Showing space of functions with lipschitz norm is complete I have a Banach space, $X$, given by all the complex valued functions $x: [-1,1] \to \mathbb{C}$ where $x(0) = 0$. And I've shown that the following defines a norm on $X$:
$$\|x\| = inf \{ \beta : |x(s)-x(t)| \le \beta |s-t| \}$$
I'm now struggling to show that this space is complete. So far I'm considering a cauchy sequence $x_n(t)$ and then I can use the norm to say that for all $s,t$,  $|x_n(s) + x_n(t) - x_m(s) + x_m(t)| \to 0$ but I'm not really sure that use that is,
Thanks
 A: The sketch of proof is always the same for these questions:


*

*if $(x_n)$ is Cauchy sequence, then for each $u$, $(x_n(u))$ is Cauchy sequence.

*define $x(u) = \lim x_n(u)$. Then show that $x$ is Lipschitz continuous (pass to the limit in the definition of the sequence).

*finally, prove that $x_n \to x$ (pass to the limit is the definition of: $(x_n(u))$ is a Cauchy sequence).

A: It also follows quite nicely from showing $x_n \rightarrow x$ first. 

1. As $(x_n)$ is Cauchy, exists $N$ such that for all $n,m \ge N$, 
$$ |(x_n - x_m)(u) - (x_n- x_m)(t) | \le \epsilon |u-t| $$ 
for al $u,t \in [-1,1]$. Letting $t = 0$ we obtain a Cauchy sequence $(x_n(u))$ allowing us to define $x(u) := \lim_{n \rightarrow \infty} x_n(u)$. By limiting $m \rightarrow \infty$, we obtain that $x_n \rightarrow x$ uniformly.   

2. Thus, 
$$| (x_n - x) |_u^t |  \le | (x_n - x_m)|_u^t | + |(x_m - x ) |_u^t | \\
 \le \epsilon |u-t| + |(x_m - x ) (u)| + |(x_m -x) (t) |   $$
By limiting $m \rightarrow \infty$ and uniform convergence derived previously, we obtain
$$ |(x_n- x)|_u^t| \le \epsilon |x-t| \Rightarrow || x_n - x ||_{lip} \le \epsilon$$ 
So by limiting $n$ to infinity, we have $|| x_n -x ||_{lip} \rightarrow 0 $. 

3. From above, we know that $x_n - x \in X$ for all $n$, and as the space is a vector space, $x = x_n + (x- x_n) \in X$. 
