Proving a Lipschitz function is continuous A function $f:D\subset \mathbb R \to \mathbb R$
is lipschitz given that there exists a $L\gt0$ such that $|f(x)-f(y)|\le L|x-y|$
I need to prove this function is then continuous. Is there a best definition of continuous functions to use for this proof given the definition of Lipschitz?
Is the epsilon delta way the only way to do it? can someone help me set it up? I really struggle with implementing that definition.
 A: Yes, every Lipschitz function is continuous. Let, $\epsilon>0$ be arbitrary, then choose $\delta=\frac{\epsilon}{L}$ we have, 
$$|f(x)-f(y)|<\epsilon\text{ whenever,  }|x-y|<\delta$$
A: You can also use the sequence-based characterization of continuity without an $\epsilon$-$\delta$ argument:


*

*$f$ is continuous on $D$, if and only if for all $x \in D$ and for all sequences $\left(x_n \right)_{n\in\mathbb{N}}$ with $\lim_{n \to \infty}x_n = x$ we have
$$\lim_{n \to \infty}f(x_n) = f(x) \Leftrightarrow |f(x_n) - f(x)|\stackrel{n \to \infty}{\longrightarrow} 0$$
Now, continuity follows from Lipschitz-continuity immediately as follows:
$$0\leq |f(x_n) - f(x)| \leq L|x_n-x| \stackrel{n \to \infty}{\longrightarrow} 0$$
A: Following @Sujit's nearly correct outline, suppose $f$ is Lipschitz continuous: 
$$|f(x) - f(y)| < L|x-y|$$
and let $\epsilon > 0$ be arbitrary and choose $\color{red}{\delta = \frac{\epsilon}{L}}$. Then
$$\begin{align}
|x-y| < \delta &\implies |x-y| < \frac{\epsilon}{L}\\
& \implies L|x-y| < \epsilon\\
& \implies \underbrace{|f(x) - f(y)| < L|x-y|}_{supposition} < \epsilon\\
& \implies |f(x) - f(y)| < \epsilon
\end{align}$$
$$ \square$$
A: Guide:


*

*We want to show that $x \to y$, then $f(x) \to f(y)$. Having this in mind, look at the Lipschitz condition again.

*If you are not comfortable with the above, perhaps let $x=y+\delta$ and then let $\delta \to 0$.
