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Is the function $f( x)=1/|x|^{1/2}$ 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 zero,am I right?

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At zero the function $f(x)$ is not Lipschitz continuous any more, it is not even simply continuous. Close to zero - yes, but not in zero.

For every constant $C$ you can find $x_1$ and $x_2$ such that $|f(x_1)-f(x_2)| > C|x_1-x_2|$. For example if $C > 1$ you can take $x_1 = \frac{1}{C}$ and $x_2 = \frac{1}{2C}$.

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The stated $x_i$ don't work; eg if C = 1+e for small positive e, $|f(x_1)-f(x_2)| < C|x_1-x_2|$. However, $x_1 = \frac{1}{C^2}$ and $x_2 = \frac{1}{2C^2}$ work. – jwpat7 Nov 21 '12 at 16:50

For any interval $[a,b]$ which doesn't contain $0$, function $f$ is Lipschitz and the constant $L = \max\limits_{[a,b]}|f'(x)|$.

For any interval $[a,b]$ which contain $0$, function $f$ is not continuous.

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My function is different, I'm sorry for the typo! – Klara Oct 15 '12 at 22:45

The function $f(x)=\frac{1}{|x|^{\frac{1}{2}}}$ is NOT Lipschitz continuous in any interval containing zero as $f(x)$ is not even defined at zero so it is not even continuous in the interval.

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My function is different, I'm sorry for the typo. – Klara Oct 15 '12 at 22:22

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