Hints on showing that a metric space is complete

Question

Let $C[0,K]$ be the space of all continuous real valued functions on $[0,K]$ for $K>0$ and $L\geq0$, equipped with the metric $d$ defined by

$$d(f,g)=\sup_{0\leq k\leq K}e^{-Lk}|f(k)-g(k)|.$$

I am trying to show that $(C[0,K],d)$ is a complete metric space.

My attempt:

Let $(f_n)$ be an arbitrary Cauchy sequence in $(C[0,K],d)$. We need to show that it has a limit in $(C[0,K],d)$.

Then for each $\epsilon>0$, there exists $N$ such that $$m,n\geq N\implies d(f_m,f_n)=\sup_{0\leq k\leq K}e^{-Lk}|f_m(k)-f_n(k)|<\epsilon$$

The part that I struggled is the part where we have to show there is an $f\in(C[0,K],d)$ such that $d(f_n,f)<\epsilon$.

Could somebody please give some hints? Thanks!

Answer 1

Hint: Find an isometry between $(C[0,K],d)$ and $C[0,K]$ (with the standard $\sup$ norm).

Answer 2

I think that there is two ways not using the isometry (I agree that this is a very good solution, surely the simplest) to prove the result.

1) Note that your distance is associated with the norm $N(f)=\sup_{0\leq k\leq K}\exp(-kL)|f(k)|$, i.e $d(f,g)=N(f-g)$. If we note by $\|f|_{\infty}$ the usual Sup-norm, use the fact that $\exp(-KL)\leq \exp(-Lk)\leq 1$ to show that $$\exp(-KL)\|f\|_{\infty}\leq N(f)\leq \|f\|_{\infty}$$ ie the two norms on $C([0,K], \mathbb{R})$ are equivalents.

2) Follow the proof for the fact that $C([0,K], \mathbb{R})$ is complete for the $\sup$-norm:

For your Cauchy sequence for the distance $d$,

a) Fix $k$, show that $f_n(k)$ is a Cauchy sequence in $\mathbb{R}$, hence convergent. Now you can define a function $f(k)$ as the pointwise limit of $f_n(k)$

b) For your $\varepsilon$, $N(\varepsilon)$, fix $k$ and in $\exp(-kL)|f_m(k)-f_n(k)|\leq \varepsilon$, let $m\to +\infty$. Deduce that $\exp(-kL)f(k)$ is continuous, hence also $f$.

c) Deduce from what you have done in b) that $d(f,f_n)\leq \varepsilon$ for $n\geq N$ and you are done.