# Does this condition imply that $f$ is locally Lipschitz or Lipschitz?

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.

• @Masacroso I disagree: here $\delta$ does not depend on $x$. I think that this is equivalent for $f$ to be Lipschitz. – Crostul Apr 15 '18 at 18:52
• @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 $x,y \in \mathbb{R}$, $|f(x) - f(y)| < M|x-y|$. – Frederic Chopin Apr 15 '18 at 19:02
• This is not the definition of Lipschitz, however it is equivalent to being Lipschitz. See my answer below. – Crostul Apr 15 '18 at 19:03

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