"The closure of the unit ball of $C^1[0, 1]$ in $C[0, 1]$" and its compactness

Question

[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.

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.

Answer 1

Ascoli-Arzelá requires says that a set $\scr F$ in $C^0([0,1])$ is compact iff it is:

1) closed, 2) bounded (or ``equibounded'') and 3) equicontinuous.

If if $L=\{f\in \cal C^1:\max\{||f'||_\infty, ||f||_\infty\} <1 \}$ is considered as a subset of $C^0$ then 1) it is equibounded because it is a subset of the unit ball in $\cal C^0$ (i.e. $||f||_\infty<1$);

2) it is equicontinuous because all $f\in L$ are Lipshitz with constant at worst $1$ (by the "mean value theorem" of calculus).

3) As written it is clearly not closed in $\cal C^0$; but the question asks for its closure in $\cal C^0$, so that comes for free.

What I am not sure of, is whether the closure in $\cal C^0$ coincides with the closure in $\cal C^1$, but I doubt it.

Answer 2

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.

Answer 3

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 $f \in B$.