Re: Running wide

From: Rob Johnson (rob_at_trash.whim.org)
Date: 06/02/04 

Date: Wed, 2 Jun 2004 17:20:18 +0000 (UTC)

It would be nice if the relevant parts of previous posts to which you
refer are quoted so that each post makes more sense, both to first time
readers as well as to refresh the memories of those who are following
the thread.

For example, in this thread, we are talking about simple, closed, convex
curves in R^2.

In article <ZGhvc.3068$wd7.1052@front-1.news.blueyonder.co.uk>,
"George Barwood" <george.barwood.removethis@blueyonder.co.uk> wrote:
>> Let f(t) be a parameterization of the curve where f'(t) does not vanish.
>
>Why is this possible?
>
>Is there a theorem that tells us that such a parametrization is always
>possible?
>
>The curve is continuous ( the image of a continuous function on [0,1] ), but
>that doesn't imply a tangent exists anywhere at all, does it?
>
>Or does convexity + finite length somehow imply the existence of a tangent?
>
>If so, I don't see how - please explain!

It is a theorem that any convex function has monotonically increasing
left and right derivatives, and at each point, the right derivative is
greater than or equal to the left derivative. It is a simple corollary
that the left and right derivatives are equal except at a countable set
of points and that the sum of the differences of the derivatives at the
countable set of points is less than or equal to the difference of the
maximum of the left derivative and the minimum of the right derivative.

A simple adaptation of this proof gives us that a simple, closed, convex
curve in R^2 has left and right tangents at each point. If the curve
circles its interior clockwise, then the arguments of these tangents are
monotonically decreasing, and the argument of the right tangent is less
than or equal to the argument of the left tangent at each point. We get
the same corollary that the left and right derivatives are equal except
at a countable set of points and that the sum of the differences of the
arguments at the countable set of points is less than or equal to 2 pi.

Since the arguments of the tangents are monotonically decreasing, they
are differentiable almost everywhere, so the curvature exists almost
everywhere. Furthermore, the length integral of the curvature plus the
sum of the differences of the arguments equals 2 pi.

--------------------------------

I had intended the differential geometry proof to be used on smooth
approximations to the curve and then limits taken. The comments I made
above regarding tangents and curvature, allow this to be done.

Rob Johnson <rob@trash.whim.org>
take out the trash before replying

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