Baby Rudin Problem 7.16

Problem 7.16: Suppose $\{f_n\}$ is an equicontinuous family of functions on a compact set $K$ and $\{f_n\}$ converges pointwise to some $f$ on $K$. Prove that $f_n \to f$ uniformly.

Now for this problem I assume that $f_n,f : K \subset \Bbb{R} \rightarrow \Bbb{R}$ with the usual euclidean metric. Even though this is not assumed in the problem, I assume this to simplify matters first.

Now I believe I have proven that $f_n$ is uniformly cauchy on $K$ as follows. By equicontinuity of the family $\{f_n\}$ I can choose $\delta> 0$ such that $|x-p_i|< \delta$ will imply that $|f_m(x) - f_m(p_i)| < \epsilon$, and $|f_n(x) - f_n(p_i)| < \epsilon$.

Consider the collection $\{B_\delta(x)\}_{x \in K}$ that clearly covers $K$, by compactness of $K$ we get that there are finitely many points $p_1,\ldots p_n \in K$ such that $\{B_\delta(p_i)\}_{i=1}^n$ is a cover for $K$. Furthermore, because $f_n \rightarrow f$ pointwise for each $x \in K$ we get a cauchy sequence of numbers, so in particular given any $\epsilon > 0$, for each $p_i$ there exists $N_i$ such that $m,n \geq N_i$ implies that $|f_m(p_i) - f_n(p_i) | < \epsilon$. Taking

$$N = \max_{1 \leq i \leq n} N_i$$

gives that $m,n\geq N$ implies that $|f_m(p_i) - f_n(p_i)| < \epsilon$ for all $i$.

Now we can finally put everything together to prove uniform cauchyness, take any $x \in K$ so that $x \in B_\delta(p_j)$ for some $1 \leq j \leq n$. Then

$$\begin{eqnarray*} |f_n(x) - f_m(x)| &\leq& |f_n(x) - f_n(p_i) | + | f_n(p_i) - f_m(p_i)| + |f_m(p_i) - f_m(x)| \\ &<& \epsilon + \epsilon + \epsilon \\ &=& 3\epsilon. \end{eqnarray*}$$

The first and last term being less than $\epsilon$ come from equicontinuity, the middle term being less than $\epsilon$ comes from the derivation just before. Now what I am thinking of doing now to prove uniform cauchyness is to take the sup on the left, is this something legal I can do? Also are there are any mistakes in the proof above?

Here is some context why I want to prove uniform cauchyness: Suppose I know that $\{f_n\}$ is uniformly cauchy. Then I know that given any $\epsilon > 0$, there exists $N \in \Bbb{N}$ such that $m,n\geq N$ implies that $|f_n(x) - f_m(x)| < \epsilon$ for all $x \in K$. Now we do this trick of fixing one of the indices. Fix $n$ to be some integer greater than $N$ and let $m\rightarrow \infty$, we see that

$$\begin{eqnarray*} |f_n - f| &=& \lim_{m\rightarrow \infty} | f_n - f_m| \\ &\leq& \epsilon \end{eqnarray*}$$

by the limit comparison test. Recall that $f$ was the pointwise limit of $\{f_n\}$. But then since $n$ was any arbitrary integer greater than $N$ we have that $f_n \rightarrow f$ uniformly.

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Your argument is great until the point where you say "Now what I am thinking of doing..."

You have already proven that $\{f_n\}$ is uniformly Cauchy since you've shown that for any $\epsilon>0$, there is an $N$ so that if $n,m>N$, $|f_m(x)-f_n(x)|<3\epsilon$ for all $x\in K$.

To show uniform convergence, we simply need to say that, because of the pointwise convergence, for any $x$, there is some $M>N$ so that if $m>M$, $|f_m(x)-f(x)|<\epsilon$. Then, for any $n>N$, we have that \begin{align} |f_n(x)-f(x)| &\le|f_n(x)-f_m(x)|+|f_m(x)-f(x)|\\ &<3\epsilon+\epsilon\\ &=4\epsilon \end{align}

That is, for any $\epsilon>0$, there is an $N$ so that if $n>N$, $|f_n(x)-f(x)|<4\epsilon$ for all $x\in K$.

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May I know what is the problem there when I take the sup on both sides? Because before that I don't think I have proven uniform cauchyness. What about the bit after the uniform cauchy stuff? I mean, the bit after the line across in my post. –  user38268 Jun 4 '12 at 13:42
There is nothing wrong with taking the sup. However, you've already shown that $\{f_n\}$ is uniformly Cauchy (that's the "for all $x\in K$" part of what I said above), so I don't see why you'd need to take the sup. –  robjohn Jun 4 '12 at 14:12
The part after the line changes gears from an $\epsilon-\delta$ proof to using limits. However, I think what I wrote in $\epsilon-\delta$ form is in the same spirit as what is below the line. –  robjohn Jun 4 '12 at 14:15
I don't think I have proven uniform cauchyness, namely because the quantifier $\forall$ $x$ in $K$ should be at the end not at the beginning no? –  user38268 Jun 5 '12 at 3:17
@BenjaminLim: The whole point of the $3\epsilon$ inequality is that you can take any $x$ (which will be in one of the $B_\delta(p_j)$) and then you have an $N$ so that if $n,m>N$, then $$|f_n(x)-f_m(x)|\le|f_n(x)-f_n(p_j)|+|f_n(p_j)-f_m(p_j)|+|f_m(p_j)-f_m(x)| < 3\epsilon$$ where $N$ does not depend on $x$. This confirms that $\{f_n\}$ is uniformly Cauchy. –  robjohn Jun 5 '12 at 5:30

Yes this looks ok. The point with the $3\varepsilon$ is, that the rhs does not depend on $x$, and then the lhs is estimated for all $x$ and sufficiently large $n,m$ simultaneously. That's the important thing to see here.

You could, alternatively, try to work with $f$ and $f_n$ right from the beginning -- the Cauchy sequence approach is correct, but I'd consider it a detour. Depends on your liking alone, though.

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There may be a way that I am not seeing, but I think that the Cauchy sequence is easiest. If we repeat the argument with only $f$ and $f_n$, we don't have the equicontinuity allowing us to get all three of the $\epsilon$s above (we don't even know that $f$ is continuous). –  robjohn Jun 4 '12 at 12:53
@robjohn As I said, I consider it a matter of personal liking. Continuity of $f$ is quite immediate with equicontinuity and pointwise convergence. Then $f$ is uniformly continous since $K$ is compact. You may consider this an equal effort in total, admittedly, or even prefer the solution of Benjamin. –  user20266 Jun 4 '12 at 13:24
If we are at the stage of proving that the pointwise convergence of an equicontinuous family is uniform, I would not take for granted that the limit is continuous (especially when proofs of the continuity of a limit typically use uniform convergence). –  robjohn Jun 4 '12 at 14:08
@robjohn I did not say you should take if for granted, but that it is easy to show. –  user20266 Jun 4 '12 at 14:37