Are you using the Euclidean norm here? Because if so then for any we have and hence . But this would mean the Lipschitz constant is . If they're asking you to show that it is at most , then something else must be going on with the norm.
Hi I have encountered the following statement:
Let be a function with Lipschitz constant L, . If we define , "it is easy to show" that has Lipschitz constant at most .
First of all it I showed that .
I used this to show
I have tried to do further manipulations with the last term to get another but was not successful.
Any ideas or references to a similar problem would be great.
Thanks
Are you using the Euclidean norm here? Because if so then for any we have and hence . But this would mean the Lipschitz constant is . If they're asking you to show that it is at most , then something else must be going on with the norm.
I am sorry my definitions should have been more precise here. I define as a function as stated above and is any norm on . Since is not necessarily an element of , is not necessarily equal to one.
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.
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 has zero expectation value with respect to the uniform measure on the sphere, . Therefore I think we can safely exclude the case of a constant .
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?
The proof wants to employ the more general
Theorem: Let be a Lipschitz function with constant ( endowed with the euclidean metric). Let be independent, mean zero, normalized in , Gaussian random variables. Then,
Then we can proove the following
Corollary: Let be a function with Lipschitz constant . Then
where is the Haar measure on and an absolute constant.
In order to extend the function f to all of and to use the above theorem we need to explicitly know the Lipschitz constant of . The proof assumes so that we only need to proove something about . This can be for sure done by defining .
Well, I can get a constant: Let and then the triangle inequality gives and by the zero average condition (if f is not identically zero) so that and this together with your last estimation in the first post gives a constant of at most . Hope this helps.