Isn't all C$^*$-algebra have this property?
$begingroup$
A$ $ C$^*!$-algebra $A$ has property $*$ if for each suquence $(x_n)_n$:
$$text{if} x_nto 0(text{weakly}) Longrightarrow
yx_nto 0 (text{in norm}) forall y! in ! ! A $$
Is there any infinite dimensional C$^*$-algebra with this property?
I can only prove that $$x_nto 0(weakly) text{is equivalent to} yx_nto 0(weakly) forall y!in !!A $$
c-star-algebras
asked Jan 2 at 16:07
DarmanDarman
538112
$endgroup$
add a comment |
$begingroup$
A$ $ C$^*!$-algebra $A$ has property $*$ if for each suquence $(x_n)_n$:
$$text{if} x_nto 0(text{weakly}) Longrightarrow
yx_nto 0 (text{in norm}) forall y! in ! ! A $$
Is there any infinite dimensional C$^*$-algebra with this property?
I can only prove that $$x_nto 0(weakly) text{is equivalent to} yx_nto 0(weakly) forall y!in !!A $$
c-star-algebras
asked Jan 2 at 16:07
DarmanDarman
538112
$endgroup$
add a comment |
$begingroup$
A$ $ C$^*!$-algebra $A$ has property $*$ if for each suquence $(x_n)_n$:
$$text{if} x_nto 0(text{weakly}) Longrightarrow
yx_nto 0 (text{in norm}) forall y! in ! ! A $$
Is there any infinite dimensional C$^*$-algebra with this property?
I can only prove that $$x_nto 0(weakly) text{is equivalent to} yx_nto 0(weakly) forall y!in !!A $$
c-star-algebras
asked Jan 2 at 16:07
DarmanDarman
538112
$endgroup$
A$ $ C$^*!$-algebra $A$ has property $*$ if for each suquence $(x_n)_n$:
$$text{if} x_nto 0(text{weakly}) Longrightarrow
yx_nto 0 (text{in norm}) forall y! in ! ! A $$
Is there any infinite dimensional C$^*$-algebra with this property?
I can only prove that $$x_nto 0(weakly) text{is equivalent to} yx_nto 0(weakly) forall y!in !!A $$
c-star-algebras
c-star-algebras
asked Jan 2 at 16:07
DarmanDarman
538112
asked Jan 2 at 16:07
DarmanDarman
538112
edited Jan 2 at 16:19
Darman
asked Jan 2 at 16:07
DarmanDarman
538112
asked Jan 2 at 16:07
DarmanDarman
538112
asked Jan 2 at 16:07
DarmanDarman
538112
538112
add a comment |
add a comment |
1 Answer
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$begingroup$
No.
This is easy when $A$ is separable. Your condition implies that every weakly convergent sequence is norm convergent (this is trivial when $A$ is unital, and if $A$ is non-unital it follows easily by using an approximate unit). When $A$ is separable, the weak topology is metrizable, so the condition $(*)$ implies that the weak topology and the norm topology agree; this only happens in finite dimension.
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No.
This is easy when $A$ is separable. Your condition implies that every weakly convergent sequence is norm convergent (this is trivial when $A$ is unital, and if $A$ is non-unital it follows easily by using an approximate unit). When $A$ is separable, the weak topology is metrizable, so the condition $(*)$ implies that the weak topology and the norm topology agree; this only happens in finite dimension.
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
add a comment |
$begingroup$
No.
This is easy when $A$ is separable. Your condition implies that every weakly convergent sequence is norm convergent (this is trivial when $A$ is unital, and if $A$ is non-unital it follows easily by using an approximate unit). When $A$ is separable, the weak topology is metrizable, so the condition $(*)$ implies that the weak topology and the norm topology agree; this only happens in finite dimension.
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
add a comment |
$begingroup$
No.
This is easy when $A$ is separable. Your condition implies that every weakly convergent sequence is norm convergent (this is trivial when $A$ is unital, and if $A$ is non-unital it follows easily by using an approximate unit). When $A$ is separable, the weak topology is metrizable, so the condition $(*)$ implies that the weak topology and the norm topology agree; this only happens in finite dimension.
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
No.
This is easy when $A$ is separable. Your condition implies that every weakly convergent sequence is norm convergent (this is trivial when $A$ is unital, and if $A$ is non-unital it follows easily by using an approximate unit). When $A$ is separable, the weak topology is metrizable, so the condition $(*)$ implies that the weak topology and the norm topology agree; this only happens in finite dimension.
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 2 at 22:38
Martin ArgeramiMartin Argerami
128k1183183
128k1183183
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
add a comment |
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
What about non separable algebras?
$endgroup$
– Aweygan
Jan 2 at 22:53
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
$begingroup$
Not immediately obvious to me, but I would be really surprised if such an algebra exists.
$endgroup$
– Martin Argerami
Jan 2 at 23:26
add a comment |
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