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

I could quote from our book but that wouldn't do much
difference for anybody of us, would it? :)

If two people use different definitions, they might come
to different conclusions.


Good point. The thing is that we're unclear by what the
claimed definition says; we find it ambigous. It goes
like this.

"If the constant K depends on x and y then the function
f \in L locally."

Now, on one hand, K does indeed depend on the choice
of x and y. On the other hand, it can be infinity if the
choice is made evily...

--
Vänligen
Konrad
---------------------------------------------------

Sleep - thing used by ineffective people
as a substitute for coffee

Ambition - a poor excuse for not having
enough sence to be lazy
---------------------------------------------------


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