Metric-spaces, equicontinuity, proof

[bobby2k]

Hi, can you please check if my proof is correct?

Proposition
Assume that a sequence of continuous functions $\{f_n\}$ converges uniformly to $f$.
$f_n: A\rightarrow B$. Also assume that $A$ is compact.
Then $\{f_n\}$ is equicontinuous.

Proof:
Given an $\epsilon$ we must find a $\delta$ so that if $x,y \in A,d_A(x,y)< \delta$, then $d_B(f_n(x),f_n(y))<\epsilon$, for all $n$, and for all $x,y$.

Given $\epsilon$. There exists an $N$ such that $d_B(f_n(x),f(x))<\epsilon/3$ for all $n\ge N$ and all $x \in A$.
Since f must be continuous and A is compact, f is uniformly continuous. So there exists a $\delta^*$ so that if $d_A(x,y)<\delta^*\rightarrow d_B(f(x),f(y))<\epsilon/3$
For each $i, i <N$, there is a $\delta_i$, so that if $d_A(x,y)<\delta_i \rightarrow d_B(f_i(x),f_i(y))<\epsilon$. This must be, because each $f_n$ must be uniformly continuous on the compact space A.
Choose $\delta = min\{\delta^*,\delta_1,\delta_2,\delta_{N-1}\}$.

Now, for any n, if $d_A(x,y)<\delta \rightarrow d_B(f_n(x),f_n(y))< \epsilon$. This is so, because if $n < N$ it obviously holds, if $n \ge N$ we have:
$d_B(f_n(x),f_n(y)) \le d_B(f_n(x),f(x))+d_B(f(x),f(y))+d_B(f(y),f_n(y))< \epsilon/3+\epsilon/3+\epsilon/3 = \epsilon$.

 

[micromass]

There are some typo's in your proof, but aside from that, the idea seems to be entirely correct.

 

[bobby2k]

There are some typo's in your proof, but aside from that, the idea seems to be entirely correct.

Thanks, do you know if the statement is correct or not?(it could actually be false) It is the "converse"(not entirely converse, some assumptions are omitted) of a statement I read in a book, so I am not entirely sure if it is correct. The original proof states that if the funcitons converges pointwise on a dense subset of A, and A is compact, B is complete, and the functions are equicontinuous, then it converges uniformly on A. So this is kind of the converse of that.

 

[micromass]

Thanks, do you know if the statement is correct or not? It is the converse of a statement I read in a book, so I am not entirely sure if it is correct.

The statement is absolutely correct. In fact, if you're familiar with function spaces like $\mathcal{C}([a,b])$ with the uniform norm, then you can prove that compactness is equivalent to being equicontinuous and being closed and being pointswise bounded. This is the Arzela-Ascoli theorem. In particular, if $f_n\rightarrow f$ uniformly, then

$$\{f_n~\vert~n\}\cup \{f\}$$

is equicontinuous. So it remains equicontinuous if we leave out $f$.

 

[bobby2k]

The statement is absolutely correct. In fact, if you're familiar with function spaces like $\mathcal{C}([a,b])$ with the uniform norm, then you can prove that compactness is equivalent to being equicontinuous and being closed and being pointswise bounded. This is the Arzela-Ascoli theorem. In particular, if $f_n\rightarrow f$ uniformly, then

$$\{f_n~\vert~n\}\cup \{f\}$$

is equicontinuous. So it remains equicontinuous if we leave out $f$.

Thanks, yes it is the "converse" of a lemma in the book, leading up to Arzela-Ascoli theorem. However I wasn't able to use that theorem on it directly, because the theorem in the book was stated only for $C(X,R^m)$, where X is compact, but in this lemma, Y can be any complete space. But beside from that if you want to show that $\{f_n\}$ is equicontinuous using the arzela-ascoli theorem, then you also have to show that that $\{f_n\}$ is compact, and I presume that is hard?

 

[micromass]

Thanks, yes it is the "converse" of a lemma in the book, leading up to Arzela-Ascoli theorem. However I wasn't able to use that theorem on it directly, because the theorem in the book was stated only for $C(X,R^m)$, where X is compact, but in this lemma, Y can be any complete space. But beside from that if you want to show that $\{f_n\}$ is equicontinuous using the arzela-ascoli theorem, then you also have to show that that $\{f_n\}$ is compact, and I presume that is hard?

No, that is actually very easy in this case if you use the right definition (= open cover definition). You have in any metric space $X$ that if $x_n\rightarrow x$, then $\{x_n~\vert~n\}\cup \{x\}$ is compact.

 

[bobby2k]

No, that is actually very easy in this case if you use the right definition (= open cover definition). You have in any metric space $X$ that if $x_n\rightarrow x$, then $\{x_n~\vert~n\}\cup \{x\}$ is compact.

Ah, cool thanks. But the proposition in the first post would then follow from Arzela-Ascoli if B was $R^n$? Because then $\{f_n\}$ is compact, and hence equicontinuous. But you agree that I can't use the theorem directly when B is any space, or even if B is complete?(A is compact no matter what.)

 

[micromass]

Ah, cool thanks. But the proposition in the first post would then follow from Arzela-Ascoli if B was $R^n$? Because then $\{f_n\}$ is compact, and hence equicontinuous. But you agree that I can't use the theorem directly when B is any space, or even if B is complete?

Do they state Arzela-Axcoli only for $B=\mathbb{R}^n$? It's true in a far greater generality. For example, there is a version of the theorem for $B$ an arbitrary metric space.

 

[bobby2k]

Do they state Arzela-Axcoli only for $B=\mathbb{R}^n$? It's true in a far greater generality. For example, there is a version of the theorem for $B$ an arbitrary metric space.

Yeah, here is the theorem:

Could you change $R^m$ with something more general here?

 

[micromass]

Yeah, here is the theorem:

Could you change $R^m$ with something more general here?

Yes, certainly. Check out wikipedia: http://en.wikipedia.org/wiki/Arzelà...ct_metric_spaces_and_compact_Hausdorff_spaces for a very general version. The proof can be found in Kelley, but it requires topology.

 

[bobby2k]

Thanks, I am taking real analysis and have not taken topology, so I must admit I do not understand much of that.
But if I want to generalize the theorem in my book. The link from Wikipedia says:

Let X be a compact Hausdorff space and Y a metric space. Then a subset F of C(X,Y) is compact in the compact-open topology if and only if it is equicontinuous, pointwise relatively compact and closed.

Now this link says that all metric spaces are Hausdorff spaces:
http://math.stackexchange.com/quest...ll-metric-spaces-are-hausdorff-spaces-correct So that is fine. The problem is the "f is pointwise relatively compact", I guess that if B= $R^m$ "bounded" implies "pointwise relatively compact". But that maybe if B is only a complete metric space, or arbirary metric space, "bounded" may not imply "pointwise relatively compact".EDIT: I see that for my original question this may not affect it. Because it is on the other side of the equivalence. So that if we start by saying that $\{f_n\}$ converge uniformly to f, and that A is compact, we must have that the limit is continuous, so that limit is in C(A,B). The open covering property says that $f_n\cup f$ is compact. And since we are on the left side of the Arzela-Ascoli(Metric->Hausdorff) we get that they all must be equicontinious, and we remove f?, så all $f_n$ are equicontinuous?

Thanks for your help.