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Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.

I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?

user146311
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2 Answers2

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The basic idea is that around $0$, the change in $f(x)$ is not bounded by a constant times the amount that we change $x$ (note that the derivative becomes arbitrarily large). More formally, for any fixed $k>0$,

$$\sqrt{\frac{1}{(k+1)^2}} = \frac{1}{k+1} > k \frac{1}{(k+1)^2}$$

which violates the Lipschitz condition.

A.S
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  • But how can this be tied to the condition $\displaystyle \frac{|f(x)-f(c)|}{|x-c|} \leq M$? – user146311 Apr 28 '14 at 06:47
  • @user146311 Let $M = k$. Then, fixing $M$, let $c = 0$ and $x = \frac{1}{(k+1)^2}$. The expression I have written falls out of this. (multiply both sides by $|x-c|$ to see this) – A.S Apr 28 '14 at 06:48
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Given any $M>0$, choose $c=0$ and $ 0<x<\frac{1}{m^{2}}$, so that $\frac{1}{\sqrt{x}}>M$. then we have: $$ \frac{|f(x)-f(c)|}{|x-c|}=\frac{|\sqrt{x}|}{|x|}=\frac{1}{\sqrt{x}}>M $$ Since $M$ was arbitrary, this shows that $F$ is not Lipschitz continuous.

Kns
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