Let $X$ be a Banach space. Show that $L=\{f:X\to\mathbb{R}: f \mbox{ is Lipschitz}, f(0) = 0\}$ with the norm
$$||f||_{Lip_0} = \sup\left\{\frac{|f(x)-f(y)|}{||x-y||}, x\neq y\in X\right\}$$
is a Banach space.
I've found Banach space of p-Lipschitz functions but I did not understand the proof given.
I have a few questions first. Which norm is $||x-y||$?
So I need to prove that every Cauchy sequence in $L$ converges to an element of $L$, right?
In other words, $\forall \epsilon>0$ there exists $n_0$ such that $m,n>n_0\implies ||f_m-f_n||_{Lip_0}<\epsilon$
$$ ||f_m-f_n||_{Lip_0} = \sup\left\{\frac{|(f_m-f_n)(x)-(f_m-f_n)(y)|}{||x-y||}, x\neq y\in X\right\} = \sup\left\{\frac{|f_m(x)-f_m(y)|}{||x-y||}+\frac{f_n(y)-f_n(x)}{||x-y||}, x\neq y\in X\right\}$$
both $f_m$ and $f_n$ are Lipschitz so they're continous, which means something I don't know what.
Best Answer
It would be a good exercise to show that $||f||_{Lip_0}$ is indeed a norm. Otherwise, here are the general steps:
Assume $f_n$ is Cauchy.
1) First step is to show pointwise convergence so we can define a limit. $$|f_m(x) - f_n(x)| = |(f_m-f_n)(x) - (f_m-f_n)(0)| \leq ||f_m-f_n||_{Lip_0} ||x||.$$ This shows that $f_m(x)$ is Cauchy and therefore has a limit, i.e., f(x).
2) It remains to show that $f$ is in $L$. It is obvious that $f(0) = 0$. It therefore only remains to show that $f$ as defined above is Lipschitz. We have:
$$|f_n(x) - f_n(y)| \leq ||f_n||_{Lip_0} ||x-y||.$$
But $f_n$ is Cauchy, hence bounded by say $M$. Hence, $$|f_n(x) - f_n(y)| \leq M ||x-y||.$$ At the limit, we have: $$|f(x) - f(y)| \leq M ||x-y||.$$ Hence $f$ is Lipschitz, and the result is proven