Global and local Lipschitz proof

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SUMMARY

The function f(x) defined as f(x)={0 for x<0, \sqrt{x} else} is not globally Lipschitz on the interval [a,b]xRn due to a discontinuity at x=0. However, f(x) is locally Lipschitz for all positive real numbers (R+) because both f(x) and its derivative f'(x) are continuous throughout this domain. The proof relies on the continuity of the function and its derivative, which can be established through limit analysis.

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  • Understanding of Lipschitz continuity
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  • Familiarity with derivatives and their continuity
  • Basic knowledge of piecewise functions
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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.

Homework Equations


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?
 
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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+.
 

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