Space of lipschitz functions form a Banach space

Question

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.

Answer 1

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

Answer 2

To complete Pebeto's answer, one needs to show that $\|f_n-f\|_{Lip_0}\to0$ as $n\to\infty$. Let $\varepsilon>0$. Then there exists $N$ such that $\|f_n-f_m\|_{Lip_0}<\frac\varepsilon2$ for all $m,n\ge N$. Fix $n\ge N$. We will be done if we can show $\|f_n-f\|_{Lip_0}\le\varepsilon$.

Fix $x,y\in X$ with $x\neq y$. There exists $M_{x,y}$ such that $$\frac{|(f_m-f)(x)-(f_m-f)(y)|}{\|x-y\|}<\frac\varepsilon2$$ for all $m\ge M_{x,y}$. Let $m=\max\{N,M_{x,y}\}$. Then

\begin{align*}\frac{|(f_n-f)(x)-(f_n-f)(y)|}{\|x-y\|} &\le \frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{\|x-y\|} + \frac{|(f_m-f)(x)-(f_m-f)(y)|}{\|x-y\|} \\ &\le \|f_n-f_m\|_{Lip_0}+\frac\varepsilon2 \\ &<\varepsilon. \end{align*} Taking suprema of both sides over $x,y$, we find $\|f_n-f\|_{Lip_0}\le\varepsilon$ for all $n\ge N$, completing the proof.