Show that $\vert \vert x \vert \vert_{\infty} \leq \vert \vert x \vert \vert_{\operatorname{Lip}}$

Question

Let $X$ be the vector space of Lipschitz-Continuous functions on $[0,1]$

Show that $\vert \vert x \vert \vert_{\infty} \leq \vert \vert x \vert \vert_{\operatorname{Lip}}$

where $\vert \vert x \vert \vert_{\operatorname{Lip}}=\vert x(0)\vert+\sup\limits_{s\neq t}\vert \frac{x(t)-x(s)}{t-s}\vert$

My idea: Let $u \in [0,1]$ arbitrary and since $x$ is a continuous function

$x(u)=x(0)+\frac{x(u)-x(0)}{u-0}u$ (**) and therefore

$\vert x(u)\vert\leq\vert x(0)\vert+\vert\frac{x(u)-x(0)}{u-0}\vert u\leq \vert x(0)\vert+\vert\frac{x(u)-x(0)}{u-0}\vert\leq \vert x(0)\vert +\sup\limits_{s\neq t}\vert \frac{x(t)-x(s)}{t-s}\vert=\vert \vert x \vert \vert_{\operatorname{Lip}}$

I think I have the right idea of the proof but I am not sure on (**) and would like to see if my proof is correct. I have not used Lipschitz-Continuity which makes me believe I am wrong. Thank you for your help.

Answer 1

Your proof is essentially fine. There are two small points I can make.

Firstly $(\ast \ast)$ should really be the expression for $u \neq 0$ since it is undefined at $u = 0$. This is no problem since $|x(0)| \leq \|x\|_{\text{Lip}}$ trivially.

Secondly, whilst it's technically true that you don't use Lipschitz continuity in the sense that all the inequalities hold without that (in the sense that the right hand side will be infinite), you have only said something meaningful when $x$ is Lipschitz since otherwise you just want to prove that $\|x\|_\infty \leq \infty$ which is trivially true.