Show that Lipchitz functions space is Banach

Let X be a Banach space. Show that $$L = \{ f:X \to \mathbb{R}: f \text{ is lipschitz and } f(0)=0 \}$$ with norm:

$$||f||_{lip}= sup \left\{ \frac{|f(x)-f(y)|}{||x-y||}; x \neq y \in X \right\}$$

is Banach space.

I want to get a sequence of cauchy in $$L$$ and show that converges, I'm trying to transform into a sequence of cauchy in $$\mathbb{R}$$ but I'm not getting.

• Try first the special case $X = [0, 1] \subset {\mathbb R}$. – avs Mar 14 at 20:11
• This looks related: math.stackexchange.com/questions/790995/…. – Martin R Mar 14 at 20:13
• I always have the same problem. I have a string of $f_n$ cauchy and conclude that for n large enough: $[f_n(x)-f_m(x)]-[f_n(y)-f_m(y)] < \epsilon ||x-y||$ With this I would like to conclude something of the type $|f_n(x)-f(x)|< \epsilon||x-y||$, but i can not. – Lucas Mar 14 at 20:20
• If you have a Cauchy sequence with this norm you can check that for each $x\in X$ the value $f_n(x)$ is also Cauchy and thus must admit a limit. Let $f$ be this pointwise limit, can you find an argument that shows $f_n$ converges to $f$ in your norm? – s.harp Mar 14 at 21:02
• The only step I'm not getting is check that for each x∈X the value fn(x) is also Cauchy. From then on I understand everything. – Lucas Mar 14 at 22:39