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 $\lim_{n\rightarrow\infty}{f_{n}} = g$ w.r.t the supnorm.

SO

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

Let xε[a,b]

Let ε>0

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

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

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

Hence.

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

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

Thus

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

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

for all x:xε[a,b]======> $\lim_{n\rightarrow\infty}{f_{n}(x)} = g(x)$.................................................. .......................................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: $n\geq N,m\geq N$=====> $|f_{n}(x)-f_{m}(x)|$<ε/2................................................. ...........................

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

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

Τherefor { $f_{n}$} 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 { $f_{n}$} converges to g w.r.t the supnorm.

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

Let $n\geq N$

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

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