To show a metric space is complete

Question

Question: Consider the set of functions L such that the functions are Lipschitz. That is: $$|f(x)- f(y)| \leq K|x-y|$$. Consider the metric $$\rho (f_1,f_2) = \sum_{j=1}^\infty 2^{-j} \text{sup}_{x \in [-j,j]} |f_1(x)-f_2(x)|$$ Show that $L$ is a complete metric space with metric $\rho$.

Attempt: Assume a Cauchy sequence. Then $$\rho(f_n,f_m)< \epsilon$$ This means $$|f_n(x)-f_m(x)|\leq \epsilon$$ Thus its Uniformly Cauchy and $f_n \rightarrow f$ uniformly and we are done.

Is this correct? I am not sure about the step where $\sum_{j=1}^\infty 2^{-j} \text{sup}_{x \in [-j,j]} |f_1(x)-f_2(x)|\leq \epsilon \implies |f_n(x)-f_m(x)|\leq \epsilon$

Answer 1

Let every $f_n$ be $K$-Lipschitz and let $(f_n)_n$ be a $\rho$-Cauchy sequence. For each $j\in \Bbb N$ and for all $m,m'$ we have $$\sup_{|x|\leq j}|f_m(x)-f_{m'}(x)|\leq 2^j\rho (f_m,f_{m'})$$ and we have $$\lim_{n\to \infty} \sup_{n\leq m<m'} \rho (f_m,f_{m'})=0.$$ So for each $j\in \Bbb N $ we have $$\lim_{n\to \infty}\sup_{n\leq m<m'}\sup_{|x|\leq j}|f_m(x)-f_{m'}(x)|=0 .$$

So the sequence $(f_n)_n$ is uniformly convergent on each $[-j,j]$ to a function $g_j:[-j,j]\to \Bbb R.$

Since (obviously) $g_j(x)=g_k(x)$ when $|x|\leq|j<k,$ we can let $f(x)=g_j(x)$ for $|x|\leq j.$ Then $(f_n)_n$ converges pointwise to $f$ on $\Bbb R$ and $(f_n)_n$ converges uniformly to $f$ on each $[-j,j].$

Since every $f_n$ is $K$-Lipschitz, it suffices to use the uniform convergence on each $[-j,j]$ to show that $f$ restricted to any $[-j,j]$ is also $K$-Lipschitz. This implies that $f$ is $K$-Lipchitz on $\Bbb R.$

And there is one more step: Show that $\lim_{n\to \infty}\rho(f,f_n)=0.$

Remark. Because of the terms $2^{-j}$ in the def'n of $\rho,$ I think there must be examples where $(f_n)_n$ does not converge uniformly to $f$ on all of $\Bbb R.$