Mixing
$C ^1 [0, 1]$ is compact? Is the closure of the unit ball of $C ^1 [0, 1]$ in $C [0, 1]$ compact?
$begingroup$
I was reading Compactness. I faced two problem .
$C ^1 [0, 1]$ is compact?
||f- g|| =[ max | f - g| for all $ x in [0, 1]$]
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 compact is not compact as $ f_n (x) = sqrt((x-1/2)^2 +1/n) $ uniformly convergent to |x-1/2| which is not differentiable at 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?
Can anyone please help me to understand this problem?
general-topology metric-spaces compactness
asked Jan 16 at 6:48
cmicmi
1,121312
$endgroup$
|
show 3 more comments
$begingroup$
I was reading Compactness. I faced two problem .
$C ^1 [0, 1]$ is compact?
||f- g|| =[ max | f - g| for all $ x in [0, 1]$]
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 compact is not compact as $ f_n (x) = sqrt((x-1/2)^2 +1/n) $ uniformly convergent to |x-1/2| which is not differentiable at 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?
Can anyone please help me to understand this problem?
general-topology metric-spaces compactness
asked Jan 16 at 6:48
cmicmi
1,121312
$endgroup$
1
$begingroup$
Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.)
$endgroup$
– Arthur
Jan 16 at 6:54
$begingroup$
@Arthur I have edited my question ..
$endgroup$
– cmi
Jan 16 at 6:59
$begingroup$
@Arthur Please have a look
$endgroup$
– cmi
Jan 16 at 7:03
$begingroup$
@Arthur I am trying to understand your hint
$endgroup$
– cmi
Jan 16 at 7:06
2
$begingroup$
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.
$endgroup$
– Arthur
Jan 16 at 7:19
|
show 3 more comments
$begingroup$
I was reading Compactness. I faced two problem .
$C ^1 [0, 1]$ is compact?
||f- g|| =[ max | f - g| for all $ x in [0, 1]$]
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 compact is not compact as $ f_n (x) = sqrt((x-1/2)^2 +1/n) $ uniformly convergent to |x-1/2| which is not differentiable at 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?
Can anyone please help me to understand this problem?
general-topology metric-spaces compactness
asked Jan 16 at 6:48
cmicmi
1,121312
$endgroup$
I was reading Compactness. I faced two problem .
$C ^1 [0, 1]$ is compact?
||f- g|| =[ max | f - g| for all $ x in [0, 1]$]
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 compact is not compact as $ f_n (x) = sqrt((x-1/2)^2 +1/n) $ uniformly convergent to |x-1/2| which is not differentiable at 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?
Can anyone please help me to understand this problem?
general-topology metric-spaces compactness
general-topology metric-spaces compactness
asked Jan 16 at 6:48
cmicmi
1,121312
asked Jan 16 at 6:48
cmicmi
1,121312
edited Jan 16 at 10:17
cmi
asked Jan 16 at 6:48
cmicmi
1,121312
asked Jan 16 at 6:48
cmicmi
1,121312
asked Jan 16 at 6:48
cmicmi
1,121312
1,121312
1
$begingroup$
Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.)
$endgroup$
– Arthur
Jan 16 at 6:54
$begingroup$
@Arthur I have edited my question ..
$endgroup$
– cmi
Jan 16 at 6:59
$begingroup$
@Arthur Please have a look
$endgroup$
– cmi
Jan 16 at 7:03
$begingroup$
@Arthur I am trying to understand your hint
$endgroup$
– cmi
Jan 16 at 7:06
2
$begingroup$
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.
$endgroup$
– Arthur
Jan 16 at 7:19
|
show 3 more comments
1
$begingroup$
Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.)
$endgroup$
– Arthur
Jan 16 at 6:54
$begingroup$
@Arthur I have edited my question ..
$endgroup$
– cmi
Jan 16 at 6:59
$begingroup$
@Arthur Please have a look
$endgroup$
– cmi
Jan 16 at 7:03
$begingroup$
@Arthur I am trying to understand your hint
$endgroup$
– cmi
Jan 16 at 7:06
2
$begingroup$
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.
$endgroup$
– Arthur
Jan 16 at 7:19
1
1
$begingroup$
Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.)
$endgroup$
– Arthur
Jan 16 at 6:54
$begingroup$
Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.)
$endgroup$
– Arthur
Jan 16 at 6:54
$begingroup$
@Arthur I have edited my question ..
$endgroup$
– cmi
Jan 16 at 6:59
$begingroup$
@Arthur I have edited my question ..
$endgroup$
– cmi
Jan 16 at 6:59
$begingroup$
@Arthur Please have a look
$endgroup$
– cmi
Jan 16 at 7:03
$begingroup$
@Arthur Please have a look
$endgroup$
– cmi
Jan 16 at 7:03
$begingroup$
@Arthur I am trying to understand your hint
$endgroup$
– cmi
Jan 16 at 7:06
$begingroup$
@Arthur I am trying to understand your hint
$endgroup$
– cmi
Jan 16 at 7:06
2
2
$begingroup$
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.
$endgroup$
– Arthur
Jan 16 at 7:19
$begingroup$
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.
$endgroup$
– Arthur
Jan 16 at 7:19
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
add a comment |
$begingroup$
Your argument is basically correct: if $f_n(x)=sqrt{left(x-frac12right)^2+frac1n}$, then, in $C[0,1]$, $lim_{ninmathbb N}=f$, with $f(x)=leftlvert x-frac12rightrvert$. But $fnotin C^1[0,1]$. 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.
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
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I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
1
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
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– José Carlos Santos
Jan 16 at 10:12
1
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
|
show 3 more comments
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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.
answered Jan 16 at 7:18
ArthurArthur
116k7116198
$endgroup$
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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?
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– Arthur
Jan 16 at 8:25
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@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.
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– Arthur
Jan 16 at 10:10
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I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
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– cmi
Jan 16 at 10:15
1
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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.
$endgroup$
– Arthur
Jan 16 at 10:25
add a comment |
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3 Answers
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3 Answers
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$begingroup$
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.
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
add a comment |
$begingroup$
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.
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
add a comment |
$begingroup$
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.
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
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.
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
edited Jan 16 at 10:21
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 7:26
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
61.7k42262
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
add a comment |
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:12
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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..
$endgroup$
– cmi
Jan 16 at 9:24
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
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
$endgroup$
– cmi
Jan 16 at 10:20
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
$begingroup$
I got it..Thanks a lot...@Kavi Rama Murthy
$endgroup$
– cmi
Jan 16 at 10:27
add a comment |
$begingroup$
Your argument is basically correct: if $f_n(x)=sqrt{left(x-frac12right)^2+frac1n}$, then, in $C[0,1]$, $lim_{ninmathbb N}=f$, with $f(x)=leftlvert x-frac12rightrvert$. But $fnotin C^1[0,1]$. 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.
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
$begingroup$
I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
1
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Jan 16 at 10:12
1
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
|
show 3 more comments
$begingroup$
Your argument is basically correct: if $f_n(x)=sqrt{left(x-frac12right)^2+frac1n}$, then, in $C[0,1]$, $lim_{ninmathbb N}=f$, with $f(x)=leftlvert x-frac12rightrvert$. But $fnotin C^1[0,1]$. 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.
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
$begingroup$
I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
1
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Jan 16 at 10:12
1
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
|
show 3 more comments
$begingroup$
Your argument is basically correct: if $f_n(x)=sqrt{left(x-frac12right)^2+frac1n}$, then, in $C[0,1]$, $lim_{ninmathbb N}=f$, with $f(x)=leftlvert x-frac12rightrvert$. But $fnotin C^1[0,1]$. 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.
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
Your argument is basically correct: if $f_n(x)=sqrt{left(x-frac12right)^2+frac1n}$, then, in $C[0,1]$, $lim_{ninmathbb N}=f$, with $f(x)=leftlvert x-frac12rightrvert$. But $fnotin C^1[0,1]$. 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.
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 16 at 7:14
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
$begingroup$
I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
1
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Jan 16 at 10:12
1
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
|
show 3 more comments
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
$begingroup$
I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
1
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Jan 16 at 10:12
1
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
$begingroup$
How can any normed linear space be compact? Not that your answer is wrong, but I am wondering why the OP constructs such a complicated example.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:31
$begingroup$
I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
I suppose that the OP 's aim was to prove that the closed unit ball is not compact.
$endgroup$
– José Carlos Santos
Jan 16 at 7:33
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
$begingroup$
But the standard proof is to consider ${x^{n}}$ right?.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 7:34
1
1
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Jan 16 at 10:12
$begingroup$
I could edit my answer, but then I would just be repeating Kavi Rama Murthy's idea: to use the sequence $(x^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Jan 16 at 10:12
1
1
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
$begingroup$
I have already answered that request.
$endgroup$
– José Carlos Santos
Jan 16 at 10:21
|
show 3 more comments
$begingroup$
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.
answered Jan 16 at 7:18
ArthurArthur
116k7116198
$endgroup$
$begingroup$
@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?
$endgroup$
– Arthur
Jan 16 at 8:25
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:10
$begingroup$
I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
$endgroup$
– cmi
Jan 16 at 10:15
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:25
add a comment |
$begingroup$
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.
answered Jan 16 at 7:18
ArthurArthur
116k7116198
$endgroup$
$begingroup$
@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?
$endgroup$
– Arthur
Jan 16 at 8:25
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:10
$begingroup$
I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
$endgroup$
– cmi
Jan 16 at 10:15
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:25
add a comment |
$begingroup$
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.
answered Jan 16 at 7:18
ArthurArthur
116k7116198
$endgroup$
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.
answered Jan 16 at 7:18
ArthurArthur
116k7116198
edited Jan 16 at 7:25
answered Jan 16 at 7:18
ArthurArthur
116k7116198
answered Jan 16 at 7:18
ArthurArthur
116k7116198
answered Jan 16 at 7:18
ArthurArthur
116k7116198
116k7116198
$begingroup$
@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?
$endgroup$
– Arthur
Jan 16 at 8:25
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:10
$begingroup$
I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
$endgroup$
– cmi
Jan 16 at 10:15
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:25
add a comment |
$begingroup$
@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?
$endgroup$
– Arthur
Jan 16 at 8:25
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:10
$begingroup$
I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
$endgroup$
– cmi
Jan 16 at 10:15
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:25
$begingroup$
@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?
$endgroup$
– Arthur
Jan 16 at 8:25
$begingroup$
@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?
$endgroup$
– Arthur
Jan 16 at 8:25
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:10
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:10
$begingroup$
I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
$endgroup$
– cmi
Jan 16 at 10:15
$begingroup$
I could not grasp your comment..here in my example the limit function itself is non $C^1$ function..
$endgroup$
– cmi
Jan 16 at 10:15
1
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:25
$begingroup$
@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.
$endgroup$
– Arthur
Jan 16 at 10:25
add a comment |
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1
$begingroup$
Unit ball in what norm? (Your example probably works for that case too, possibly with minor tweaks.)
$endgroup$
– Arthur
Jan 16 at 6:54
$begingroup$
@Arthur I have edited my question ..
$endgroup$
– cmi
Jan 16 at 6:59
$begingroup$
@Arthur Please have a look
$endgroup$
– cmi
Jan 16 at 7:03
$begingroup$
@Arthur I am trying to understand your hint
$endgroup$
– cmi
Jan 16 at 7:06
2
$begingroup$
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.
$endgroup$
– Arthur
Jan 16 at 7:19