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Let $n,m\in\mathbb{N}$, such that $E=(\mathbb{R}^n,\|\cdot\|_E)$ and $F=(\mathbb{R}^m,\|\cdot\|_F)$ with $\|\cdot\|_E$ and $\|\cdot\|_F$ two norms (we do not care which one since $n,m\in\mathbb{N}$).

I have a little doubt about the "lipschitzness" of a mapping $f$. Assuming $f:E\rightarrow F$, differentiable over $E$ (thus continuous over $E$). If $C\subset E$ is compact, then $f$ is $k$-lipschitz for a given $k>0$ over $C$, right ?

If $n=m=1$, I know that is true, but I didn't find the same result in multivariate fashion (maybe because it is too obvious) and I did not go further on this question since it's been a while I do not have pratice math. Since $\mathbb{R}^n$ has very good topological properties, I think that this result is true. Am I wrong ?

Thanks !

Edit : Nice counter examples below !

The aforementioned statement is right only if $f$ is $C^1$.

ar_ma_cs
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  • hmm, I have some issues with this. First of all, if you say $f$ is $k-$ lipschitz for a given $k>0$, then this is, for sure, not correct. What you may hope for is that there exists a constant $k>0$ such that $f$ is $k-$ lipschitz. Make sure you understand the difference. If it is not clear to you, try to prove the one dimensional case for $k= 1/n$ and let $n\rightarrow \infty$. Then, showing Lipschitz continuity with differentiability alone is, very likely, not true. You will need the boundedness of the derivative. Depending on the domain of definition the geometry of the domain of definition – Thomas Nov 03 '22 at 20:03
  • may also play a role, and the set $C$ should be connected...Apart from that, searching here may already help you. Look, e.g., at this thread: https://math.stackexchange.com/questions/73368/a-continuously-differentiable-map-is-locally-lipschitz – Thomas Nov 03 '22 at 20:04
  • Thank you so much for all your anwsers, i need derivatives bounded, the example of @Theo is clear for me. – ar_ma_cs Nov 03 '22 at 20:53

1 Answers1

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It’s not true, even for $m = n = 1$. Note, for a differentiable function, Lipschitzness is equivalent to having a bounded derivative. So, we just need a differentiable function with an unbounded derivative. Let: $$f(x) = \begin{cases}x^{5/3} \sin\left(\frac1x\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0.\end{cases}$$ The derivative exists at $x \neq 0$, as it is a product and composition of differentiable functions. Using the usual derivative rules, we get $$f’(x) = \frac{5}{3}x^{2/3} \sin\left(\frac{1}{x}\right) - x^{-1/3}\cos\left(\frac{1}{x}\right).$$ For $x = 0$, we use the limit definition: $$f’(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} h^{2/3} \sin\left(\frac{1}{h}\right),$$ where the above expression of $h$ is sandwiched between $\pm h^{2/3}$, both of which tend to $0$ as $h \to 0$. Thus, $f’(0) = 0$. So, the function is defined everywhere, including every compact set containing $0$ in its interior.

We just need to show that $f’$ is unbounded around $0$. If we take $x_n = \frac{1}{2n\pi}$, then $x_n \to 0$ and $$f’(x_n) = \frac{5}{3}x_n^{2/3} \sin(2\pi n) - \sqrt[3]{2\pi n}\cos(2\pi n) = -\sqrt[3]{2\pi n} \to -\infty$$ as $n \to -\infty$. Thus, $f$ cannot be Lipschitz on any compact set containing $0$ in its interior, e.g. $[-1, 1]$.

Theo Bendit
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  • Nice example, but honestly, I assume an answer showing lipschitz continuity for $C^1$ functions or functions with bounded derivative would have been more interesting to the OP. A counterexample is actually already given in the answer given by Christian Blatter to which I provided a link in my comment to the question. Just my 2 cent. – Thomas Nov 03 '22 at 20:18
  • @Thomas I think I’ll leave it up. Blatter’s comment doesn’t do the details, and I’m not convinced that this isn’t helpful to the OP. It certainly answers the question as written. – Theo Bendit Nov 03 '22 at 20:25
  • Got it ! Thank you @Theo – ar_ma_cs Nov 03 '22 at 20:54
  • @Theo Bandit But, if $f$ is at least $C^1$, then it seems to be good (i.e f differentiable over a compact + at least $C^1$) , since the "at least $C^1$" property vanishes the problem of derivative non continuity illustrated with f' and the sequence $x_n$. Right ? – ar_ma_cs Nov 03 '22 at 21:19
  • And if $f$ is $C^1$ over compact $C$, then $f'$ is continuous over $C$, thus reaches its bounds, thus $f'$ bounded, thus $f$ lipschitz over $C$ – ar_ma_cs Nov 03 '22 at 21:35
  • @ar_ma_cs Yes, if $f$ is $C^1$, then $f’$ will be bounded, and hence $f$ will be Lipschitz. – Theo Bendit Nov 03 '22 at 22:13