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Is the function $f(x)=|x|^{1/2}$ Lipschitz continuous near $0$? If yes, find a constant for some interval containing $0$.

I think the answer is yes since I can find $L=1$ that satisfies Lipschitz continuity criteria in a interval close to $0$, am I right?

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    It is clearly Lipschitz over any interval which excludes zero. But for an interval containing zero, one way of choosing the Lipschitz constant is to take the supremum of the absolute value of the derivative. but for an interval containing zero, ¿What is that supremum in this case?2012-10-15

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Let $x$, $y$ be numbers in $(0, \delta)$.

$ \left|\left(|y|^{1/2} - |x|^{1/2}\right)\left(|y|^{1/2} + |x|^{1/2}\right)\right| = \left||y| - |x|\right| = |y - x| $

Therefore: $ |f(y) - f(x)| = \frac{|y - x|}{\left||y|^{1/2} + |x|^{1/2}\right|} \tag{1} $

If $f(x)=|x|^{1/2}$ is Lipschitz continuous, we can find $K > 0$ so that:

$ |f(y) - f(x)| \le K|y - x| \tag{2} $

Put (1) and (2) together to get:

$ \frac{1}{K} \le \left||y|^{1/2} + |x|^{1/2}\right| $

By making $x$ and $y$ approach $0$, we can make the RHS as small as we desire. Thus, we have a contradiction.

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    That, or the remark that $g(x)=(f(2x)-f(x))/|x|$ defines an unbounded function $g$.2013-03-23