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I stumbled upon this answer here while studying the proposition that if $f: \mathbb R^n \to \mathbb R^n$ is $C^1$ then $f$ is locally Lipschitz.

The answer in the link applies Taylor's theorem.

And I was wondering if one could alternatively argue using the Fundamental Theorem of Calculus. Concretely,

I wrote this proof and was wondering if someone could tell me if it is correct?

Proof:

Let $f: \mathbb R^n \to \mathbb R^n$ be $C^1$ and let $x_0 \in \mathbb R^n$. Since $f$ is $C^1$ its derivative $f'$ is continuous. Hence $f'$ attains a maximum $L$ on $\overline{B(x_0,1)}$. Then since

$$ f(x) - f(x_0) \stackrel{FTC}{=} \int_{x_0}^x f'(t) dt$$

we have

$$ \|f(x) - f(x_0)\| = \left \| \int_{x_0}^x f'(t) dt \right \| \le \int_{x_0}^x \|f'(t)\| dt \le L \|x-x_0\|$$

hence $f$ is Lipschitz continuous on $B(x_0, 1)$.

Community
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a student
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  • This doesn’t make much sense unless $n =1$; otherwise what do you mean by prime notation? – Jordan Green Jun 21 '15 at 03:10
  • You can fix this by restricting $f$ to the line through $x$ and $x_0$, and applying FTC there. Also, the answer you refer to uses FTC. –  Jun 21 '15 at 03:11
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    @Idisagree Oh, good point. Thank you for your comment. If you post it in an answer then I could upvote and accept since it helped me. – a student Jun 21 '15 at 04:48

2 Answers2

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Since $f\in C^1$, there exists a $\delta>0$, such that for $\|\overrightarrow{x}-\overrightarrow{x_0}\|<\delta$, there is $$ \left\|\dfrac{f(\overrightarrow{x})-f(\overrightarrow{x_0})}{\overrightarrow{x}-\overrightarrow{x_0}}-f'(\overrightarrow{x_0})\right\|<1 $$ Since $$ \left\|\dfrac{f(\overrightarrow{x})-f(\overrightarrow{x_0})}{\overrightarrow{x}-\overrightarrow{x_0}}-f'(\overrightarrow{x_0})\right\|\geqslant\left\|\dfrac{f(\overrightarrow{x})-f(\overrightarrow{x_0})}{\overrightarrow{x}-\overrightarrow{x_0}}\right\|-\left\|f'(\overrightarrow{x_0})\right\| $$ There is $$ \left\|f(\overrightarrow{x})-f(\overrightarrow{x_0})\right\|<(\|f'(\overrightarrow{x_0})\|+1)\|\overrightarrow{x}-\overrightarrow{x_0}\|=L|\overrightarrow{x}-\overrightarrow{x_0}\| $$ So $f$ is locally Lipschitz.

Eugene Zhang
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Well, you can choose a convex neigborhood $V$ of $x\in R^n $. Then, for every $y\in V$ you can define $\lambda(t)=f(x+ty)$, with $t \in [0, 1]$. Compute the $\lambda'(t)$, use the chain rule and the mean value inequality for vector valued functions with domain in $R$.

Micael
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