The following proof is a rather detailed one.

Let c[a,b] denote the space of continuous ,real valued function, over [a,b].

To show that c[a,b] is complete w.r.t the supnorm we must show that:

Given a cauchy sequence of continuous functions over [a,b] we must show that ,there exists a function ,g belonging to c[a,b] and $\displaystyle \lim_{n\rightarrow\infty}{f_{n}} = g$ w.r.t the supnorm.

SO

Let {$\displaystyle f_{n}$} be a cauchy sequence

Let xε[a,b]

Let ε>0

SINCE {$\displaystyle f_{n}$}is cauchy in c[a,b] there exists N belonging to natural Nos and such that:

for all n.m: $\displaystyle n\geq N$ and $\displaystyle m\geq N$====> sup{$\displaystyle |f_{n}(x)-f_{m}(x)|$:xε[a,b]}< ε.

Because $\displaystyle |f_{n}(x)-f_{m}(x)|$ε{$\displaystyle |f_{n}(x)-f_{m}(x)|$:xε[a,b]}======>$\displaystyle |f_{n}(x)-f_{m}(x)|$<ε

Hence.

{for all ε>0 there exists NεN AND SUCH THAT ,for all n,m: $\displaystyle n\geq N,m\geq N$=====> $\displaystyle |f_{n}(x)-f_{m}(x)|<\epsilon$}................................................. ...............................1

Now (1)====> {$\displaystyle f_{n}(x)$} is a cauchy sequence in real Nos R ====> THERE EXISTS a unique yεR AND $\displaystyle \lim_{n\rightarrow\infty}{f_{n}(x)} = y$

Thus

for all x :xε[a,b]====>THERE EXISTS a unique yεR AND $\displaystyle \lim_{n\rightarrow\infty}{f_{n}(x)} = y$

That defines a function g:[a,b]------->R and such that:

for all x:xε[a,b]======>$\displaystyle \lim_{n\rightarrow\infty}{f_{n}(x)} = g(x)$.................................................. .......................................2

So up to now we have proved that there exists ,g:[a,b]------>R.

Now we need to show that gεc[a,b] i.e g is continuous over [a,b]

Since we have the theorem that states :uniform convergence of continuous functions to a function implies that the function is continuous we must show uniform convergence.

Let ε>0

by (1) there exists NεN AND such that:

for all n,m:$\displaystyle n\geq N,m\geq N$=====>$\displaystyle |f_{n}(x)-f_{m}(x)|$<ε/2................................................. ...........................

and if $\displaystyle n\geq N$ and xε[a,b] by (2) we have $\displaystyle \lim_{n\rightarrow\infty}{f_{n}(x)} = g(x)$ ======>g(x) is an accumulation point=======>there exists $\displaystyle m\geq N$ and $\displaystyle |f_{m}(x)-g(x)|$<ε/2

and hence $\displaystyle |f_{n}(x)-f_{m}(x)|$<ε/2 and $\displaystyle |f_{m}(x)-g(x)|$<ε/2 ====> $\displaystyle |f_{n}(x)-g(x)|$<ε (by the triangular inequality) .

Τherefor {$\displaystyle f_{n}$} converges uniformly to g and so gεc[a,b]

Sofar we have proved:there exists a gεc[a,b].

So it remains the proof that {$\displaystyle f_{n}$} converges to g w.r.t the supnorm.

Let ε>0
by uniform convergence ,there exists NεN AND such that ,
for all n,x: $\displaystyle n\geq N$ and xε[a,b]======>$\displaystyle |f_{n}(x)-g(x)|$<ε/2.

Let $\displaystyle n\geq N$

because yε{$\displaystyle |f_{n}(x)-g(x)|$:xε[a,b]}======>y<ε/2=====> the set {$\displaystyle |f_{n}(x)-g(x)|$:xε[a,b]} is bounded above by ε/2=====>sup{$\displaystyle |f_{n}(x)-g(x)|$:xε[a.b]}$\displaystyle \leq\frac{\epsilon}{2}$<ε

AND so c[a,b] is complete w.r.t the supnorm