Lipschitz continuity implies continuity

Question

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)?

Answer 1

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.

Answer 2

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

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?

Answer 3

Assume that the $f$ is not continuous. Hence, it must be that for some $c$:

$$\lim _{n\to \infty} |f(x_n) - f(c)| \not = 0 $$

where $x_n \to c$ as $n\to \infty$ . Hence it must be:

$$\lim _{n\to \infty} \dfrac{|f(x_n) - f(c)|}{|x_n - c|} = \infty $$

However this contradicts that $f$ is Lipschitz. Because for all $x_n$, it must be $|f(x_n) - f(c)| \leq L | x_n-c|$ for some real constant $L\ge 0$. As $n\to\infty$, $L$ converges to $\infty$, hence it is not constant.

Contradiction implies that $f$ must be continuous.