# Thread: Proof - space of continuous functions with sup norm metric is complete

1. ## Proof - space of continuous functions with sup norm metric is complete

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 $(f_n)$ 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

$||f_n - g|| \leq ||f_n - f_m|| + ||f_m - g||$

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?

2. Originally Posted by RanDom
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 $(f_n)$ 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

$||f_n - g|| \leq ||f_n - f_m|| + ||f_m - g||$

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?

RanDom ,there are many continuous function spaces which are complete .

State more clearly which one you consider in your post??

3. Originally Posted by xalk
RanDom ,there are many continuous function spaces which are complete .

State more clearly which one you consider in your post??
He said "sup norm" ^^

which, if I'm not mistaking, means uniformly continuous

4. Originally Posted by xalk
RanDom ,there are many continuous function spaces which are complete .

State more clearly which one you consider in your post??
the space of functions which are continuous on an interval, say [0,1]

With the sup norm / L infinity metric... i.e

$||f-g|| = sup\{|f(x)-g(x)|:x \in C([0,1])\}$

5. Let X be a closed interval [a,b] in $\mathbb{Re}$. Let $C(X, \mathbb{R})$ be a metric space with a sup-norm metric, where $C(X, \mathbb{Re})$ denotes the family of continuous, bounded, real-valued functions with domain X and the metric $\rho$ for $C(X, \mathbb{R})$ is defined by $\rho(f,g)=||f - g|| = sup\{|f(x) - g(x)| : x \in X\}$, for $f,g \in C(X, \mathbb{Re})$.

We shall show that a Cauchy sequence $\{f_n\}_{n=1}^{\infty}$ in $C(X, \mathbb{R})$ has a limit $g:X \rightarrow \mathbb{Re}$ in $C(X, \mathbb{R})$, where $\{f_n\}_{n=1}^{\infty}$ should converge uniformly to g. Since the sequence $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequence, given $\epsilon > 0$, there is an N such that for all m,n > N, we have $||f_m - f_n || < \epsilon$. It follows that the sequence $\{f_n(x)\}_{n=1}^{\infty}$ is a Cauchy sequence of real numbers. Since $\mathbb{Re}$ is complete, the sequence converges $f_m(x_0) \rightarrow g(x_0)$ where $x_0 \in X$ as $m \rightarrow \infty$. In a similar manner, we assign to each $t \in X$ a unique $g(t)$. Thus, $\{f_n(x)\}_{n=1}^{\infty}$ converges to g(x) uniformly on $X$.

We remain to show that the limit function g is bounded and continuous so that g is indeed in $C(X, \mathbb{R})$.

We first show that g is bounded. Let M be a positive integer for which $n \geq M$ implies $\rho(f_n, g) = || f_n - g || < 1$. Since $f_M$ is bounded and g(x) and $f_M(x)$ differ by no more than 1 for all x in X, g should be bounded also.
We next show that g is continuous. Let $x_0$ be an arbitrary point of X and $\epsilon$ a positive number. Let N be a positive integer such that if $n \geq N$ then $\rho(f_n,g)=|| f_n - g || < \frac{\epsilon}{3}$. Since $f_N$ is continuous at $x_0$, there is a positive number $\delta$ such that if $| x - x_0 | < \delta$, then $| f_N (x) - f_N(x_0) | < \frac{\epsilon}{3}$. Thus, for x in X satisfying $| x - x_0 | < \delta$,

$| g(x) - g(x_0) | \leq | g(x) - f_N(x) | + | f_N(x) - f_N(x_0) | + | f_N(x_0) - g(x_0) | < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$.

Thus the limit function g is continuous. It follows that $( C(X, \mathbb{R}), \rho)$ is a complete metric space (I'll leave the remaining steps to you, such as checking the above metric $\rho$ is indeed a metric in $C(X, \mathbb{R})$).

6. A CONTINOUS function over a combact set in a metric or normed space is bounbed.

So we do not have to show that the function is bounded and continuous

7. Originally Posted by xalk
A CONTINOUS function over a combact set in a metric or normed space is bounbed.

So we do not have to show that the function is bounded and continuous
The domain of continuous functions can further be generalized rather than $[a,b] \subset \mathbb{Re}$ 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 $[a,b] \subset \mathbb{Re}$ 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 $C(X, \mathbb{Re})$ is bounded and we don't even need to show that g is continuous (If the $f_m$'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?

8. Right ,uniform convergence of the sequence of the continuous functions { $f_{n}$},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 { $f_{n}$} 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 $n\geq N$,then $|f_{n}(x)-g(x)|<\epsilon$

and not

Given ε>0 then there exists N ,SUCH that:

For all ,n and for all,x if $n\geq N$ and xε[a,b],then $|f_{n}(x)-g(x)|<\epsilon$

9. Originally Posted by xalk
Right ,uniform convergence of the sequence of the continuous functions { $f_{n}$},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 { $f_{n}$} 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 $n\geq N$,then $|f_{n}(x)-g(x)|<\epsilon$

and not

Given ε>0 then there exists N ,SUCH that:

For all ,n and for all,x if $n\geq N$ and xε[a,b],then $|f_{n}(x)-g(x)|<\epsilon$
I think I tried the proof that you mentioned (read a construction part of g). If I am not mistaken, after I read your reply, I found that your understanding of the above problem is not good enough.

Why don't you post your proof?

10. Originally Posted by xalk
Right ,uniform convergence of the sequence of the continuous functions { $f_{n}$},over [a,b], to the function,g implies that g is continuous over [a,b]

But we are not done yet it remains to be proved that { $f_{n}$} converges to the function ,g w.r.t the supnorm
That is correct. What aliceinwonderland has proved is that the sequence $(f_n(x))$ has a pointwise limit g(x). We also know that $(f_n)$ is Cauchy for the sup norm $\|.\|_\infty$. To complete the proof that $f_n\to g$ in the sup norm, you need to argue as follows.

Let $\varepsilon>0$. The Cauchy condition tells you that there exists N such that $|f_n(x)-f_m(x)|<\varepsilon$ for all m,n>N and for all x in the space. For each fixed x, let $m\to\infty$ in that inequality. That tells you that $|f_n(x) - g(x)|\leqslant\varepsilon$ (for all n>N and for all x). In other words, $n>N\:\Rightarrow\: \|f_n-g\|_\infty \leqslant\varepsilon$. This shows that $\|f_n-g\|_\infty\to0$, as required.

11. well, using $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequence in $( C(X, \mathbb{Re}), \rho)$ , I showed that $\{f_n(x)\}_{n=1}^{\infty}$ converges to g(x) for all x in $X$( the way how g is constructed is shown in the above post ). From this, I conclude that $lim_{n \rightarrow \infty}f_n = g$ with respect to $\rho$ in $( C(X, \mathbb{Re}), \rho)$ which implies that $f_n$ converges uniformly to g on X. I should have provided the details of the proof that $lim_{n \rightarrow \infty}f_n = g$ with respect to $\rho$ in $( C(X, \mathbb{Re}), \rho)$.

Thanks for showing the missing steps of the proof, Opalg.

12. Originally Posted by Opalg

Let $\varepsilon>0$. The Cauchy condition tells you that there exists N such that $|f_n(x)-f_m(x)|<\varepsilon$ for all m,n>N and for all x in the space. For each fixed x, let $m\to\infty$ in that inequality. That tells you that $|f_n(x) - g(x)|\leqslant\varepsilon$ (for all n>N and for all x). .

You mean that:

$\lim_{m\rightarrow\infty}{(|f_{m}(x)-f_{n}(x)|)} = |f_{n}(x)-g(x)|$ ??

13. Originally Posted by xalk
You mean that:

$\lim_{m\rightarrow\infty}{(|f_{m}(x)-f_{n}(x)|)} = |f_{n}(x)-g(x)|$ ??
Yes.

14. Originally Posted by Opalg
That is correct. What aliceinwonderland has proved is that the sequence $(f_n(x))$ has a pointwise limit g(x). We also know that $(f_n)$ is Cauchy for the sup norm $\|.\|_\infty$. To complete the proof that $f_n\to g$ in the sup norm, you need to argue as follows.

Let $\varepsilon>0$. The Cauchy condition tells you that there exists N such that $|f_n(x)-f_m(x)|<\varepsilon$ for all m,n>N and for all x in the space. For each fixed x, let $m\to\infty$ in that inequality. That tells you that $|f_n(x) - g(x)|\leqslant\varepsilon$ (for all n>N and for all x). In other words, $n>N\:\Rightarrow\: \|f_n-g\|_\infty \leqslant\varepsilon$. This shows that $\|f_n-g\|_\infty\to0$, as required.
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 $\{f_n\}_{n=1}^{\infty}$ to g in $C(X, \mathbb{R})$ w.r.t sup norm rather than $f_n\to g$ in $C(X, \mathbb{Re})$ w.r.t sup norm. It is because $lim_{m \rightarrow \infty}f_m = g$ is implictly used in $|f_n(x) - g(x)|\leqslant\varepsilon$ (for all n>N and for all x). The essential construction of g such that $lim_{n \rightarrow \infty}f_n = g$ in $C(X, \mathbb{Re})$ w.r.t sup norm has already been shown.

The uniform convergence of $\{f_n\}_{n=1}^{\infty}$ to g w.r.t sup norm is not essential part of the proof because I already showed that g is a limit of $\{f_n\}_{n=1}^{\infty}$ in $C(X, \mathbb{Re})$ and continuous.

As you know, if we exclude the steps of showing that g is continuous in $C(X, \mathbb{R})$, then we need to show the uniform convergence of $\{f_n\}_{n=1}^{\infty}$ to g in $C(X, \mathbb{R})$ w.r.t sup norm (If g is the uniform convergence of continuous functions $\{f_n\}$ in $C(X,\mathbb{Re})$, then g is continuous.) This is acheived either by arguing the proof you have shown or prove that $lim_{n \rightarrow \infty}f_n = g$ in $C(X, \mathbb{Re})$ w.r.t sup norm iff $\{f_n\}_{n=1}^{\infty}$ converges uniformly to g on X.

Please correct me if I am wrong.

15. Originally Posted by aliceinwonderland
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 $\{f_n\}_{n=1}^{\infty}$ to g in $C(X, \mathbb{R})$ w.r.t sup norm rather than $f_n\to g$ in $C(X, \mathbb{Re})$ w.r.t sup norm.
There is no difference between those things. Uniform convergence is the same thing as convergence in the sup norm

Originally Posted by aliceinwonderland
It is because $lim_{m \rightarrow \infty}f_m = g$ is implicitly used in $|f_n(x) - g(x)|\leqslant\varepsilon$ (for all n>N and for all x). The essential construction of g such that $lim_{n \rightarrow \infty}f_n = g$ in $C(X, \mathbb{Re})$ w.r.t sup norm has already been shown.
There are two different sorts of convergence for $f_n\to g$: pointwise convergence, and uniform (or sup norm) convergence. What is implicitly used in the proof of $|f_n(x) - g(x)|\leqslant\varepsilon$ (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.

Originally Posted by aliceinwonderland
The uniform convergence of $\{f_n\}_{n=1}^{\infty}$ to g w.r.t sup norm is not essential part of the proof because I already showed that g is a limit of $\{f_n\}_{n=1}^{\infty}$ in $C(X, \mathbb{Re})$ and continuous.
On the contrary, the uniform convergence is an essential part of the proof. Without it, only pointwise convergence has been proven.

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