The following is a question from my homework assignment:
A function $f : [1, 1] → \mathbb{R}$ is said to be a Lipschitz function if there is a constant $L$ such that $$|f(s) − f(t)| \leq L|s − t|$$ for all $s, t ∈ [−1, 1]$.
For such a function we define $Lip(f)$ to be the smallest constant $L$ such that the above inequality holds. Let $Z$ be the real vector space consisting of all Lipschitz functions on $[−1, 1]$ and $X$ be the subspace of $Z$ consisting of those $f$ for which $f(0) = 0$.
$(i)$ Show that setting $||f||_{Lip} = Lip(f)$ defines a norm on $X$. Does the same formula define a norm on $Z$?
My attempt We basically have to prove the $3$ axioms which are needed for a norm.
$(N0)$ The norm is obviously finite by definition;
$(N1)$ It's also non-negative
$(N2)$ Scaling is also obvious
$(N3)$ The triangle inequality follows from summing up the $2$ definitions of Lipschitz continuity and applying the triangle inequality for the modulus.
My problem is that I've never used that fact that $f(0)=0$ and the norm would seem to be applicable to $Z$, too. However I read a bit about the Lipschitz norm in general and it always includes $|f(0)|$ as a term so this makes me question the validity of my solution.
Any advice is appreciated.
Edit1 I've also checked $||f||_{Lip}=0 \implies f=0$
Edit2 Thanks to the kind person in the comments section I realised where the mistake was