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[I really want to apologize if this problem looks a little too long.]

The problem :

This is taken from here [Question. 3.7 (c)] and it says...

Prove or disprove the comapctness of the closure of the unit ball of $C^1[0, 1]$ in $C[0, 1]$.

What have I tried?

I think it's pretty clear that the unit ball in $C^1[0,1]$ is $B= \{f \in C^{1}[0,1]: \lVert f \rVert \leq 1\}$. Let us denote $K= \operatorname{cl}_{C[0,1]}(B)$, the closure of $B$ in $C[0,1]$.

I really didn't have much clue how to start working on this problem but finally thought about using the sequential approach. If $a$ and $b$ are two points such that $0<a<b<1$ then I think there can be functions $f_a$ and $f_b$ having the property that $\lVert f_a-f_b \rVert = 1$ which are sufficiently smooth to be in $C^1$. Now as it is quite tedious to construct these functions explicitly, I scanned a hand-drawn picture of what I think they might look like.

Drawing my intuition

Here $f_a$ and $f_b$ take zero values in almost all of the interval $[0,1]$ and jumps up at $a$ and $b$ respectively. Also, as is clear from the figure, they don't assume non-zero values simultaneously. Now, my argument is clear: however close the two points $a$ and $b$ come to each other (remaining distinct) there will always be functions like $f_a$ and $f_b$. They may get steeper and steeper but will never lose their $C^1$-ness.

So if I consider the sequence $(\frac{1}{n})$ in $[0,1]$, I will get a sequence $(f_{\frac{1}{n}})$ in $B (\subseteq K)$ where for $m \neq n$ we will have $\lVert f_{\frac{1}{m}} - f_{\frac{1}{n}}\rVert = 1$. Hence the sequence $(f_{\frac{1}{n}})$ cannot have a convergent subsequence, thereby proving that $K$ is not sequentally compact and hence not compact.

But I'm not very sure about all these. A friend of mine told me that this same problem has a positive answer and that made me sufficiently confused.

So, here comes my question...

In the above argument, where have I gone wrong? What is the real answer? And how to prove it?

Thanks a million for reading my extra-long question. And thanks a zillion for any help that you can offer.

share|cite|improve this question
What are $C^1 [0,1]$ and $C[0,1]$? – Benjamin Dickman Jan 17 '13 at 6:51
@B.D : $C[0,1]$ is set of all real continuous functions on $[0,1]$ and $C^1[0,1]$ is the set of all continuously differentiable real functions on $[0,1]$. – Sayantan Jan 17 '13 at 7:01
Your friend is right. Do you know the Arzela-Ascoli theorem? – Martin Jan 17 '13 at 7:04
Your argument is valid if you put the sup-norm on $C^1[0,1]$. However, the $C^1$-norm also requires a bound on the derivative. Specifically, both $f$ and $f'$ are bounded by $1$ for $f$ in the $C^1$-unit ball. This excludes your examples $f_a$ and $f_b$ because their derivatives have to become too steep when $a$ and $b$ are getting too close to each other. The Arzela-Ascoli theorem is the tool to use to prove compactness of $K$: the functions in $B$ are pointwise bounded and equicontinuous (since you have a bound on the derivative) and A-A tells you that $K = \mathrm{cl}B$ is then compact. – Martin Jan 17 '13 at 8:27
I understand the feeling, thanks for posting anyway (and no worries: both your question and your answer are really good!). It is explicitly encouraged to answer your own question. – Martin Jan 17 '13 at 17:15
up vote 2 down vote accepted

This is basically a rehash of what Martin put in his comments following my question.

With the norm that I had in mind while posting this problem (the sup-norm $\lVert \rVert _{\infty}$ on $C^1[0,1]$), the argument which I originally laid out, works well. Therefore, $K$ is not compact in $(C^1[0,1], \lVert \rVert _{\infty})$.

But, as Martin kindly pointed out, the usual norm on $C^1[0,1]$ is not the sup-norm but the norm defined by $\lVert f \rVert _{C^1[0,1]}= \operatorname{max}\{\lVert f \rVert_{\infty}, \lVert f' \rVert_{\infty}\}$. If we consider this norm then, the unit ball changes to $B' = \{f \in C^1[0,1]: \lVert f \rVert_{C^1[0,1]} \leq 1\} = \{f \in C^1[0,1]: \lVert f \rVert_{\infty} \leq 1, \lVert f' \rVert_{\infty} \leq 1\} $. Here my original argument fails, because as the functions $f_{\frac{1}{n}}$ get very steep with each increasing $n$, their derivatives become unbounded. Hence they cannot belong in $B'$.

Evebtually, by an application of Arzelà–Ascoli theorem in $(C^1[0,1], \lVert \rVert_{C^1[0,1]})$, the set $K$ turns out to be compact. But since I don't have much experience about that theorem (and about stuff like uniform boundedness, equicontinuity etc.), I won't risk any half-hearted attempt to prove this claim.

share|cite|improve this answer
Great, thanks! Let me add: one reason people use the $C^1$-norm I gave is that $C^1$ is complete with respect to that norm. This is a way to express a fact you probably know: Let $f_n$ be continuously differentiable functions and suppose that $f_n \to f$ uniformly (or only pointwise) and $f_{n}' \to g$ uniformly where $f$ and $g$ are continuous. Then $f$ is continuously differentiable and $f' = g$. // The Arzela-Ascoli is a useful compactness criterion that shows up a lot. It is definitely worth learning it and see how it works. There is certainly no need to spell the argument out here. – Martin Jan 17 '13 at 17:12
@Martin : I guess I saw that result somewhere in real analysis course. And I'll venture into A-A theorem when I'll have time. Thanks for encouraging me to answer my own question, it was worth a try. Regards. :-) – Sayantan Jan 17 '13 at 17:53

The argument for the positive answer is a straightforward application of the Arzela-Ascoli theorem. Recall that a set $B \subset C[0,1]$ is sequentially compact in the uniform ($\|\cdot\|_{\infty}$) topology on $C[0,1]$ iff it is equicontinuous and uniformly bounded. I will first define these terms and will then show they hold in this case.

Equicontinuity: for all $\epsilon > 0$ there exists $\delta > 0$ such that for any $f \in B$ and any $x,y \in [0,1]$, we have $|x-y| < \delta$ implies $|f(x) - f(y)| < \epsilon$. Note that we're demanding that the modulus of continuity may be uniformly chosen across $B$.

Uniform Boundedness: There is an $M > 0$ such that for any $f \in B$, $\|f\|_{\infty} \leq M$.

Indeed, $B = \{f \in C^1[0,1] \mid \|f\|_{\infty} \vee \|f'\|_{\infty} \leq 1\}$ and so immediately we have uniform boundedness for $M = 1$. For equicontinuity, recall by the mean value theorem that for any $f \in B$, $$ |f(x) - f(y)| \leq \|f'\|_{\infty}|x-y| \leq |x - y| $$ which means that for the $\epsilon$ challenge, we may take $\delta = \epsilon$ for any $f \in B$.

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Note however that the $\|\cdot\|_{\cdot}$-limit of members of $B$ may fail to be differentiable! This is a big problem when you're trying to prove the unstable/stable manifold theorem, for instance. – A Blumenthal Jan 18 '13 at 2:31
Hmm, I don't see how you've shown equicontinuity here. Namely, the original condition for inclusion in $B$ was $\lVert f \rVert _{C^1[0,1]}= \operatorname{max}\{\lVert f \rVert_{\infty}, \lVert f' \rVert_{\infty}\}$, not $\|f\|_{\infty} \vee \|f'\|_{\infty} \leq 1$, as you used. Yes, if we go by your norm, then that follows, but it doesn't seem to be the case that that's the norm on $C^1[0,1]$. Or am I missing something? – Ryker Sep 14 '14 at 3:19

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