1
$\begingroup$

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 $B(x_0, 1)$.

$\endgroup$
  • $\begingroup$ This doesn’t make much sense unless $n =1$; otherwise what do you mean by prime notation? $\endgroup$ – Jordan Green Jun 21 '15 at 3:10
  • $\begingroup$ 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. $\endgroup$ – user147263 Jun 21 '15 at 3:11
  • 1
    $\begingroup$ @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. $\endgroup$ – a student Jun 21 '15 at 4:48
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.