From: "David C. Ullrich" <ullrich@math.okstate.edu>
Subject: Re: Approximation of Lipschitzian functions
Date: Sun, 26 Sep 1999 14:29:23 -0500
Newsgroups: sci.math
Keywords: ...by C^1 functions having the same Lipschitz constant

TTL wrote:

> TTL <tlim@gmu.edu> wrote in message news:7sk8t4$2l5@portal.gmu.edu...
>
> > If f :R-->R is lipschitzian with lipschitz constant L, given epsilon >0,
> > does there exist a C^1 function g such that
> > |f(x)-g(x)|< epsilon and the lipschitz constant of g is <=
> > L - epsilon ?
>
> Oop, I mean  |M-L|<epsilon, where M is the lipschitz constant of g.

    You can get |f - g| < epsilon with the Lipschitz constant of g <= L,
in fact.

> > I believe the answer is yes. But I need to see a reference for the proof.

    Well, do you know how to prove that you can approximate continuous
functions uniformly by smooth functions? The standard technique for
doing that does this. Assuming that you understand some of this stuff
or the question wouldn't arise:

    Convolve f with a smooth approximate identity. That gives you a
smooth g with |f - g| < epsilon, and it's easy to see that the Lipschitz
constant of g is <= L. Say the approximate identity is K. You have

g(x) = integral(f(x-t) K(t) dt) ,

hence

|g(x) - g(y)| <= integral(|f(x-t) - f(y-t)| K(t) dt)
                  <= L |x-y| integral(K)
                    = L |x-y| .

> Thanks. (Note: this is not a homework problem.)

Better not be...