Mixing
If $Esubset X^{*}$ is bounded, then so is its weak* closure
$begingroup$
If $X$ is a Banach space and $Esubset X^{*}$ is norm-bounded, I've shown that its weak* closure is also norm-bounded using Alaoglu's theorem. But perhaps using Alaoglu's theorem is not necessary?
(I've shown that if $Fsubset X$ is norm-bounded, then its weak closure is also norm-bounded...)
functional-analysis banach-spaces topological-vector-spaces
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space and $Esubset X^{*}$ is norm-bounded, I've shown that its weak* closure is also norm-bounded using Alaoglu's theorem. But perhaps using Alaoglu's theorem is not necessary?
(I've shown that if $Fsubset X$ is norm-bounded, then its weak closure is also norm-bounded...)
functional-analysis banach-spaces topological-vector-spaces
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space and $Esubset X^{*}$ is norm-bounded, I've shown that its weak* closure is also norm-bounded using Alaoglu's theorem. But perhaps using Alaoglu's theorem is not necessary?
(I've shown that if $Fsubset X$ is norm-bounded, then its weak closure is also norm-bounded...)
functional-analysis banach-spaces topological-vector-spaces
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
$endgroup$
If $X$ is a Banach space and $Esubset X^{*}$ is norm-bounded, I've shown that its weak* closure is also norm-bounded using Alaoglu's theorem. But perhaps using Alaoglu's theorem is not necessary?
(I've shown that if $Fsubset X$ is norm-bounded, then its weak closure is also norm-bounded...)
functional-analysis banach-spaces topological-vector-spaces
functional-analysis banach-spaces topological-vector-spaces
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
edited Mar 1 '15 at 22:00
Aubrey
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
asked Mar 1 '15 at 21:34
AubreyAubrey
699416
699416
add a comment |
add a comment |
3 Answers
3
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oldest
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I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R={phiin X^*: |phi|le R}$ is weak*-closed; indeed, any bounded set is contained in such a ball.
Take any $phi in X^*setminus B_R$. By definition of the norm, there is a unit vector $uin X$ such that $|phi(u)|>R$. Let $epsilon = |phi(u)|-R$. The set
$$U = {psiin X^* : |psi(u)-phi(u)|<epsilon}$$
is weak* open, contains $phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.
$endgroup$
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
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I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded.
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
$endgroup$
add a comment |
$begingroup$
If you are comfortable with the language of nets, this is not too hard to prove.
Since $E$ is bounded, suppose all $fin E$ satisfies $||f|| le C$ for some constant $C$. Let $f'$ be in the weak$^*$ closure of $E$. Then these exists a net $langle f_{alpha}rangle$ such that $f_{alpha}to f'$ in a weak$^*$ manner, i.e. $f_{alpha}(x)to f'(x)$ for all $xin X$.
For any fixed $x$ satisfying $||x|| = 1$, we have $||f_{alpha}(x)||le ||f_{alpha}|| le C$. Since $f_{alpha}(x)to f'(x)$, we have $||f'(x)||le C$. By definition of the norm of a functional, $||f'||le C$. This shows that the weak$^*$ closure of $E$ is also bounded (by the same constant!).
Remark: OP stated that he/she managed to prove that if $Fsubseteq X$ is norm-bonded, then so is the weak-closure. I would have thought that the proof is similar and in fact harder in this latter case (having to use the isometry of $X$ into $X^{**}$).
answered Jan 19 at 3:37
suncup224suncup224
1,186716
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R={phiin X^*: |phi|le R}$ is weak*-closed; indeed, any bounded set is contained in such a ball.
Take any $phi in X^*setminus B_R$. By definition of the norm, there is a unit vector $uin X$ such that $|phi(u)|>R$. Let $epsilon = |phi(u)|-R$. The set
$$U = {psiin X^* : |psi(u)-phi(u)|<epsilon}$$
is weak* open, contains $phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.
$endgroup$
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
add a comment |
$begingroup$
I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R={phiin X^*: |phi|le R}$ is weak*-closed; indeed, any bounded set is contained in such a ball.
Take any $phi in X^*setminus B_R$. By definition of the norm, there is a unit vector $uin X$ such that $|phi(u)|>R$. Let $epsilon = |phi(u)|-R$. The set
$$U = {psiin X^* : |psi(u)-phi(u)|<epsilon}$$
is weak* open, contains $phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.
$endgroup$
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
add a comment |
$begingroup$
I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R={phiin X^*: |phi|le R}$ is weak*-closed; indeed, any bounded set is contained in such a ball.
Take any $phi in X^*setminus B_R$. By definition of the norm, there is a unit vector $uin X$ such that $|phi(u)|>R$. Let $epsilon = |phi(u)|-R$. The set
$$U = {psiin X^* : |psi(u)-phi(u)|<epsilon}$$
is weak* open, contains $phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.
$endgroup$
I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R={phiin X^*: |phi|le R}$ is weak*-closed; indeed, any bounded set is contained in such a ball.
Take any $phi in X^*setminus B_R$. By definition of the norm, there is a unit vector $uin X$ such that $|phi(u)|>R$. Let $epsilon = |phi(u)|-R$. The set
$$U = {psiin X^* : |psi(u)-phi(u)|<epsilon}$$
is weak* open, contains $phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.
answered Mar 8 '15 at 6:49
user147263
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
add a comment |
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
$begingroup$
Isn't it obvious from Alaoglu's theorem that $B_R$ is weak*-compact and hence weak*-closed?
$endgroup$
– pikachuchameleon
Mar 17 '16 at 20:50
add a comment |
$begingroup$
I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded.
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
$endgroup$
add a comment |
$begingroup$
I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded.
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
$endgroup$
add a comment |
$begingroup$
I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded.
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
$endgroup$
I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded.
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
answered Mar 1 '15 at 22:23
AubreyAubrey
699416
699416
add a comment |
add a comment |
$begingroup$
If you are comfortable with the language of nets, this is not too hard to prove.
Since $E$ is bounded, suppose all $fin E$ satisfies $||f|| le C$ for some constant $C$. Let $f'$ be in the weak$^*$ closure of $E$. Then these exists a net $langle f_{alpha}rangle$ such that $f_{alpha}to f'$ in a weak$^*$ manner, i.e. $f_{alpha}(x)to f'(x)$ for all $xin X$.
For any fixed $x$ satisfying $||x|| = 1$, we have $||f_{alpha}(x)||le ||f_{alpha}|| le C$. Since $f_{alpha}(x)to f'(x)$, we have $||f'(x)||le C$. By definition of the norm of a functional, $||f'||le C$. This shows that the weak$^*$ closure of $E$ is also bounded (by the same constant!).
Remark: OP stated that he/she managed to prove that if $Fsubseteq X$ is norm-bonded, then so is the weak-closure. I would have thought that the proof is similar and in fact harder in this latter case (having to use the isometry of $X$ into $X^{**}$).
answered Jan 19 at 3:37
suncup224suncup224
1,186716
$endgroup$
add a comment |
$begingroup$
If you are comfortable with the language of nets, this is not too hard to prove.
Since $E$ is bounded, suppose all $fin E$ satisfies $||f|| le C$ for some constant $C$. Let $f'$ be in the weak$^*$ closure of $E$. Then these exists a net $langle f_{alpha}rangle$ such that $f_{alpha}to f'$ in a weak$^*$ manner, i.e. $f_{alpha}(x)to f'(x)$ for all $xin X$.
For any fixed $x$ satisfying $||x|| = 1$, we have $||f_{alpha}(x)||le ||f_{alpha}|| le C$. Since $f_{alpha}(x)to f'(x)$, we have $||f'(x)||le C$. By definition of the norm of a functional, $||f'||le C$. This shows that the weak$^*$ closure of $E$ is also bounded (by the same constant!).
Remark: OP stated that he/she managed to prove that if $Fsubseteq X$ is norm-bonded, then so is the weak-closure. I would have thought that the proof is similar and in fact harder in this latter case (having to use the isometry of $X$ into $X^{**}$).
answered Jan 19 at 3:37
suncup224suncup224
1,186716
$endgroup$
add a comment |
$begingroup$
If you are comfortable with the language of nets, this is not too hard to prove.
Since $E$ is bounded, suppose all $fin E$ satisfies $||f|| le C$ for some constant $C$. Let $f'$ be in the weak$^*$ closure of $E$. Then these exists a net $langle f_{alpha}rangle$ such that $f_{alpha}to f'$ in a weak$^*$ manner, i.e. $f_{alpha}(x)to f'(x)$ for all $xin X$.
For any fixed $x$ satisfying $||x|| = 1$, we have $||f_{alpha}(x)||le ||f_{alpha}|| le C$. Since $f_{alpha}(x)to f'(x)$, we have $||f'(x)||le C$. By definition of the norm of a functional, $||f'||le C$. This shows that the weak$^*$ closure of $E$ is also bounded (by the same constant!).
Remark: OP stated that he/she managed to prove that if $Fsubseteq X$ is norm-bonded, then so is the weak-closure. I would have thought that the proof is similar and in fact harder in this latter case (having to use the isometry of $X$ into $X^{**}$).
answered Jan 19 at 3:37
suncup224suncup224
1,186716
$endgroup$
If you are comfortable with the language of nets, this is not too hard to prove.
Since $E$ is bounded, suppose all $fin E$ satisfies $||f|| le C$ for some constant $C$. Let $f'$ be in the weak$^*$ closure of $E$. Then these exists a net $langle f_{alpha}rangle$ such that $f_{alpha}to f'$ in a weak$^*$ manner, i.e. $f_{alpha}(x)to f'(x)$ for all $xin X$.
For any fixed $x$ satisfying $||x|| = 1$, we have $||f_{alpha}(x)||le ||f_{alpha}|| le C$. Since $f_{alpha}(x)to f'(x)$, we have $||f'(x)||le C$. By definition of the norm of a functional, $||f'||le C$. This shows that the weak$^*$ closure of $E$ is also bounded (by the same constant!).
Remark: OP stated that he/she managed to prove that if $Fsubseteq X$ is norm-bonded, then so is the weak-closure. I would have thought that the proof is similar and in fact harder in this latter case (having to use the isometry of $X$ into $X^{**}$).
answered Jan 19 at 3:37
suncup224suncup224
1,186716
answered Jan 19 at 3:37
suncup224suncup224
1,186716
answered Jan 19 at 3:37
suncup224suncup224
1,186716
answered Jan 19 at 3:37
suncup224suncup224
1,186716
1,186716
add a comment |
add a comment |
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