Discuss compactness of the set of $L^2$ bounded functions
$begingroup$
Discuss weak and strong compactness of the following subsets of $L^2(0,1)$:
$A={uin L^2(0,1):||u||_{L^2}le1}.$
I know some theorems which might be helpful, but I don't know if I applied them correctly.
The set $A$ is a closed unit ball of the Banach space $L^2(0,1)$, which is an infinite dimensional normed space, so by Riesz theorem $A$ is not strongly compact. Moreover, since $L^2(0,1)$ is reflexive, by Banach-Alaoglu theorem its closed unit ball $A$ is weakly compact.
Is this correct?
functional-analysis hilbert-spaces compactness
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
$endgroup$
add a comment |
$begingroup$
Discuss weak and strong compactness of the following subsets of $L^2(0,1)$:
$A={uin L^2(0,1):||u||_{L^2}le1}.$
I know some theorems which might be helpful, but I don't know if I applied them correctly.
The set $A$ is a closed unit ball of the Banach space $L^2(0,1)$, which is an infinite dimensional normed space, so by Riesz theorem $A$ is not strongly compact. Moreover, since $L^2(0,1)$ is reflexive, by Banach-Alaoglu theorem its closed unit ball $A$ is weakly compact.
Is this correct?
functional-analysis hilbert-spaces compactness
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
$endgroup$
1
$begingroup$
Well, in this case $n=1$. You will also need the Poincaré's inequality to control the $L^2$ norm in terms of the $L^2$ norm of the derivative.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 10:00
2
$begingroup$
Riesz' lemma does the job, but for Hilbert spaces it's even simpler. Just take any orthonormal sequence $(e_n)$. Since $| e_n-e_m|=1$ for $nneq m$, it can't have a convergent subsequence.
$endgroup$
– MaoWao
Dec 4 '18 at 13:18
$begingroup$
@MaoWao very good example thank you, so thats prove that the set is not strongly compact, what about weak compact? Can you use the orthonormal sequence in this case too?
$endgroup$
– sound wave
Dec 4 '18 at 13:24
2
$begingroup$
You can, although I don't like it that much. An orthonormal basis incuces an isomorphism $L^2(0,1)toell^2$. On bounded subsets of $ell^2$, weak convergence is equivalent to pointwise convergence (since finitely supported sequences are dense in $ell^2$). To extract a pointwise convergent subsequence is elementary, just apply the usual diagonal argument.
$endgroup$
– MaoWao
Dec 4 '18 at 13:29
add a comment |
$begingroup$
Discuss weak and strong compactness of the following subsets of $L^2(0,1)$:
$A={uin L^2(0,1):||u||_{L^2}le1}.$
I know some theorems which might be helpful, but I don't know if I applied them correctly.
The set $A$ is a closed unit ball of the Banach space $L^2(0,1)$, which is an infinite dimensional normed space, so by Riesz theorem $A$ is not strongly compact. Moreover, since $L^2(0,1)$ is reflexive, by Banach-Alaoglu theorem its closed unit ball $A$ is weakly compact.
Is this correct?
functional-analysis hilbert-spaces compactness
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
$endgroup$
Discuss weak and strong compactness of the following subsets of $L^2(0,1)$:
$A={uin L^2(0,1):||u||_{L^2}le1}.$
I know some theorems which might be helpful, but I don't know if I applied them correctly.
The set $A$ is a closed unit ball of the Banach space $L^2(0,1)$, which is an infinite dimensional normed space, so by Riesz theorem $A$ is not strongly compact. Moreover, since $L^2(0,1)$ is reflexive, by Banach-Alaoglu theorem its closed unit ball $A$ is weakly compact.
Is this correct?
functional-analysis hilbert-spaces compactness
functional-analysis hilbert-spaces compactness
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
edited Dec 4 '18 at 13:56
sound wave
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
asked Dec 3 '18 at 1:04
sound wavesound wave
1618
1618
1
$begingroup$
Well, in this case $n=1$. You will also need the Poincaré's inequality to control the $L^2$ norm in terms of the $L^2$ norm of the derivative.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 10:00
2
$begingroup$
Riesz' lemma does the job, but for Hilbert spaces it's even simpler. Just take any orthonormal sequence $(e_n)$. Since $| e_n-e_m|=1$ for $nneq m$, it can't have a convergent subsequence.
$endgroup$
– MaoWao
Dec 4 '18 at 13:18
$begingroup$
@MaoWao very good example thank you, so thats prove that the set is not strongly compact, what about weak compact? Can you use the orthonormal sequence in this case too?
$endgroup$
– sound wave
Dec 4 '18 at 13:24
2
$begingroup$
You can, although I don't like it that much. An orthonormal basis incuces an isomorphism $L^2(0,1)toell^2$. On bounded subsets of $ell^2$, weak convergence is equivalent to pointwise convergence (since finitely supported sequences are dense in $ell^2$). To extract a pointwise convergent subsequence is elementary, just apply the usual diagonal argument.
$endgroup$
– MaoWao
Dec 4 '18 at 13:29
add a comment |
1
$begingroup$
Well, in this case $n=1$. You will also need the Poincaré's inequality to control the $L^2$ norm in terms of the $L^2$ norm of the derivative.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 10:00
2
$begingroup$
Riesz' lemma does the job, but for Hilbert spaces it's even simpler. Just take any orthonormal sequence $(e_n)$. Since $| e_n-e_m|=1$ for $nneq m$, it can't have a convergent subsequence.
$endgroup$
– MaoWao
Dec 4 '18 at 13:18
$begingroup$
@MaoWao very good example thank you, so thats prove that the set is not strongly compact, what about weak compact? Can you use the orthonormal sequence in this case too?
$endgroup$
– sound wave
Dec 4 '18 at 13:24
2
$begingroup$
You can, although I don't like it that much. An orthonormal basis incuces an isomorphism $L^2(0,1)toell^2$. On bounded subsets of $ell^2$, weak convergence is equivalent to pointwise convergence (since finitely supported sequences are dense in $ell^2$). To extract a pointwise convergent subsequence is elementary, just apply the usual diagonal argument.
$endgroup$
– MaoWao
Dec 4 '18 at 13:29
1
1
$begingroup$
Well, in this case $n=1$. You will also need the Poincaré's inequality to control the $L^2$ norm in terms of the $L^2$ norm of the derivative.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 10:00
$begingroup$
Well, in this case $n=1$. You will also need the Poincaré's inequality to control the $L^2$ norm in terms of the $L^2$ norm of the derivative.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 10:00
2
2
$begingroup$
Riesz' lemma does the job, but for Hilbert spaces it's even simpler. Just take any orthonormal sequence $(e_n)$. Since $| e_n-e_m|=1$ for $nneq m$, it can't have a convergent subsequence.
$endgroup$
– MaoWao
Dec 4 '18 at 13:18
$begingroup$
Riesz' lemma does the job, but for Hilbert spaces it's even simpler. Just take any orthonormal sequence $(e_n)$. Since $| e_n-e_m|=1$ for $nneq m$, it can't have a convergent subsequence.
$endgroup$
– MaoWao
Dec 4 '18 at 13:18
$begingroup$
@MaoWao very good example thank you, so thats prove that the set is not strongly compact, what about weak compact? Can you use the orthonormal sequence in this case too?
$endgroup$
– sound wave
Dec 4 '18 at 13:24
$begingroup$
@MaoWao very good example thank you, so thats prove that the set is not strongly compact, what about weak compact? Can you use the orthonormal sequence in this case too?
$endgroup$
– sound wave
Dec 4 '18 at 13:24
2
2
$begingroup$
You can, although I don't like it that much. An orthonormal basis incuces an isomorphism $L^2(0,1)toell^2$. On bounded subsets of $ell^2$, weak convergence is equivalent to pointwise convergence (since finitely supported sequences are dense in $ell^2$). To extract a pointwise convergent subsequence is elementary, just apply the usual diagonal argument.
$endgroup$
– MaoWao
Dec 4 '18 at 13:29
$begingroup$
You can, although I don't like it that much. An orthonormal basis incuces an isomorphism $L^2(0,1)toell^2$. On bounded subsets of $ell^2$, weak convergence is equivalent to pointwise convergence (since finitely supported sequences are dense in $ell^2$). To extract a pointwise convergent subsequence is elementary, just apply the usual diagonal argument.
$endgroup$
– MaoWao
Dec 4 '18 at 13:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Concerning point A, you solved it correctly via "abstract nonsense". However, I think that it is always better to use concrete examples when possible.
In this case, a concrete example of a sequence contained in $A$ that has no converging subsequences is
$$
f_n(x)=sqrt n f(nx), $$
where $f$ is any nonzero element of $A$.
Here I omitted an important piece of information.
The function $f$ is defined on $[0, 1]$, but we implicitly consider that $$f(x)=0,qquad text{if }xnotin [0,1].$$
In particular, for all $xne 0$, it holds that $nx>1$ for all sufficiently large $n$, and so $f(nx)=0$ eventually. Therefore
$$
lim_{nto infty} f_n(x)= 0,$$
and so $f_nto 0$ pointwise almost everywhere.
Now the change of variable formula for integrals yields $$tag{1}|f_n|_{L^2}=|f|_{L^2},qquad forall nge 1;$$
so $f_nin A$ for all $n$. Now suppose for a contradiction that there exists $gin L^2$ and a subsequence $f_{k_n}$ such that $$|f_{k(n)}-g|_{L^2}to 0.$$
By (1), it must be that $|g|_{L^2}=|f|_{L^2}ne 0$; thus, $$gne 0.$$ However, any sequence that converges in $L^2$ has a subsequence that converges pointwise almost everywhere, so there exists a sub-sub-sequence $f_{k(h(n))}$ such that
$$
f_{k(h(n))}to g,qquad text{almost everywhere.}$$
And this is a contradiction, for $f_{k(h(n))}to 0$ almost everywhere by the remark in the colored box, and so it would imply that $g=0$.
$endgroup$
1
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
1
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
1
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
1
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
1
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
|
show 6 more comments
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1 Answer
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$begingroup$
Concerning point A, you solved it correctly via "abstract nonsense". However, I think that it is always better to use concrete examples when possible.
In this case, a concrete example of a sequence contained in $A$ that has no converging subsequences is
$$
f_n(x)=sqrt n f(nx), $$
where $f$ is any nonzero element of $A$.
Here I omitted an important piece of information.
The function $f$ is defined on $[0, 1]$, but we implicitly consider that $$f(x)=0,qquad text{if }xnotin [0,1].$$
In particular, for all $xne 0$, it holds that $nx>1$ for all sufficiently large $n$, and so $f(nx)=0$ eventually. Therefore
$$
lim_{nto infty} f_n(x)= 0,$$
and so $f_nto 0$ pointwise almost everywhere.
Now the change of variable formula for integrals yields $$tag{1}|f_n|_{L^2}=|f|_{L^2},qquad forall nge 1;$$
so $f_nin A$ for all $n$. Now suppose for a contradiction that there exists $gin L^2$ and a subsequence $f_{k_n}$ such that $$|f_{k(n)}-g|_{L^2}to 0.$$
By (1), it must be that $|g|_{L^2}=|f|_{L^2}ne 0$; thus, $$gne 0.$$ However, any sequence that converges in $L^2$ has a subsequence that converges pointwise almost everywhere, so there exists a sub-sub-sequence $f_{k(h(n))}$ such that
$$
f_{k(h(n))}to g,qquad text{almost everywhere.}$$
And this is a contradiction, for $f_{k(h(n))}to 0$ almost everywhere by the remark in the colored box, and so it would imply that $g=0$.
$endgroup$
1
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
1
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
1
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
1
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
1
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
|
show 6 more comments
$begingroup$
Concerning point A, you solved it correctly via "abstract nonsense". However, I think that it is always better to use concrete examples when possible.
In this case, a concrete example of a sequence contained in $A$ that has no converging subsequences is
$$
f_n(x)=sqrt n f(nx), $$
where $f$ is any nonzero element of $A$.
Here I omitted an important piece of information.
The function $f$ is defined on $[0, 1]$, but we implicitly consider that $$f(x)=0,qquad text{if }xnotin [0,1].$$
In particular, for all $xne 0$, it holds that $nx>1$ for all sufficiently large $n$, and so $f(nx)=0$ eventually. Therefore
$$
lim_{nto infty} f_n(x)= 0,$$
and so $f_nto 0$ pointwise almost everywhere.
Now the change of variable formula for integrals yields $$tag{1}|f_n|_{L^2}=|f|_{L^2},qquad forall nge 1;$$
so $f_nin A$ for all $n$. Now suppose for a contradiction that there exists $gin L^2$ and a subsequence $f_{k_n}$ such that $$|f_{k(n)}-g|_{L^2}to 0.$$
By (1), it must be that $|g|_{L^2}=|f|_{L^2}ne 0$; thus, $$gne 0.$$ However, any sequence that converges in $L^2$ has a subsequence that converges pointwise almost everywhere, so there exists a sub-sub-sequence $f_{k(h(n))}$ such that
$$
f_{k(h(n))}to g,qquad text{almost everywhere.}$$
And this is a contradiction, for $f_{k(h(n))}to 0$ almost everywhere by the remark in the colored box, and so it would imply that $g=0$.
$endgroup$
1
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
1
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
1
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
1
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
1
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
|
show 6 more comments
$begingroup$
Concerning point A, you solved it correctly via "abstract nonsense". However, I think that it is always better to use concrete examples when possible.
In this case, a concrete example of a sequence contained in $A$ that has no converging subsequences is
$$
f_n(x)=sqrt n f(nx), $$
where $f$ is any nonzero element of $A$.
Here I omitted an important piece of information.
The function $f$ is defined on $[0, 1]$, but we implicitly consider that $$f(x)=0,qquad text{if }xnotin [0,1].$$
In particular, for all $xne 0$, it holds that $nx>1$ for all sufficiently large $n$, and so $f(nx)=0$ eventually. Therefore
$$
lim_{nto infty} f_n(x)= 0,$$
and so $f_nto 0$ pointwise almost everywhere.
Now the change of variable formula for integrals yields $$tag{1}|f_n|_{L^2}=|f|_{L^2},qquad forall nge 1;$$
so $f_nin A$ for all $n$. Now suppose for a contradiction that there exists $gin L^2$ and a subsequence $f_{k_n}$ such that $$|f_{k(n)}-g|_{L^2}to 0.$$
By (1), it must be that $|g|_{L^2}=|f|_{L^2}ne 0$; thus, $$gne 0.$$ However, any sequence that converges in $L^2$ has a subsequence that converges pointwise almost everywhere, so there exists a sub-sub-sequence $f_{k(h(n))}$ such that
$$
f_{k(h(n))}to g,qquad text{almost everywhere.}$$
And this is a contradiction, for $f_{k(h(n))}to 0$ almost everywhere by the remark in the colored box, and so it would imply that $g=0$.
$endgroup$
Concerning point A, you solved it correctly via "abstract nonsense". However, I think that it is always better to use concrete examples when possible.
In this case, a concrete example of a sequence contained in $A$ that has no converging subsequences is
$$
f_n(x)=sqrt n f(nx), $$
where $f$ is any nonzero element of $A$.
Here I omitted an important piece of information.
The function $f$ is defined on $[0, 1]$, but we implicitly consider that $$f(x)=0,qquad text{if }xnotin [0,1].$$
In particular, for all $xne 0$, it holds that $nx>1$ for all sufficiently large $n$, and so $f(nx)=0$ eventually. Therefore
$$
lim_{nto infty} f_n(x)= 0,$$
and so $f_nto 0$ pointwise almost everywhere.
Now the change of variable formula for integrals yields $$tag{1}|f_n|_{L^2}=|f|_{L^2},qquad forall nge 1;$$
so $f_nin A$ for all $n$. Now suppose for a contradiction that there exists $gin L^2$ and a subsequence $f_{k_n}$ such that $$|f_{k(n)}-g|_{L^2}to 0.$$
By (1), it must be that $|g|_{L^2}=|f|_{L^2}ne 0$; thus, $$gne 0.$$ However, any sequence that converges in $L^2$ has a subsequence that converges pointwise almost everywhere, so there exists a sub-sub-sequence $f_{k(h(n))}$ such that
$$
f_{k(h(n))}to g,qquad text{almost everywhere.}$$
And this is a contradiction, for $f_{k(h(n))}to 0$ almost everywhere by the remark in the colored box, and so it would imply that $g=0$.
edited Dec 3 '18 at 20:00
community wiki
3 revs
Giuseppe Negro
1
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
1
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
1
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
1
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
1
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
|
show 6 more comments
1
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
1
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
1
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
1
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
1
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
1
1
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
$begingroup$
@soundwave: That contradicts the very definition of compactness.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:13
1
1
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
$begingroup$
@soundwave: I really think you should try harder; these are questions with an immediate answer. Think about equation (1).
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 13:25
1
1
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
$begingroup$
@soundwave: Fair enough, that's a rightful question and I'll answer it with a edit.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:12
1
1
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
$begingroup$
@soundwave: No. I truly mean what I wrote, I double-checked.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 17:25
1
1
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
$begingroup$
I wrote it. If the argument of $f$ is outside $[0, 1]$ then $f$ vanishes.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 18:26
|
show 6 more comments
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Well, in this case $n=1$. You will also need the Poincaré's inequality to control the $L^2$ norm in terms of the $L^2$ norm of the derivative.
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– Giuseppe Negro
Dec 3 '18 at 10:00
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Riesz' lemma does the job, but for Hilbert spaces it's even simpler. Just take any orthonormal sequence $(e_n)$. Since $| e_n-e_m|=1$ for $nneq m$, it can't have a convergent subsequence.
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– MaoWao
Dec 4 '18 at 13:18
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@MaoWao very good example thank you, so thats prove that the set is not strongly compact, what about weak compact? Can you use the orthonormal sequence in this case too?
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– sound wave
Dec 4 '18 at 13:24
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You can, although I don't like it that much. An orthonormal basis incuces an isomorphism $L^2(0,1)toell^2$. On bounded subsets of $ell^2$, weak convergence is equivalent to pointwise convergence (since finitely supported sequences are dense in $ell^2$). To extract a pointwise convergent subsequence is elementary, just apply the usual diagonal argument.
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– MaoWao
Dec 4 '18 at 13:29