Global and local Lipschitz proof

Homework Statement

f(x)={0 for x<0, $\sqrt{x}$ else}

a) Is f(x) globally Lipschitz? Explain

b) Find the area for which f(x) is locally Lipschitz.

The Attempt at a Solution

a) f(x) is not globally Lipschitz in x on [a,b]xRn since there is a discontinuity at x=0.

b) I would think that f(x) is locally Lipschitz for all real positive numbers, R+, but how do I go about proving it?

Answers and Replies

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Ok, f(x) and f'(x) are continuous throughout R+. This can be proven by limits. So therefore f(x) is by definition locally Lipschitz in x on R+.