# Is the function $f(x)=|x|^{1/2}$ Lipschitz continuous?

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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¿Is this homework? If so, check the homework tag! It is clearly Lipschitz over any interval which excludes zero. But for an interval containg zero, one way oc choosing the Lipschits 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? –  kjetil b halvorsen Oct 15 '12 at 23:23
Exact duplicate: math.stackexchange.com/q/214290 –  user44798 Oct 15 '12 at 23:36
@gee That question was for a different function ($f(x) = 1/|x|^{1/2}$) until Klara edited it an hour ago. I've rolled back the edit. –  Ayman Hourieh Oct 15 '12 at 23:48

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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Thank you for your reply, Is the contradiction that K has to go to infinity to satisfy the inequality for x, y close to zero? What is a reasonable value for K to say that a function in general is Lipschitz cont.? –  Klara Oct 16 '12 at 0:06
@Klara The contradiction is that no matter what $K$ we pick, we can always find $x, y > 0$ that don't satisfy the Lipschitz continuity condition. In general, $K$ must be a finite non-negative number. –  Ayman Hourieh Oct 16 '12 at 0:10
thank you for your help! –  Klara Oct 16 '12 at 1:21
That, or the remark that $g(x)=(f(2x)-f(x))/|x|$ defines an unbounded function $g$. –  Did Mar 23 '13 at 7:53