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$x$ is a $3$d vector. This is what I have so far, don't know if it is enough to prove , that $f(x) = 1/|x|$ is not Lipschitz-continuous on $|x|<1$: First we have to show, that for all $L>0$ there exist $x,y$ with $|x|<1$, $|y|<1$ such that $|f(x) - f(y)| > L \cdot |x-y|$. Let $0<a<1/(|x|^2\cdot L )$ and $y=a\cdot x$ and we choose $x$ such that $|x|>1/L$ (now it is unclear here why $|x|<1$.) Then we have: $|y| = a \cdot |x| < |x|/(|x|^2\cdot L) = 1/(|x|\cdot L) < 1$ and in particular $1/(|x|\cdot|y|) > L$. Then we have:

$|f(x)-f(y)| = |1/|x|-1/|y|| = ||x|-|y||/(|x|\cdot|y|) = |x-y|/(|x|\cdot|y|) > L |x-y|$.

Now I am sure I am missing some detail to complete the exercise, but don't know where to look at.

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    $\begingroup$ Sure you got the statement of the problem right? Doesn't make much sense, because $1/||x|$ is not even defined for $|x|<1$... $\endgroup$ Commented May 16, 2016 at 17:55
  • $\begingroup$ The problem is to find out if $f(x) = 1/|x|$ is Lipschitz continuous on ${x: |x| < 1}$, $f:R^3->R$ and if so to give the Lipschitz constant. $\endgroup$
    – user276611
    Commented May 16, 2016 at 17:58
  • $\begingroup$ @stackExchangeUser: I think David's point is that $1/0$ is undefined. $\endgroup$
    – parsiad
    Commented May 16, 2016 at 17:59
  • $\begingroup$ Yes, I know... But that's how it is written in the book. I guess it should be $0<|x|<1$. $\endgroup$
    – user276611
    Commented May 16, 2016 at 18:00

3 Answers 3

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Make your life easier by choosing simple $x$ and $y$, say $x:=(a,0,0)$ and $y:=(b,0,0)$, with both $a$ and $b$ between $0$ and $1$. Then $$ \left|f(x)-f(y)\right|=\left|\frac1a-\frac1b\right|={\left|b-a\right|\over ab}>\frac1a\left|b-a\right|=\frac1a\left|x-y\right|,\tag1 $$ and with suitable choice of $a$ you can make the RHS of (1) as large as you want.

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  • $\begingroup$ A suitable choice would be $a > 1/L$ for some $L> 0$, right? $\endgroup$
    – user276611
    Commented May 16, 2016 at 18:19
  • $\begingroup$ Given $L$, take $a<1/L$ -- we want $a$ to be small! $\endgroup$
    – grand_chat
    Commented May 16, 2016 at 18:20
  • $\begingroup$ Ok, yes, that's what I actually meant. Thanks! $\endgroup$
    – user276611
    Commented May 16, 2016 at 18:21
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I think you are on the right way. You got $$ |f(x)-f(y)| = |x-y|/(|x|\cdot|y|) $$ and $1/(|x|\cdot|y|) $ is by sure unbounded for $ x,y\in D:=\{z\in\mathbb R^n\setminus(0): |z|\leq 1\}$. So $$ |f(x)-f(y)| \leq L|x-y|\quad \forall x,y\in D $$ cannot hold for any positive $L$.

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Assume $f(x)= \frac{1}{\Vert x \Vert}$ is lipschitz for $0< \Vert x \Vert <1$. Then there exists $L>0$ such that

$$ L \Vert x - y \Vert \geq \vert f(x) - f(y) \vert = \frac{\vert \ \Vert x \Vert - \Vert y \Vert \ \vert}{\Vert x \Vert \cdot \Vert y \Vert}.$$

Not choose $2y=x$, then you get

$$ L \Vert x \Vert \geq \frac{\frac{1}{2}\Vert x \Vert}{\frac{1}{2} \Vert x \Vert^2} = \frac{1}{ \Vert x \Vert}$$

and therefore

$$ L \geq \frac{1}{\Vert x \Vert^2}.$$

This gives you a contradiction.

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  • $\begingroup$ I think you mean $y = 2x$. How do you know that $ 0 < |y| < 1$? $\endgroup$
    – user276611
    Commented May 16, 2016 at 18:17
  • $\begingroup$ Thanks for pointing out that my calculation is wrong. I meant $2y=x$ (precisely to avoid the issue that you wrote). I'll correct my computations. $\endgroup$ Commented May 18, 2016 at 18:48

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