# Metric-spaces, equicontinuity, proof

1. Mar 15, 2014

### 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:
$d_B(f_n(x),f_n(y)) \le d_B(f_n(x),f(x))+d_B(f(x),f(y))+d_B(f(y),f_n(y))< \epsilon/3+\epsilon/3+\epsilon/3 = \epsilon$.

Last edited: Mar 15, 2014
2. Mar 15, 2014

### micromass

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

3. Mar 15, 2014

### bobby2k

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.

Last edited: Mar 15, 2014
4. Mar 15, 2014

### micromass

Staff Emeritus
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 $f$.

5. Mar 15, 2014

### bobby2k

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?

6. Mar 15, 2014

### micromass

Staff Emeritus
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.

7. Mar 15, 2014

### bobby2k

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.)

8. Mar 15, 2014

### micromass

Staff Emeritus
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.

9. Mar 15, 2014

### bobby2k

Yeah, here is the theorem:

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

10. Mar 15, 2014

### micromass

Staff Emeritus
11. Mar 15, 2014

### 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:
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?