Difference of Closed Convex sets in Banach Space
$begingroup$
Let $A$ be a closed, convex, set in a Banach space $X$, and let $B$ be a closed, bounded, convex set in $X$. Assume that $A cap B = emptyset$. Set $C = A- B$. Prove that $C$ is closed, and convex.
So proving $C$ is convex is not too hard, however I am having issues proving it is closed. According to existing convex geometry literature, this is a well-known result, that requires the assumption of $B$ being bounded. However, I am not sure how to use it. I assume the proof would begin by picking $c_n in C$ such that $c_n to c$, $c in X$, and showing that in fact that $c in C$. We would write:
$$
c_n = a_n - b_n
$$
and somehow show that $c = a -b$ for some $a,b in C$, presumably $a = lim_{n to infty} a_n$, $b = lim_{n to infty} b_n$. But the existence of the limit for $c$ says nothing about the limits of $a_n,b_n$ (it can be deduced that $a_n$ is norm bounded using the norm bound on $b_n$), but I am truly at a loss on how to proceed. A hint would be appreciated.
functional-analysis convex-analysis
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
$endgroup$
add a comment |
$begingroup$
Let $A$ be a closed, convex, set in a Banach space $X$, and let $B$ be a closed, bounded, convex set in $X$. Assume that $A cap B = emptyset$. Set $C = A- B$. Prove that $C$ is closed, and convex.
So proving $C$ is convex is not too hard, however I am having issues proving it is closed. According to existing convex geometry literature, this is a well-known result, that requires the assumption of $B$ being bounded. However, I am not sure how to use it. I assume the proof would begin by picking $c_n in C$ such that $c_n to c$, $c in X$, and showing that in fact that $c in C$. We would write:
$$
c_n = a_n - b_n
$$
and somehow show that $c = a -b$ for some $a,b in C$, presumably $a = lim_{n to infty} a_n$, $b = lim_{n to infty} b_n$. But the existence of the limit for $c$ says nothing about the limits of $a_n,b_n$ (it can be deduced that $a_n$ is norm bounded using the norm bound on $b_n$), but I am truly at a loss on how to proceed. A hint would be appreciated.
functional-analysis convex-analysis
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
$endgroup$
$begingroup$
Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer.
$endgroup$
– SmileyCraft
Jan 19 at 18:52
$begingroup$
So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it.
$endgroup$
– rubikscube09
Jan 19 at 18:55
add a comment |
$begingroup$
Let $A$ be a closed, convex, set in a Banach space $X$, and let $B$ be a closed, bounded, convex set in $X$. Assume that $A cap B = emptyset$. Set $C = A- B$. Prove that $C$ is closed, and convex.
So proving $C$ is convex is not too hard, however I am having issues proving it is closed. According to existing convex geometry literature, this is a well-known result, that requires the assumption of $B$ being bounded. However, I am not sure how to use it. I assume the proof would begin by picking $c_n in C$ such that $c_n to c$, $c in X$, and showing that in fact that $c in C$. We would write:
$$
c_n = a_n - b_n
$$
and somehow show that $c = a -b$ for some $a,b in C$, presumably $a = lim_{n to infty} a_n$, $b = lim_{n to infty} b_n$. But the existence of the limit for $c$ says nothing about the limits of $a_n,b_n$ (it can be deduced that $a_n$ is norm bounded using the norm bound on $b_n$), but I am truly at a loss on how to proceed. A hint would be appreciated.
functional-analysis convex-analysis
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
$endgroup$
Let $A$ be a closed, convex, set in a Banach space $X$, and let $B$ be a closed, bounded, convex set in $X$. Assume that $A cap B = emptyset$. Set $C = A- B$. Prove that $C$ is closed, and convex.
So proving $C$ is convex is not too hard, however I am having issues proving it is closed. According to existing convex geometry literature, this is a well-known result, that requires the assumption of $B$ being bounded. However, I am not sure how to use it. I assume the proof would begin by picking $c_n in C$ such that $c_n to c$, $c in X$, and showing that in fact that $c in C$. We would write:
$$
c_n = a_n - b_n
$$
and somehow show that $c = a -b$ for some $a,b in C$, presumably $a = lim_{n to infty} a_n$, $b = lim_{n to infty} b_n$. But the existence of the limit for $c$ says nothing about the limits of $a_n,b_n$ (it can be deduced that $a_n$ is norm bounded using the norm bound on $b_n$), but I am truly at a loss on how to proceed. A hint would be appreciated.
functional-analysis convex-analysis
functional-analysis convex-analysis
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
asked Jan 19 at 18:35
rubikscube09rubikscube09
1,270719
1,270719
$begingroup$
Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer.
$endgroup$
– SmileyCraft
Jan 19 at 18:52
$begingroup$
So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it.
$endgroup$
– rubikscube09
Jan 19 at 18:55
add a comment |
$begingroup$
Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer.
$endgroup$
– SmileyCraft
Jan 19 at 18:52
$begingroup$
So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it.
$endgroup$
– rubikscube09
Jan 19 at 18:55
$begingroup$
Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer.
$endgroup$
– SmileyCraft
Jan 19 at 18:52
$begingroup$
Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer.
$endgroup$
– SmileyCraft
Jan 19 at 18:52
$begingroup$
So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it.
$endgroup$
– rubikscube09
Jan 19 at 18:55
$begingroup$
So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it.
$endgroup$
– rubikscube09
Jan 19 at 18:55
add a comment |
1 Answer
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oldest
votes
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If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $bin B$ and $c+bin A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.
That $B$ is bounded is really necessary for this theorem. Consider the following example.
Let $X=mathbb{R}^2$ and $A={(x,y):xgeq0,xygeq1}$ and $B={(x,y):xgeq0,xyleq-1}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B={(x,y):y>0}$ is not closed.
That $X$ is reflexive is also necessary for this theorem. Consider the following example.
Let $X=ell^1$ and $A={xin X:x_ngeq0,sum x_n=1}$. Define $Tin B(X)$ by $(Tx)_n=(1+frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0notin A-B$. However $e_n-T(e_n)to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is also necessary. Consider the following example.
Let $X=ell^2$ and let $A$ be the closed unit ball and let $B={(1+frac1n)e_n}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0notin A-B$. However $e_n-(1+frac1n)e_nto0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $bin B$ and $c+bin A$ and $c=c+b-b$.
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Nice counterexample!
$endgroup$
– gerw
Jan 19 at 19:44
$begingroup$
Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
$endgroup$
– rubikscube09
Jan 19 at 19:48
1
$begingroup$
@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
$endgroup$
– SmileyCraft
Jan 19 at 19:51
1
$begingroup$
You have a bit of a typo, but that is a brilliant question worthy of its own post :)
$endgroup$
– SmileyCraft
Jan 19 at 20:05
1
$begingroup$
Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
$endgroup$
– rubikscube09
Jan 19 at 20:20
|
show 2 more comments
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$begingroup$
If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $bin B$ and $c+bin A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.
That $B$ is bounded is really necessary for this theorem. Consider the following example.
Let $X=mathbb{R}^2$ and $A={(x,y):xgeq0,xygeq1}$ and $B={(x,y):xgeq0,xyleq-1}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B={(x,y):y>0}$ is not closed.
That $X$ is reflexive is also necessary for this theorem. Consider the following example.
Let $X=ell^1$ and $A={xin X:x_ngeq0,sum x_n=1}$. Define $Tin B(X)$ by $(Tx)_n=(1+frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0notin A-B$. However $e_n-T(e_n)to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is also necessary. Consider the following example.
Let $X=ell^2$ and let $A$ be the closed unit ball and let $B={(1+frac1n)e_n}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0notin A-B$. However $e_n-(1+frac1n)e_nto0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $bin B$ and $c+bin A$ and $c=c+b-b$.
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Nice counterexample!
$endgroup$
– gerw
Jan 19 at 19:44
$begingroup$
Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
$endgroup$
– rubikscube09
Jan 19 at 19:48
1
$begingroup$
@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
$endgroup$
– SmileyCraft
Jan 19 at 19:51
1
$begingroup$
You have a bit of a typo, but that is a brilliant question worthy of its own post :)
$endgroup$
– SmileyCraft
Jan 19 at 20:05
1
$begingroup$
Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
$endgroup$
– rubikscube09
Jan 19 at 20:20
|
show 2 more comments
$begingroup$
If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $bin B$ and $c+bin A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.
That $B$ is bounded is really necessary for this theorem. Consider the following example.
Let $X=mathbb{R}^2$ and $A={(x,y):xgeq0,xygeq1}$ and $B={(x,y):xgeq0,xyleq-1}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B={(x,y):y>0}$ is not closed.
That $X$ is reflexive is also necessary for this theorem. Consider the following example.
Let $X=ell^1$ and $A={xin X:x_ngeq0,sum x_n=1}$. Define $Tin B(X)$ by $(Tx)_n=(1+frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0notin A-B$. However $e_n-T(e_n)to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is also necessary. Consider the following example.
Let $X=ell^2$ and let $A$ be the closed unit ball and let $B={(1+frac1n)e_n}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0notin A-B$. However $e_n-(1+frac1n)e_nto0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $bin B$ and $c+bin A$ and $c=c+b-b$.
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Nice counterexample!
$endgroup$
– gerw
Jan 19 at 19:44
$begingroup$
Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
$endgroup$
– rubikscube09
Jan 19 at 19:48
1
$begingroup$
@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
$endgroup$
– SmileyCraft
Jan 19 at 19:51
1
$begingroup$
You have a bit of a typo, but that is a brilliant question worthy of its own post :)
$endgroup$
– SmileyCraft
Jan 19 at 20:05
1
$begingroup$
Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
$endgroup$
– rubikscube09
Jan 19 at 20:20
|
show 2 more comments
$begingroup$
If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $bin B$ and $c+bin A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.
That $B$ is bounded is really necessary for this theorem. Consider the following example.
Let $X=mathbb{R}^2$ and $A={(x,y):xgeq0,xygeq1}$ and $B={(x,y):xgeq0,xyleq-1}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B={(x,y):y>0}$ is not closed.
That $X$ is reflexive is also necessary for this theorem. Consider the following example.
Let $X=ell^1$ and $A={xin X:x_ngeq0,sum x_n=1}$. Define $Tin B(X)$ by $(Tx)_n=(1+frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0notin A-B$. However $e_n-T(e_n)to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is also necessary. Consider the following example.
Let $X=ell^2$ and let $A$ be the closed unit ball and let $B={(1+frac1n)e_n}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0notin A-B$. However $e_n-(1+frac1n)e_nto0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $bin B$ and $c+bin A$ and $c=c+b-b$.
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
$endgroup$
If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $bin B$ and $c+bin A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.
That $B$ is bounded is really necessary for this theorem. Consider the following example.
Let $X=mathbb{R}^2$ and $A={(x,y):xgeq0,xygeq1}$ and $B={(x,y):xgeq0,xyleq-1}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B={(x,y):y>0}$ is not closed.
That $X$ is reflexive is also necessary for this theorem. Consider the following example.
Let $X=ell^1$ and $A={xin X:x_ngeq0,sum x_n=1}$. Define $Tin B(X)$ by $(Tx)_n=(1+frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0notin A-B$. However $e_n-T(e_n)to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is also necessary. Consider the following example.
Let $X=ell^2$ and let $A$ be the closed unit ball and let $B={(1+frac1n)e_n}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0notin A-B$. However $e_n-(1+frac1n)e_nto0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $bin B$ and $c+bin A$ and $c=c+b-b$.
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
edited Jan 19 at 19:51
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
answered Jan 19 at 19:24
SmileyCraftSmileyCraft
3,591517
3,591517
$begingroup$
Nice counterexample!
$endgroup$
– gerw
Jan 19 at 19:44
$begingroup$
Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
$endgroup$
– rubikscube09
Jan 19 at 19:48
1
$begingroup$
@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
$endgroup$
– SmileyCraft
Jan 19 at 19:51
1
$begingroup$
You have a bit of a typo, but that is a brilliant question worthy of its own post :)
$endgroup$
– SmileyCraft
Jan 19 at 20:05
1
$begingroup$
Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
$endgroup$
– rubikscube09
Jan 19 at 20:20
|
show 2 more comments
$begingroup$
Nice counterexample!
$endgroup$
– gerw
Jan 19 at 19:44
$begingroup$
Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
$endgroup$
– rubikscube09
Jan 19 at 19:48
1
$begingroup$
@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
$endgroup$
– SmileyCraft
Jan 19 at 19:51
1
$begingroup$
You have a bit of a typo, but that is a brilliant question worthy of its own post :)
$endgroup$
– SmileyCraft
Jan 19 at 20:05
1
$begingroup$
Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
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– rubikscube09
Jan 19 at 20:20
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Nice counterexample!
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– gerw
Jan 19 at 19:44
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Nice counterexample!
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– gerw
Jan 19 at 19:44
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Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
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– rubikscube09
Jan 19 at 19:48
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Thank you, I understand the use of Alaoglu's theorem to extract a weak limit $b$, however would this have to phrased in the language of nets because compact $not implies$ sequentially compact in general?
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– rubikscube09
Jan 19 at 19:48
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@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
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– SmileyCraft
Jan 19 at 19:51
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@rubikscube09 Good point. I looked it up, and it follows from the Eberlein Smulian theorem.
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– SmileyCraft
Jan 19 at 19:51
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You have a bit of a typo, but that is a brilliant question worthy of its own post :)
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– SmileyCraft
Jan 19 at 20:05
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You have a bit of a typo, but that is a brilliant question worthy of its own post :)
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– SmileyCraft
Jan 19 at 20:05
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Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
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– rubikscube09
Jan 19 at 20:20
$begingroup$
Nevermind, it's rather silly. Weakly closed implies strongly closed, so $mathrm{cl}(K) subset mathrm{cl-wk}(K)$ as $mathrm{cl-wk}(K)$ is a closed set containing $K$ . Meanwhile, strongly closed + convex implies weakly closed, so $mathrm{cl-wk}(K) subset mathrm{cl}(K)$ because $mathrm{cl}(K)$ is a weakly closed set containing $K$.
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– rubikscube09
Jan 19 at 20:20
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Are you sure $X$ is a Banach space? Off the top of my head, I can come up with a proof for reflexive spaces, using the fact that the unit ball is weakly compact. I can also come up with a counterexample if $B$ is not bounded. However, for arbitrary Banach spaces I do not yet see the answer.
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– SmileyCraft
Jan 19 at 18:52
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So the book I am using (Brezis'-Functional Analysis) actually does this in the context of Hilbert spaces! I realize now that he perhaps would intend on me decomposing $c_n = P_A(c_n) - P_B(c_n)$ and then passing to the limit (not sure if this works either because it does not use boundedness). But if you have a more general proof (in reflexive spaces) I would by all means be happy to see it.
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– rubikscube09
Jan 19 at 18:55