Lipschitz trouble

See Lipshitz

Rap

This doesn't seem to be true. If g(x)=x then g(x) is a Lipschitz function for all values of x. f(x) would then be x^2 which is not a first order Lipschitz function. Am I missing something here?

That is, as I understand, correct. A Lipshitz function will be uniformly increasing over an interval, consequently if g(x)=xf(x) & f(x)=x g(x) does not increase uniformly over an interval, and is not a Lipshitz function.

Rap

If I'm not mistaken, sin(x) is Lipshitz, but does not increase uniformly over all intervals.

markr, I think you're right Sine is a Lipschitz functions. If X=0 and Y=pi/2 then
|sin(0)-sin(pi)|=|1-(-1)|=2<|0-pi|=Pi

Similarly the same thing can be said about cosine.

so would xsin(x)
|0sin(0)-(pi)sin(pi)|=|0-pi(-1)|=pi<+|0-pi|=pi

However if f(x)=x^n then
|a^n-b^n| is not <=M|a-b| for any n>1 nor would it be true for tan, or csc.

Wonder what else would be Lipschitz?

I am sorry but just saying Lipschitz makes me giggle--its automatic as in grade school I had a friend whose name was Weischitz (pronounced Wye-sights) usually after Dick let new teachers mispronounce it---once.


Rap

The link you provided gives an excellent test. If the first derivative is bounded then it is a Lipschitz function and M is the highest value the derivative achieves. You could also find out where certain functions fail the test. Ln x is bounded by M=1 for x>=1 for example. From tbis you could also say that the sum of two Lipschitz functions is another Lipschitz function.

On a separate note, your new avatar is hysterical.

lipschitz
i understand f and g are on a finite interval in real space R. then there is an upper max of x in thiś interval. if f and g are well behaved on this interval, then norm f(x)-g(y) .lt. k norm (x-y) and k is max x over this interval for any x,y in this intervall this is lipschitz by def.. if the interval is infinite it will never contract

thanks