Is the following set a compact set?
Let $A$ be defined as
$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$
I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?
general-topology functional-analysis compactness complete-spaces locally-compact-groups
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:44
MathCracky
445212
add a comment |
Let $A$ be defined as
$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$
I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?
general-topology functional-analysis compactness complete-spaces locally-compact-groups
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:44
MathCracky
445212
Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55
How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10
add a comment |
Let $A$ be defined as
$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$
I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?
general-topology functional-analysis compactness complete-spaces locally-compact-groups
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:44
MathCracky
445212
Let $A$ be defined as
$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$
I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?
general-topology functional-analysis compactness complete-spaces locally-compact-groups
general-topology functional-analysis compactness complete-spaces locally-compact-groups
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:44
MathCracky
445212
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:44
MathCracky
445212
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
edited Nov 25 '18 at 17:37
Yiorgos S. Smyrlis
62.5k1383162
62.5k1383162
asked Nov 25 '18 at 9:44
MathCracky
445212
asked Nov 25 '18 at 9:44
MathCracky
445212
asked Nov 25 '18 at 9:44
MathCracky
445212
445212
Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55
How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10
add a comment |
Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55
How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10
Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55
Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55
How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10
How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10
add a comment |
2 Answers
2
active
oldest
votes
The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.
But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.
I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
2
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
add a comment |
The answer is NO.
Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.
But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.
I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
2
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
add a comment |
The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.
But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.
I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
2
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
add a comment |
The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.
But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.
I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.
But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.
I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
edited Nov 25 '18 at 10:03
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
answered Nov 25 '18 at 9:58
José Carlos Santos
150k22121221
150k22121221
2
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
add a comment |
2
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
2
2
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
– Yanko
Nov 25 '18 at 10:44
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
– MaoWao
Nov 25 '18 at 17:58
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
– José Carlos Santos
Nov 25 '18 at 18:04
add a comment |
The answer is NO.
Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
add a comment |
The answer is NO.
Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
add a comment |
The answer is NO.
Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
The answer is NO.
Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
answered Nov 25 '18 at 10:55
Yiorgos S. Smyrlis
62.5k1383162
62.5k1383162
add a comment |
add a comment |
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Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55
How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10