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I have a question about the following statement. I saw this in this paper but the proof in this paper seems somewhat unreasonable to me. So I tried to give more rigorous proof, but I have no idea to prove this statement.

The paper is:

A.A. Zevin, Minimal periods of solutions of Lipschitizian differential equations in a vector space with an arbitrary norm

In this paper, there is a lemma which says:

Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is Lipschitz continuous with constant $L$. Let $F(x) = Df(x) f(x)$. (The author assumes that $f$ is differentiable, without loss of generality. He mentioned that Lipschitz continuous is differentiable almost everywhere, but I can't understand why this assumption is still valid.) Then $F$ is Lipschitz continuous with constant $L^2$.

edited: $f$ is a differentiable function.

I tried to use the fact that the norm of $Df(x)$ is bounded by $L$. (I think that the proof of Rademacher's theorem verifies this.)

Can anyone give me a hint about the proof or help me understand the proof of this lemma given in the paper?

This is my first question in this community, so any suggestions about the format of my question are also welcome.

Thank you.

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    If $n=m=1$ and $f (x) = |x|+1$, then $F = f' f $ is not even continuous. – 2017-01-18
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    @JohnMa I think the differentiability of $f$ should be assumed. The author assumes it before he states the lemma, although I can't understand about it. Thank you for your comment. – 2017-01-18
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    Remark: $Df(x)$ is a linear map $\mathbb R^n\to\mathbb R^m$, $f(x)$ is an element of $\mathbb R^m$, so $Df(x)\cdot f(x)$ a priori makes no sense. They probably mean $Df(x)^T f(x)$. The proof in the paper certainly is very suspect! – 2017-01-18

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The Theorem is false.

Start with $g(x)=1+x^{4/3}$ and let $G:\Bbb R\to \Bbb R$ be a smooth function so that $G$ is constant $1$ on $[-1,1]$ and $0$ outside of $[-2,2]$. Our function $f$ will be

$$f=g\cdot G$$

$f$ is Lipschitz continuous. This follows from it being a continuously differentiable function $\Bbb R\to\Bbb R$ with bounded derivative (boundedness of the derivative follows from compactness of the support). But the construction $F=f' f$ gives for $x\in(-1,1)$: $$F(x)=\frac43 x^{1/3}(1+x^{4/3})$$ for which there exists no Lipschitz constant in any neighbourhood of $0$. Suppose you have a Lipschitz constant $L$, then for $x>0$, $x<1$ you get:

$$Lx≥F(x)-F(0)=\frac43 x^{1/3}(1+x^{4/3})≥\frac43x^{1/3}$$ and it follows $$L≥\frac43 x^{-2/3}$$ for all such $x$, which is impossible since the expression on the right is unbounded.

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    Thanks for your answer. Should I throw the paper away..? ............. – 2017-01-20
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    Some comments: I read parts of the paper, the author is interested in maps $f$ $\Bbb R^n\to \Bbb R^n$, so the previous comment about needing to take a transpose is not relevant. The author is also only interested in such functions that result in periodic solutions to the equation $\dot x= f(x)$ so one dimensional examples are certainly out. I think the example I gave can be modified in a suitable way to also find a counterexample to the lemma with these conditions. – 2017-01-20
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    As to what to do with the paper, that depends on why you were interested in it. If you need the results you should ask your advisor whether you should try to prove it yourself/find counter-examples to the main result of the paper @user381027. – 2017-01-20
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    Well, I want to introduce some results to people in my academic club (I'm an undergraduate student) about the problem regarding the minimal period of Lipschitzian DEs. I think I should look for other papers first. Thanks for your help. – 2017-01-20