0
$\begingroup$

Say $\vert f(x)-f(y)\vert \le L\vert x-y\vert$. How to prove the following: $\forall \lim_{n \to \infty} x_n = x_0 \wedge x_0\in \Bbb{R}$: $\lim_{n \to \infty}f(x_n) = f(x_0)$?

In other words: how to prove that Lipschitz-continuity implies regular continuity(without using differentiation or similar methods)?

$\endgroup$
1
$\begingroup$

$x_n\rightarrow x \Leftrightarrow \forall \varepsilon >0 \,\exists N_0: n\ge N_0 \Rightarrow |x-x_n|\le \varepsilon$

Now if $f$ is Lipshitz and $x_n\rightarrow x$ this implies that

$$|f(x)-f(x_n)|\le L|x-x_n|\le L\varepsilon$$ for such a sequence.

Can you finish the proof with this information?

$\endgroup$
  • $\begingroup$ No, unfortunately. I'm thinking of $\lim_{n \to \infty} \vert f(x_0)-f(x_n)\vert \le \lim_{n \to \infty} \vert x_0-x_n \vert $ But I don't think it's gonna lead me further... $\endgroup$ – Conny Dago Dec 6 '18 at 20:50
  • $\begingroup$ Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument. $\endgroup$ – Matt A Pelto Dec 7 '18 at 6:50
1
$\begingroup$

hint

$$|f(x_n)-f(x_0)|\le L|x_n-x_0|$$

$$\implies$$

$$f(x_0)-L|x_n-x_0|\le f(x_n)\le f(x_0)+L|x_n-x_0|$$

now squeeze.

$\endgroup$
0
$\begingroup$

Assuming $L>0$ (otherwise the function $f$ is constant which is trivial):

Let $\{x_n\}_{n=1}^\infty$ be a sequence of real numbers such that $\lim_{n \to \infty} x_n=x_0$.

Let $\varepsilon>0$ be given, and define $\varepsilon':=\min\{\varepsilon, \frac{\varepsilon}L\}$. Since $x_n \to x_0$ as $n \to \infty$ and $\varepsilon'>0$, we may find a positive integer $N$ so that $|x_n-x_0|<\varepsilon'$ whenever $n \geq N$. Thus we have $$ |\, f(x_n)-f(x_0)|\leq L|x_n-x_0| < L\varepsilon' \leq \varepsilon \text{ whenever } n \geq N.$$

Therefore, $f$ is sequentially continuous on $\mathbb R$.

$\endgroup$

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.