Space of lipschitz functions form a Banach space

Solution 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