0
$\begingroup$

Consider a Lipschitz function from $\mathbb{R}\to\mathbb{R}$. Can we say that $\lim_{x\to\infty}\frac{f(x)}{|x|}=0$. Can we also say that $f$ is differentiable. Continuity is quite evident. But linear order is hard to come by. Any hints? Thanks beforehand.

$\endgroup$
  • 1
    $\begingroup$ @user539887 thanks. edited the post $\endgroup$ – vidyarthi Dec 10 '18 at 7:30
0
$\begingroup$

$f(x)=x$ is a counterexample for the first part. Lipschitz functions need not be differentiable at every point but they are differentiable almost everywhere.

$\endgroup$
0
$\begingroup$

No, consider the function $f(x)=|x|$ which is Lipschitz in $\mathbb{R}$ (with constant $1$), not differentiable at $0$, and $$\lim_{x\to\pm\infty}\frac{f(x)}{|x|}=\pm 1.$$

$\endgroup$
  • $\begingroup$ @vidyarthi Are you still interested in this question? $\endgroup$ – Robert Z Jan 16 at 17:34

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.