Lipschitz continuity implies continuity

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)?

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

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?

• 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... – Conny Dago Dec 6 '18 at 20:50
• 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. – Matt A Pelto Dec 7 '18 at 6:50

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.

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 $$\mathbb R$$.