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Problem: If $g(x)=\sqrt{x}$ for $x\in[0,1]$,show that there does not exist a constant K such that $|g(x)|\leq K|x|$ for all $x\in[0,1]$. Conclude that the uniformly continuous $g$ is not a Lipschitz function.

I am stuck on this problem but I tried using contradiction to do this problem. Suppose that there exists a $K>0$ such that $|g(x)|\leq K|x|= \sqrt{x}\leq Kx\implies x\leq K^2x^2\implies 1\leq K^2x$ where $x\in(0,1]$. Then $\frac{1}{\sqrt{x}}\leq K$. From here I can't seem to find a contradiction. A hint would help thanks.a

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Since your K must be uniform, your last line says that K has to be infinite, which is absurd

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  • $\begingroup$ Ah yes thanks. Since $x$ gets smaller and smaller it pushes $k$ to infinity. $\endgroup$ – user60887 Dec 6 '13 at 19:57
  • $\begingroup$ -----welcome ----- $\endgroup$ – aflous Dec 6 '13 at 20:01
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Hint: the last inequality must hold for all $x\in (0,1]$, even for $x =\frac{1}{2K^2}$.

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  • $\begingroup$ Don't you mean $x=\frac{1}{4K^2}$? $\endgroup$ – user60887 Dec 6 '13 at 20:00
  • $\begingroup$ doesn't matter since $\sqrt 2 > 1$ $\endgroup$ – aflous Dec 6 '13 at 20:02
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Hint

$$1\leq K^2x \Leftrightarrow x \geq \frac{1}{K^2}$$

Can you find some $x \in (0,1]$ smaller than $\frac{1}{K^2}$?

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