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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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Could you correct the rhs of the $f(x)$, pls? – Berci Oct 15 '12 at 15:38
I fixed it, hopefully the way it was intended. – Harald Hanche-Olsen Oct 15 '12 at 15:44
@ Harald Thank you,I am just a beginner at Tex writing.Now it is right. – Klara Oct 15 '12 at 15:46
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Oct 15 '12 at 15:49
I rolled back your edit since you posted another question for the new function. The answers are for the current function. Changing the question now will confuse everyone. – Ayman Hourieh Oct 15 '12 at 23:47

3 Answers 3

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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