From: artfldodgr@my-deja.com
Subject: Re: differentiable surjections
Date: Sat, 24 Jun 2000 03:06:16 GMT
Newsgroups: sci.math
Summary: [missing]

In article <01bfdc4a$dcde25a0$e7ce8ad1@daves-dell>,
  "David C. Ullrich" <ullrich@math.okstate.edu> wrote:
>
>
> macavity <mxyptlkNOmxSPAM@hotmail.com.invalid> wrote in article
> <2a6077c8.595ab979@usw-ex0106-047.remarq.com>...
> [...]
> >
> > Yes, coming to think of it, I have never come across an
> > extension of Rolle into any dimension other than 1.  But never
> > thought it (or an appropriate reformulation - based on the total
> > derivative) would not hold - the arguments involved being rather
> > basic.
>
> 	They're rather basic but also rather specific to R. How
> do you prove Rolle's theorem? WLOG f(a) = f(b) = 0;
> now if f is not identically zero look at a global extremum
> in (a,b); f' must vanish at that point.
>
> 	The most reasonable interpretation of "extremum"
> for a function from R^n to itself might be an extremum of
> |f|. But there's no reason all the partials of all the components
> of f have to vanish at such a point.
>
> 	The slogan I was taught years ago was that MVT
> fails for R^n-valued functions, although "all" of its
> consequences remain true (eg bounds on the partial
> derivative imply Lipschitz estimates.)

One explanation for this is the Mean Value Inequality for f: R^n --> R^m,
one form of which is the following: If f is a C^1 mapping of a convex
set U (contained in R^n) into R^m, and if x and y are two points of U,
then |f(x)-f(y)| <= M*|x-y|, where M is the sup of |f'(z)| (operator
norm) as z varies over the line segment connecting x to y.  For a proof
see Blue Rudin.

--A.


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From: david_ullrich@my-deja.com
Subject: Re: differentiable surjections
Date: Sat, 24 Jun 2000 17:12:18 GMT
Newsgroups: sci.math
Summary: [missing]

In article <8j18j1$2vp$1@nnrp1.deja.com>,
  artfldodgr@my-deja.com wrote:
> In article <01bfdc4a$dcde25a0$e7ce8ad1@daves-dell>,
>   "David C. Ullrich" <ullrich@math.okstate.edu> wrote:
[...]
> >
> > 	The slogan I was taught years ago was that MVT
> > fails for R^n-valued functions, although "all" of its
> > consequences remain true (eg bounds on the partial
> > derivative imply Lipschitz estimates.)
>
> One explanation for this is the Mean Value Inequality for f: R^n --> R^m,
> one form of which is the following: If f is a C^1 mapping of a convex
> set U (contained in R^n) into R^m, and if x and y are two points of U,
> then |f(x)-f(y)| <= M*|x-y|, where M is the sup of |f'(z)| (operator
> norm) as z varies over the line segment connecting x to y.

    A person might regard this as more an _instance_ of the
above than an "explanation" for it. Never mind.

> For a proof see Blue Rudin.

    Or choose a unit vector v,  let F(t) = <v ,f(t*x + (1-t)*y)>
and apply the MVT to F to deduce that |<v, f(x) - f(y)>| <= M*|x-y|,
then conclude that since this holds for all v we must have
|f(x) - f(y)| <= M*|x-y| .
     Which must be the proof in B. Rudin - he does that a lot.
I mention it just to point out that the proofs of these
consequences of MVT in contexts where MVT does not hold often
simply _use_ the one-variable MVT.

> --A.
>
> Sent via Deja.com http://www.deja.com/
> Before you buy.
>


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