How to show first property of norm?

Question

I have a normed space X which is the Space of all real-valued Lipschitz continuous functions on [0,1]. The norm is. $$ {\lVert x\rVert}_{lip} =\lvert x(0)\rvert + sup_{s \ne t} \lvert { {x(s) -x(t)} \over {s -t}} \rvert $$ I have a trouble to prove first norm property $$ {\lVert x\rVert}_{lip} = 0 \iff x = 0 $$ It can be shown that $ x(0) = 0$ easily but How can I show that $ x(s) = 0$ and $x(t) = 0$ iff $ {\lVert x\rVert}_{lip} = 0$

Answer 1

Assume $||x||_{Lip} = 0$, for every $\epsilon > 0$ it holds $$\Big| \frac{x(s) - x(t)}{s-t}\Big| < \epsilon$$ for every $s \neq t$. Taking $t = 0$ we get $|x(s)| < \epsilon |s| \leq \epsilon$ for every $s \in (0,1]$. Since $\epsilon$ is arbitrary $x(s) = 0$ for every $s \in [0,1]$