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Explain how $x\sin(1/x)$ is not Lipschitz. I know that $\sin(1/x)$ is not Lipschitz and I can show this by proving it is not uniformly continuous, but I'm not sure how to go about things to $x\sin(1/x)$. Thanks!

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$x\sin\frac1x$ is differentiable on $\mathbb R\setminus\{0\}$. Is its derivative bounded? –  Henning Makholm Apr 25 '12 at 0:08
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The derivative is unbounded as x --> 0 so it is not Lipschitz. –  Mark Apr 25 '12 at 0:43
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2 Answers 2

Or even more explicitly, let $x_n=\frac{1}{\frac{\pi}{2}+2 \pi n}$, $x_n'=\frac{1}{\frac{3\pi}{2}+2 \pi n}$. Then $\sin \frac{1}{x_n} = 1$, $\sin \frac{1}{x_n'} = -1$, and a rather tedious calculation shows $$\frac{|x_n+x_n'|}{| x_n-x_n'|} = 2(2n+1)$$ which shows that it is not Lipschitz.

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Hint: Let $f(x)=x\sin x$. Consider the points $x_n={1\over n\pi}$ and $y_n={1\over n\pi+\pi/2}$. Here, $|f(x_n)-f(y_n)|$ is of order $1/n$ while $|x_n-y_n|$ is of order $1/n^2$.

Of course, it would be a bit easier to show that $f'$ is unbounded, as Henning suggests.

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