Can someone point me to a detailed proof of this, or give it?
The stuff I can find online alludes to the triangle inequality, but I can't seem to join the dots...
if is a Cauchy sequence of continuous functions I need to show that they converge (uniformly) to some function g (so that g is continuous), right? So the triangle inequality gives
the first term on the right is arbitrarily small for n & m big enough... but what about the second one? Or am I going about this all wrong?
Let X be a closed interval [a,b] in . Let be a metric space with a sup-norm metric, where denotes the family of continuous, bounded, real-valued functions with domain X and the metric for is defined by , for .
We shall show that a Cauchy sequence in has a limit in , where should converge uniformly to g. Since the sequence is a Cauchy sequence, given , there is an N such that for all m,n > N, we have . It follows that the sequence is a Cauchy sequence of real numbers. Since is complete, the sequence converges where as . In a similar manner, we assign to each a unique . Thus, converges to g(x) uniformly on .
We remain to show that the limit function g is bounded and continuous so that g is indeed in .
We first show that g is bounded. Let M be a positive integer for which implies . Since is bounded and g(x) and differ by no more than 1 for all x in X, g should be bounded also.
We next show that g is continuous. Let be an arbitrary point of X and a positive number. Let N be a positive integer such that if then . Since is continuous at , there is a positive number such that if , then . Thus, for x in X satisfying ,
.
Thus the limit function g is continuous. It follows that is a complete metric space (I'll leave the remaining steps to you, such as checking the above metric is indeed a metric in ).
The domain of continuous functions can further be generalized rather than only. In this case, we can impose restrictions that functions are bounded on X in order to define a metric space with a supremum metric for continuous functions with a domain X.
I found that if the closed interval is assigned as X, the problem is further simplified as you pointed out. We do not need to show that a continuous function defined on is bounded and we don't even need to show that g is continuous (If the 's are continuous on [a,b] and the convergence is uniform, the limit function g is continuous on [a, b]. This theorem is found on p37 of Erwin's introductory functional analysis or other Calculus books).
Any other thing that does not seem to right?
Right ,uniform convergence of the sequence of the continuous functions { },over [a,b], to the functrion,g implies that g is continuous over [a,b]
But we are not done yet it remains to be proved that { } converges to the function ,g w.r.t the supnorm
Also in the first part of your proof you have proved pointwise convergence and not uniform,since you have proved:
For all xε[a.b] and ε>0,there exists an N ,such that :
for all ,n if ,then
and not
Given ε>0 then there exists N ,SUCH that:
For all ,n and for all,x if and xε[a,b],then
That is correct. What aliceinwonderland has proved is that the sequence has a pointwise limit g(x). We also know that is Cauchy for the sup norm . To complete the proof that in the sup norm, you need to argue as follows.
Let . The Cauchy condition tells you that there exists N such that for all m,n>N and for all x in the space. For each fixed x, let in that inequality. That tells you that (for all n>N and for all x). In other words, . This shows that , as required.
well, using is a Cauchy sequence in , I showed that converges to g(x) for all x in ( the way how g is constructed is shown in the above post ). From this, I conclude that with respect to in which implies that converges uniformly to g on X. I should have provided the details of the proof that with respect to in .
Thanks for showing the missing steps of the proof, Opalg.
I thought I finished this problem, but there were new posts in the thread so I read your post again.
I think the above proof shows the uniform convergence of to g in w.r.t sup norm rather than in w.r.t sup norm. It is because is implictly used in (for all n>N and for all x). The essential construction of g such that in w.r.t sup norm has already been shown.
The uniform convergence of to g w.r.t sup norm is not essential part of the proof because I already showed that g is a limit of in and continuous.
As you know, if we exclude the steps of showing that g is continuous in , then we need to show the uniform convergence of to g in w.r.t sup norm (If g is the uniform convergence of continuous functions in , then g is continuous.) This is acheived either by arguing the proof you have shown or prove that in w.r.t sup norm iff converges uniformly to g on X.
Please correct me if I am wrong.
There is no difference between those things. Uniform convergence is the same thing as convergence in the sup norm
There are two different sorts of convergence for : pointwise convergence, and uniform (or sup norm) convergence. What is implicitly used in the proof of (for all n>N and for all x) is the pointwise convergence, which has already been established. Converence in the sup norm has not already been shown at that stage.
On the contrary, the uniform convergence is an essential part of the proof. Without it, only pointwise convergence has been proven.