Re: Is 1/(x-a)^2 locally Lipschitz continuous?

Konrad Viltersten wrote:

I understand it can't be globally Lipschitz, because of the
unbounded derivative at the singularity. However, my wife
and i fight about whether it's locally Lipschitz.

I say it is too. She claims foolishly it's not.
Who's the man here?

Isn't it way, way more than just locally Lipschitz?
For example, from basic calculus it's easy to
see that its derivative is continuous at each
x not equal to a, and hence it has a continuous
derivative in the neighborhood of each point in
its domain, a much stronger property than being
locally Lipschitz.

In fact, I don't see why it's not analytic (more
than just C-infinity) in some neighborhood of
each point in its domain, but maybe I'm missing
something obvious.

Dave L. Renfro

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