Space of Lipschitz continuous functions is complete

Question

Let $X$ be set of functions $f:[-1,1]\to \mathbb{C}$ such that $f(0)=0$ and there exists $\alpha>0$ such that

$$ |f(t)-f(s)|\le \alpha |t-s| $$

for all $t,s\in [-1,1]$. Equip $X$ with the norm:

$$ ||f||=\inf \{\alpha:|f(t)-f(s)|\le \alpha |t-s|\} $$

Show that $||\cdot||$ is indeed a norm and that $X$ is complete with respect to this norm

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Showing that $||\cdot||$ is a norm wasn't terribly hard, but I'm stuck with completeness part. So if $x_n$ is Cauchy in $X$, then for given $\epsilon>0$ there is $N$ such that $m,n\ge N$ then

$$ \inf\{\alpha:|(x_n-x_m)(t)-(x_n-x_m)(s)| \le \alpha|t-s|\} <\epsilon $$

in particular

$$ |(x_n-x_m)(t)-(x_n-x_m)(s)| \le \epsilon|t-s| $$

taking $s=0$ then we get

$$ |(x_n-x_m)(t)|\le \epsilon|t|\le \epsilon $$

So $x_n(t)$ is Cauchy in $\mathbb{C}$ and so convergens to $x(t)\in \mathbb{C}$.

It's clear that $x(0)=0$, since it is a limit of sequence of $0$. Further, taking limit $m\to \infty$

$$ |(x_n-x)(t)-(x_n-x)(s)|\le \epsilon |t-s| $$

So it follows $||x_n-x||\le \epsilon$ given $n\ge N$ which shows that $x_n\to x$ in $X$.

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Then I've got some questions: is my reasoning correct up to this point? Also, how could I show that in fact $x\in X$?

Answer 1

A cauchy sequence is bounded (in norm), so

$$\|x_n\| \le C$$

for some $C$, which means

$$|x_n(s) - x_n(t)|\le C|t-s|$$

for all $s, t, n$. Take $n\to \infty$ gives

$$|x(s) - x(t)|\le C|t-s|$$

and so $x\in X$.

Note that this paragraph should be mentioned before you wrote "So it follows $\|x_n-x\| \le\epsilon$"

Otherwise your proof is good.

Answer 2

I think your proof is fine.

From your last inequations,

$$|x(t)-x(s)|-|x_n(t)-x_n(s)|\leq|(x_n−x)(t)−(x_n−x)(s)|\leq \epsilon|t−s|,$$ for some sufficient large fixed $n$. Adding $|x_n(t)-x_n(s)|$ on the both sides gives $$|x(t)-x(s)|\leq\epsilon|t-s|+|x_n(t)-x_n(s)|\leq \epsilon|t-s|+\alpha|t-s|\leq(\epsilon+\alpha)|t-s|,$$ for some $\alpha>0$. Therefore, $x\in X$.