Proving a Lipschitz function is continuous

Solution 1:

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 $$\begin{align} |x-y| < \delta &\implies |x-y| < \frac{\epsilon}{L}\\ & \implies L|x-y| < \epsilon\\ & \implies \underbrace{|f(x) - f(y)| < L|x-y|}_{supposition} < \epsilon\\ & \implies |f(x) - f(y)| < \epsilon \end{align}$$ $$ \square$$

Solution 2:

Yes, every Lipschitz function is continuous. Let, $\epsilon>0$ be arbitrary, then choose $\delta=\frac{\epsilon}{L}$ we have, $$|f(x)-f(y)|<\epsilon\text{ whenever, }|x-y|<\delta$$

Solution 3:

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: $$0\leq |f(x_n) - f(x)| \leq L|x_n-x| \stackrel{n \to \infty}{\longrightarrow} 0$$

Solution 4:

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 $\delta \to 0$.