TITLE: Space of lipschitz functions form a Banach space
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.