# Proof of Arzela's theorem

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## Homework Statement

So, as the title suggests, I'd like to check a few ideas, since I didn't yet complete the proof.

Theorem. Let X be compact, and let fn be a sequence of functions in C(X, R^k). If the collection {fn} is pointwise bounded and equicontinuous, then the sequence fn has a uniformly convergent subsequence.

## The Attempt at a Solution

First, I am trying a proof for k = 1, I'll comment on the proof of the other cases after I prove this.

Let x be in X. Then the sequence {fn(x) : n is in N} is bounded, and hence has a convergent subsequence, whose limit we'll deonte with Lx, so fn(x) --> Lx. Define the function f : X --> R with f(x) = Lx. I see a way to prove that fn --> f uniformly, but to prove this, I'd need to prove that f is continuous.

Of course, there's no reason to believe it is, although I got a hunch (probably wrong).

And of course, I know I need not prove continuity of a function to which fn converges uniformly, since this function must be continuous, by the uniform limit theorem.

But the concept of the proof probably includes an idea of how to define a specific function to which fn converges uniformly.

I don't require any hints yet, but I only want to see if there is something to this idea I've lain out above (i.e. defining f in this specific way, although by standard means I can't show continuity, since the condition depends on the integers related to convergence of the functions fn)

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OK, I won't give any hints yet, but I'll comment on your attempt.

You have not been very rigorous with your notation, and this doesn't bother me normally, but here it causes a small problem.

You say: for every x, there exists a subsequence $$f_{k_n}$$ such that blablabla. The problem here is that for every x, we potentially have a different subsequence.

For example: say that for $$x_0$$, we have the subsequence $$f_{2n}$$. And for $$x_1$$, we have the subsequence $$f_{2n+1}$$. Then there is no way to put these subsequences together...

I hope I am clear on this, but this is the biggest problem with your proof. If you can solve this, then I believe your proof works. But I think that solving the proof might be a little difficult...

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Yes, I'm aware my notation wasn't quite consistent... OK, I'll think about it a bit more and see if I get anywhere.

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Hmm, something potentially simple just occured to me, could we apply Lemma 45.3. here to X and C(X, fn(X)), since compactness of X and continuity of fn implies compactness of fn(X)? Could we, instead of continuous functions from X to R simply look at continuous functions of X to fn(X), i.e. if f is in the collection {fn}, consider the restriction of the codomain f : X --> f(X)? Does this make any sense?

Although, I'm not sure if it would help, since we could get total boundedness of {fn] by this Lemma here, which would imply boundedness, but I'm not sure if this would imply existence of a convergent subsequence, and uniformity is another cup of tea then...and, of course, we wouldn't use pointwise boundedness here! (at least not at first glance)

Good suggestion. However you put $$f_n(X)=Y$$ in lemma 45.3, this means that for every n, we have a different space Y. But lemma 45.3 doesn't allow for a dependence of n.

You could solve this by putting $$Y=\bigcup_{n}{f_n(X)}$$, but then this wouldn't be a compact space anymore.

Arzela's theorem is often called the theorem of Ascoli-Arzela. This suggests that you have to use Ascoli's theorem somewhere... I don't think you can do this exercise without it...

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Hmm, OK...

Yes, I already noticed an implication of Ascoli's theorem I could use here, I'll think about it.

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Well, with the help of Ascoli's theorem, this shouldn't be too hard, unless I'm mistaken...

Since {fn} is equicontinuous and pointwise bounded, it has compact closure, and hence Cl({fn}) is sequentially compact. So, fn has a convergent subsequence fnk. The only remaining part is to show uniform convergence of fnk...

Yes, that's the easiest way to show it. One question however:

So, fn has a convergent subsequence fnk.
You say that fnk converges. Under what metric does it converge?

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Yes, that's the easiest way to show it. One question however:

You say that fnk converges. Under what metric does it converge?
Well, under the uniform metric, since we're looking at the function space C(X, R^k)..

Yes, so fnk converges under the uniform metric. Doesn't that imply that fnk is uniform convergent?

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Yes, so fnk converges under the uniform metric. Doesn't that imply that fnk is uniform convergent?
Ahh, of course, yes! By the very definition of the uniform metric!

Since if fnk --> f and if ε > 0 is given, take N such that for n >= N we have ρ(fnk, f) = sup {d(fnk(x), f(x)) : x is in X} < ρ (d is the bounded metric here).

Correct! So that proves Arzela's theorem Homework Helper
Correct! So that proves Arzela's theorem OK!

Btw, I didn't consider Ascoli's theorem immediately because I actually didn't go through its proof yet...

Actually, could you say why Ascoli's theorem is important, i.e. give some examples? When you find the time for it...

The problem is, when you're no mathematician, you always have to think hard about why theorems are significant...And in the end you only conclude something very general. For example, I conclude that the theorem is important since it gives conditions on compactness of a specific class of function spaces, which then implies sequential compactness (I assume this is important, sequences of functions seem important and wide-used in general)...

The problem is, when you're no mathematician, you always have to think hard about why theorems are significant...And in the end you only conclude something very general. For example, I conclude that the theorem is important since it gives conditions on compactness of a specific class of function spaces, which then implies sequential compactness (I assume this is important, sequences of functions seem important and wide-used in general)...
This is the right conlusion. When given a set of functions, the Ascoli-Arzela theorems allow you to pick a subsequence which converges uniform. This is a practice that is really useful in mathematics.

The sad thing is actually that there is no way that you can see now, why some theorems are important. When I first studied the Ascoli-Arzela theorem, I had no idea why it could be of any importance to. And I think that this is fairly normal: the applications of this theorem are often quite deep. Thus it will require a lot of background knowledge to actually see a useful application of the Ascoli-Arzela theorem (and actually this holds for most theorems of topology).

I'll give you some applications, but I can't fully explain them in detail:
- finding solutions to integral equations
- the existence of Brownian motion
- some results in functional analysis: if you study functional analysis from Kreyszig, then you will use Ascoli-Arzela some times. Usually in place where you need to prove compactness.

If you don't like the proof of Ascoli, then I don't think there might be any harm in skipping it (you will see a more general proof anyway within a few chapters). But I would remember the existence of Ascoli-Arzela, since you will see it again Sorry if this wasn't very helpful It sketches some of the most important consequences of Ascoli-Arzela.

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This is the right conlusion. When given a set of functions, the Ascoli-Arzela theorems allow you to pick a subsequence which converges uniform. This is a practice that is really useful in mathematics.

The sad thing is actually that there is no way that you can see now, why some theorems are important. When I first studied the Ascoli-Arzela theorem, I had no idea why it could be of any importance to. And I think that this is fairly normal: the applications of this theorem are often quite deep. Thus it will require a lot of background knowledge to actually see a useful application of the Ascoli-Arzela theorem (and actually this holds for most theorems of topology).

I'll give you some applications, but I can't fully explain them in detail:
- finding solutions to integral equations
- the existence of Brownian motion
- some results in functional analysis: if you study functional analysis from Kreyszig, then you will use Ascoli-Arzela some times. Usually in place where you need to prove compactness.

If you don't like the proof of Ascoli, then I don't think there might be any harm in skipping it (you will see a more general proof anyway within a few chapters). But I would remember the existence of Ascoli-Arzela, since you will see it again Sorry if this wasn't very helpful On the contrary, it was very helpful. This is exactly what I need to know at this stage - a brief outline of topics where the theorem arises.