1
$\begingroup$

Suppose there are constants $\delta > 0$ and $M < \infty$ such that for all $x \in \mathbb{R}$, $|f(x+t)−f(x)| \leq M|t|$ for all $t \in (−\delta,\delta)$. Then is $f$ locally Lipschitz or Lipschitz on $\mathbb{R}$? And if so, why?

Thank you.

$\endgroup$
  • 1
    $\begingroup$ @Masacroso I disagree: here $\delta$ does not depend on $x$. I think that this is equivalent for $f$ to be Lipschitz. $\endgroup$ – Crostul Apr 15 '18 at 18:52
  • $\begingroup$ @Crostul I agree with you that this is not the definition of locally Lipschitz. But is it the definition of Lipschitz? If I recall correctly, $f$ being Lipschitz means that there is an $M < \infty$ such that for all $x,y \in \mathbb{R}$, $|f(x) - f(y)| < M|x-y|$. $\endgroup$ – Frederic Chopin Apr 15 '18 at 19:02
  • $\begingroup$ This is not the definition of Lipschitz, however it is equivalent to being Lipschitz. See my answer below. $\endgroup$ – Crostul Apr 15 '18 at 19:03
3
$\begingroup$

$f$ is Lipschitz.

To prove this, consider arbitrary $x,y \in \Bbb R$ with $x<y$. Divide the interval $[x,y]$ into $N$ small subintervals $[x_i,x_{i+1}]$ (with $x_{i+1}-x_i < \delta$, $x_0=x$ and $x_N=y$).

Then

$$|f(y)-f(x)| = |f(x_N)-f(x_0)| \le \sum_{i=0}^{N-1} |f(x_{i+1})-f(x_i)| \le M \sum_{i=0}^{N-1} |x_{i+1}-x_i| = M|y-x|$$

$\endgroup$
1
$\begingroup$

$f : X \to Y$ is locally Lipschitz if and olny if $$\forall x \in X,\, \exists \delta > 0,\, \exists M > 0 \,\forall t,\, |t| < \delta \Rightarrow |f(x+t) - f(x)| \le M |t|$$ In the definition $\delta$ and $M$ could depend on $x$. In your question they do not depend on $x$ which is stronger.

$\endgroup$

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.