get in each quadrant a formula $\arctan \frac{y}{x} + C$ or $\operatorname{arccot} \frac{x}{y} + C$ (where $C$ is a multiple of $\pi$ depending on the quadrant and the chosen branch of the argument), and its real derivative is given by the partial derivatives – Daniel Fischer Sep 26 '14 at 11:28
  • 1
    $\frac{\partial}{\partial x}\arg(x+iy) =\frac{-y}{x^2+y^2}$ and $\frac{\partial}{\partial y}\arg(x+iy) =\frac{x}{x^2+y^2}$. We can also work with $\arg z= \operatorname{Im}\log z$, and obtain $$\frac{\partial}{\partial x}\arg z=\operatorname{Im}\left(\frac{\partial}{\partial z}\log z\cdot \frac{\partial z}{\partial x}\right)= \operatorname{Im}\left(\frac{1}{z}\cdot 1\right)= \operatorname{Im} \frac{\overline{z}}{\lvert z\rvert^2}= \frac{-y}{x^2+y^2}$$ and similarly for $\frac{\partial}{\partial y}$. – Daniel Fischer Sep 26 '14 at 11:30
  • 1
    So the (real) derivative of the argument is given by the matrix $$D\arg(x,y) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right).$$ – Daniel Fischer Sep 26 '14 at 11:30