Extending Lipschitz continuous functions from the sphere (S^{n-1}) to R^n

**leonardo** · June 22nd 2011, 02:56 AM

Hi I have encountered the following statement:

Let $f: S^{n-1}\rightarrow \mathbb{R}$ be a function with Lipschitz constant L, $|f(x)-f(y)|\leq L|x-y|$ . If we define $g=|x|f(\frac{x}{|x|})$ , "it is easy to show" that $g$ has Lipschitz constant at most $3L$ .

First of all it I showed that $|f(\frac{x}{|x|})-f(\frac{y}{|y|})|\leq \left\{\begin{array}{cc} \frac{2L}{|x|}|x-y| \\ \frac{2L}{|y|}|x-y| \end{array}\right.$ .

I used this to show
$||x|f(\frac{x}{|x|})-|y| f(\frac{y}{|y|}) |\leq |x| |f(\frac{x}{|x|})-f(\frac{y}{|y|})| + |f(\frac{y}{|y|})| ||x|-|y| |$

$=2L|x-y|+|f(\frac{y}{|y|})| ||x|-|y||$

I have tried to do further manipulations with the last term to get another $L|x-y|$ but was not successful.

Any ideas or references to a similar problem would be great.

Thanks

**hatsoff** · June 22nd 2011, 04:09 AM

Are you using the Euclidean norm here? Because if so then for any $x,y\in S^{n-1}$ we have $|x|=|y|=1$ and hence $f=g$ . But this would mean the Lipschitz constant is $L$ . If they're asking you to show that it is at most $3L$ , then something else must be going on with the norm.

**leonardo** · June 22nd 2011, 04:49 AM

I am sorry my definitions should have been more precise here. I define $g$ as a function $g:\mathbb{R}^n \rightarrow \mathbb{R}$ as stated above and $|x|$ is any norm on $\mathbb{R}^n$ . Since $x$ is not necessarily an element of $S^{n-1}$ , $|x|$ is not necessarily equal to one.

**Jose27** · June 22nd 2011, 02:25 PM

Could you give the reference where this appears? Is the constant 3L necessary for whatever argument that follows? For example this is false when f is a constant, but maybe removing these cases makes this work.

**leonardo** · June 23rd 2011, 12:29 AM

First of all I'd be glad to provide the reference, it is "Asymptotic Theory of Finite Dimensional Normed Spaces" by V. Milman and G. Schechtmann. The statement can be found in the appendix. It arises in the context of a proof for the concentration of measure on the unit sphere. They require that $f$ has zero expectation value with respect to the uniform measure $\mu$ on the sphere, $\int f \, \mathrm{d}\mu =0$ . Therefore I think we can safely exclude the case of a constant $f=C$ .

**Jose27** · June 23rd 2011, 10:22 PM

Unfortunately I don't have access to the book right now. Regarding the problem I belive that the condition that it has zero average must play a key role (if this is true at all), because in general, take x=ay with a>1 then |g(x)-g(y)|=c|x-y| where c=f(x/|x|)=f(y/|y|), and since adding constants doesn't affect the Lipschitz constant of a function we can make c arbitrarily large; this, of course, breaks down in case we have zero average but I don't know how to use it (I tried using Rademacher's theorem on Lipschitz functions and then estimating the constant by polygonal paths and arguing by density, but I couldn't reach something useful).

Edit: If it isn't too much trouble could you post the theorem you're proving as stated in the book?

**leonardo** · June 24th 2011, 01:47 AM

The proof wants to employ the more general

Theorem: Let $F: \mathbb{R}^n\leftarrow \mathbb{R}$ be a Lipschitz function with constant $L$ ( $\mathbb{R}^n$ endowed with the euclidean metric). Let $x_1,\dot,x_n$ be independent, mean zero, normalized in $\mathrm{L}^2$ , Gaussian random variables. Then,

$P(|F(x_1,\dots,x_n)-E\left[F(x_1,\dots,x_n)\right]|>C)\leq 2 e^{\frac{-2C^2}{\pi^2 L^2}}.$

Then we can proove the following

Corollary: Let $f: S^{n-1}\rightarrow \mathbb{R}$ be a function with Lipschitz constant $L$ . Then

$\mu(|f-\int f\,\mathrm{d}\mu|>C)\leq 4e^{-\frac{\delta C^2 n}{L^2}}$

where $\mu$ is the Haar measure on $S^{n-1}$ and $\delta>0$ an absolute constant.

In order to extend the function f to all of $\mathbb{R}^n$ and to use the above theorem we need to explicitly know the Lipschitz constant of $g(x)=|x|f(\frac{x}{|x|})$ . The proof assumes $\int f\,\mathrm{d}\mu=0$ so that we only need to proove something about $P(|\tilde{f}|>C)$ . This can be for sure done by defining $\tilde{f}=f-\int f\,\mathrm{d}\mu$ .

**Jose27** · June 24th 2011, 10:18 AM

Well, I can get a $4L$ constant: Let $K=\max_{\mathbb{S}^{n-1}}f$ and $k=\min_{\mathbb{S}^{n-1}}f$ then the triangle inequality gives $K-k \leq 2L$ and by the zero average condition (if f is not identically zero) $k<0<K$ so that $K<2L$ and this together with your last estimation in the first post gives a constant of at most $4L$ . Hope this helps.

**leonardo** · June 26th 2011, 03:00 AM

Originally Posted by Jose27

Well, I can get a $4L$ constant: Let $K=\max_{\mathbb{S}^{n-1}}f$ and $k=\min_{\mathbb{S}^{n-1}}f$ then the triangle inequality gives $K-k \leq 2L$ and by the zero average condition (if f is not identically zero) $k<0<K$ so that $K<2L$ and this together with your last estimation in the first post gives a constant of at most $4L$ . Hope this helps.

Thank you very much for pointing this out. This yields an explicit bound on the Lipschitz constant of $g$ , which is good.

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