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