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I was reading about compactness. I faced two problems.

Is $ C ^ 1 [ 0 , 1 ] $ compact?

$$ \| f - g \| = \max _ { x \in [0, 1] } | f ( x ) - g ( x ) | $$ This norm is given with the two spaces - $ C ^ 1 [ 0 , 1 ] $ and the unit ball of $ C ^ 1 [ 0 , 1 ] $. $ C ^ 1 [ 0 , 1 ] $ is the space of all differentiable functions on $ [ 0 , 1 ] $.

Is the closure of the unit ball of $ C ^ 1 [ 0 , 1 ] $ in $ C [ 0 , 1 ] $ compact?

I think $ C ^ 1 [ 0 , 1 ] $ is not compact as $ f _ n ( x ) = \sqrt { \left( x - \frac 1 2 \right) ^ 2 + \frac 1 n } $ uniformly convergent to $ \left| x - \frac 1 2 \right| $ which is not differentiable at $ \frac 1 2 $.

Is my argument correct?

I have no idea about the closure of the unit ball of $ C ^ 1 [ 0 , 1 ] $. Is this a compact space?

Mohsen Shahriari
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cmi
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    Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.) – Arthur Jan 16 '19 at 06:54
  • @Arthur I have edited my question .. – cmi Jan 16 '19 at 06:59
  • @Arthur Please have a look – cmi Jan 16 '19 at 07:03
  • @Arthur I am trying to understand your hint – cmi Jan 16 '19 at 07:06
  • @Arthur Can you please tell me will the unit ball of $C^1 [0 , 1] $ be compact? – cmi Jan 16 '19 at 07:08
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    Calm down, please. You don't have to ask me four times in nine minutes. Once is enough. There are other people on this site who can help you too (like José below), and while I probably spend too much time on this site, I do have things to do elsewhere, so you can't expect me to respond immediately. – Arthur Jan 16 '19 at 07:19
  • @Arthur I am sorry..I did not want to disturb you ..I just tried to say that I have edited my question please have a look..and can you please tell if the second set is compact? I wanted to think on my own after you ensure me that the set is compact. I should have told these three sentences in a single comment. – cmi Jan 16 '19 at 09:17
  • I have made a little mistake...Actually the second set was the closure of the unit ball of $C^1[0,1]$ in $C [0, 1]$I am really sorry. Can you please edit your answer accordingly?@Kavi Rama Murthy – cmi Jan 16 '19 at 10:19

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You are making things complicated. $C^{1} [0,1]$ is not compact because the sequence of constant functions $\{1,2,\cdots\}$ has no convergent subsequence. [No normed linear space is compact because on unboundedness]. The closure of the unit ball in $C^{1}[0,1]$ w.r.t the sup norm is nothing but the closed unit ball of $C[0,1]$, thanks to Weierstrass Approximation Theorem. The unit ball of $C[0,1]$ is not compact because $\{x^{n}\}$ has no convergent subsequence.

Kavi Rama Murthy
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  • Actually I have recently read some where that sequence of functions for the counter example of the statement -- continuously differentiable sequence of functions uniformly convergent to a continuously differentiable function. That's why this example hit my mind when I encountered this problem . Btw thank you for your smart and incisive answer.. – cmi Jan 16 '19 at 09:12
  • Actually Sir I was quite sure that the second set is not compact.But when I saw the answer key is saying that the set is not compact , I got confused. This question came in NBHM 2007 phd screening test..and the answer key has been also given by them. You seem to be an Indian .You may know the exam..That's why I am telling you in detail.. – cmi Jan 16 '19 at 09:24
  • I have made a little mistake...Actually the second set was the closure of the unit ball of $C^1[0,1]$ in $C [0, 1]$I am really sorry. Can you please edit your answer accordingly?@Kavi Rama Murthy – cmi Jan 16 '19 at 10:20
  • I got it..Thanks a lot...@Kavi Rama Murthy – cmi Jan 16 '19 at 10:27
  • You mentioned "The closure of the unit ball in $C^1[a,b]$ w.r.t the sup norm is nothing but the closed unit ball of $C[0,1]$..." but I believe this is not true. Please take a look at this. But your claim is true when you consider the whole space instead of balls, i.e. $C^1[0,1]$ is dense in $C[0,1]$; maybe that's what you meant to say. – Luke Sep 23 '20 at 21:38
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    @Luke My claim is true. If $f$ is in the closed unit ball of $C[0,1]$ then there exist polynomials $p_n \to f$ uniformly. Let $c_n=\frac {|f|} {|p_n|}$. Then $c_np_n$ are smooth functions converging uniformly to $f$ and they are in the closed unit ball of $C^{1}[0,1]$ w.r.t sup norm. I think you are using some other norm on $C^{1}[0,1]$ instead of the sup norm. – Kavi Rama Murthy Sep 23 '20 at 23:19
  • @KaviRamaMurthy Oops. Yeah, I was using $C^1$ norm instead of the sup-norm, and it's clear from your answer, so my bad. I should've read more carefully. Thanks for the clarification! – Luke Sep 24 '20 at 18:00
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Your argument is basically correct: if $f_n(x)=\sqrt{\left(x-\frac12\right)^2+\frac1n}$, then, in $C[0,1]$, $\lim_{n\in\mathbb N}=f$, with $f(x)=\left\lvert x-\frac12\right\rvert$. But $f\notin C^1[0,1]$. Therefore, the unit ball of $C^1[0,1]$ is not a closed subset of $C[0,1]$. But, if it was compact, it would be a closed subset.

It follows that $C^1[0,1]$ is not compact either.

José Carlos Santos
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Your example works fine to prove that $C^1[0,1]$ is not compact: it is a sequence of $C^1$ functions which converges (uniformly) to a function which is not $C^1$, meaning the sequence has no convergent subsequences in $C^1$. (I think the author meant for you to pick an easier sequence like $f_n=n$ or something, but that's not really important.)

Now note that $\|f_n\|=\sqrt{\frac14+\frac1n}$. This means that from $n=2$ on, these functions are in the unit ball. So, setting $g_n=f_{n+1}$, we have that $g_n$ is a sequence of functions in the unit ball.

The convergence properties of $g_n$ are exactly the same as those of $f_n$, meaning it has no convergent subsequence in $C^1[0,1]$, and therefore no convergent subsequence in the unit ball, proving that the unit ball isn't compact.

Arthur
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    @KaviRamaMurthy If you notice, I did suggest the exact same example that you did. And I thought I would give feedback on the attempt the OP made, rather than throw it all away and just answer the problem from scratch. He had found a function which worked, so why not boost their confidence and tell them that they have actually basically solved the problem already? – Arthur Jan 16 '19 at 08:25
  • @cmi No need. It still works. taking the closure of a subset of $C^1$ doesn't add non-$C^1$ functions, so the limit of $f_n$ (and of $g_n$) is still not in your set, and the set is therefore not compact. – Arthur Jan 16 '19 at 10:10
  • I could not grasp your comment..here in my example the limit function itself is non $C^1$ function.. – cmi Jan 16 '19 at 10:15
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    @cmi ... which shows that within $C^1$, the function doesn't converge at all (and it doesn't have any convergent subsequence either). Whether you close the unit ball or not, that's not going to change. – Arthur Jan 16 '19 at 10:25