Show Lipschitz and Uniform Continuity of f(x)=xp on [a,b]

[tazthespaz]

Let f(x)=x^p Show that f is Lipschitz on every closed sub-interval [a,b] of (0,inf). For which values of p is f uniformly continuous.



So, we know that the map f is said to be Lipschitz iff there is a constant M s.t. |f(p)-f(q)|<=M|p-q|. And we were given the hint to use the Mean Value Theorem from calculus, which states f'(c)=(f(b)-f(a))/(b-a)



I said choose (x,y) existing in [a,b], then |f(x)-f(y)|=|x^p-y^p|<=M|x-y|. But I am completely stuck here. I'm not sure how to define what M should be.

 

[ystael]

Think of it geometrically: a Lipschitz constant $$M$$ is an upper bound for the largest (absolute value) slope of any line cutting the graph of $$f$$ in two points. The mean value theorem tells you how to estimate that largest slope.