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Bounding a positive element from below using a dense subset
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Let $A$ be a $C^*$-algebra. Let $a$ be a nonzero positive element of $A$. Suppose that $A$ equals the closed span of a subset $B$ of $A$, where $B$ is closed under the $*$ operation, linear combinations, and products. Is it possible to find $b in B$ such that $b$ is positive, nonzero, and $a geq b$ (i.e. $a-b$ is positive)?
operator-algebras c-star-algebras
asked Jul 31 at 17:04
Lucas
928
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up vote
1
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favorite
Let $A$ be a $C^*$-algebra. Let $a$ be a nonzero positive element of $A$. Suppose that $A$ equals the closed span of a subset $B$ of $A$, where $B$ is closed under the $*$ operation, linear combinations, and products. Is it possible to find $b in B$ such that $b$ is positive, nonzero, and $a geq b$ (i.e. $a-b$ is positive)?
operator-algebras c-star-algebras
asked Jul 31 at 17:04
Lucas
928
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be a $C^*$-algebra. Let $a$ be a nonzero positive element of $A$. Suppose that $A$ equals the closed span of a subset $B$ of $A$, where $B$ is closed under the $*$ operation, linear combinations, and products. Is it possible to find $b in B$ such that $b$ is positive, nonzero, and $a geq b$ (i.e. $a-b$ is positive)?
operator-algebras c-star-algebras
asked Jul 31 at 17:04
Lucas
928
Let $A$ be a $C^*$-algebra. Let $a$ be a nonzero positive element of $A$. Suppose that $A$ equals the closed span of a subset $B$ of $A$, where $B$ is closed under the $*$ operation, linear combinations, and products. Is it possible to find $b in B$ such that $b$ is positive, nonzero, and $a geq b$ (i.e. $a-b$ is positive)?
operator-algebras c-star-algebras
asked Jul 31 at 17:04
Lucas
928
edited Jul 31 at 19:17
asked Jul 31 at 17:04
Lucas
928
asked Jul 31 at 17:04
Lucas
928
asked Jul 31 at 17:04
Lucas
928
928
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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No. Let $A=C[-1,1]$, and $a(t)=e^-1/t^2$ (of course, with $a(0)=0$). Take $B=mathbb C[x]$, the polynomials. If $pin B$ and $0leq pleq a$, so $p(0)=0$. But then
$$
|p'(0)|=lim_hleft|fracp(h)-p(0)hright|=lim_hleft|fracp(h)hright|leqlim_hfraca(h)h=0.
$$
And then we can iterate this reasoning to get $p^(k)(0)=0$ for all $k$; thus $p=0$.
answered Jul 31 at 23:48
Martin Argerami
115k1071164
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up vote
1
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In general, it seems like no. Consider $A = mathbbC$. For any $lambda in mathbbC setminus 0$, $langle lambda rangle = mathbbC$. However, $lambda$ can be chosen so that $B = lambda$ does not even contain a positive element.
answered Jul 31 at 18:56
AGF
204
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No. Let $A=C[-1,1]$, and $a(t)=e^-1/t^2$ (of course, with $a(0)=0$). Take $B=mathbb C[x]$, the polynomials. If $pin B$ and $0leq pleq a$, so $p(0)=0$. But then
$$
|p'(0)|=lim_hleft|fracp(h)-p(0)hright|=lim_hleft|fracp(h)hright|leqlim_hfraca(h)h=0.
$$
And then we can iterate this reasoning to get $p^(k)(0)=0$ for all $k$; thus $p=0$.
answered Jul 31 at 23:48
Martin Argerami
115k1071164
add a comment |Â
up vote
3
down vote
accepted
No. Let $A=C[-1,1]$, and $a(t)=e^-1/t^2$ (of course, with $a(0)=0$). Take $B=mathbb C[x]$, the polynomials. If $pin B$ and $0leq pleq a$, so $p(0)=0$. But then
$$
|p'(0)|=lim_hleft|fracp(h)-p(0)hright|=lim_hleft|fracp(h)hright|leqlim_hfraca(h)h=0.
$$
And then we can iterate this reasoning to get $p^(k)(0)=0$ for all $k$; thus $p=0$.
answered Jul 31 at 23:48
Martin Argerami
115k1071164
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No. Let $A=C[-1,1]$, and $a(t)=e^-1/t^2$ (of course, with $a(0)=0$). Take $B=mathbb C[x]$, the polynomials. If $pin B$ and $0leq pleq a$, so $p(0)=0$. But then
$$
|p'(0)|=lim_hleft|fracp(h)-p(0)hright|=lim_hleft|fracp(h)hright|leqlim_hfraca(h)h=0.
$$
And then we can iterate this reasoning to get $p^(k)(0)=0$ for all $k$; thus $p=0$.
answered Jul 31 at 23:48
Martin Argerami
115k1071164
No. Let $A=C[-1,1]$, and $a(t)=e^-1/t^2$ (of course, with $a(0)=0$). Take $B=mathbb C[x]$, the polynomials. If $pin B$ and $0leq pleq a$, so $p(0)=0$. But then
$$
|p'(0)|=lim_hleft|fracp(h)-p(0)hright|=lim_hleft|fracp(h)hright|leqlim_hfraca(h)h=0.
$$
And then we can iterate this reasoning to get $p^(k)(0)=0$ for all $k$; thus $p=0$.
answered Jul 31 at 23:48
Martin Argerami
115k1071164
edited 1 hour ago
answered Jul 31 at 23:48
Martin Argerami
115k1071164
answered Jul 31 at 23:48
Martin Argerami
115k1071164
answered Jul 31 at 23:48
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
up vote
1
down vote
In general, it seems like no. Consider $A = mathbbC$. For any $lambda in mathbbC setminus 0$, $langle lambda rangle = mathbbC$. However, $lambda$ can be chosen so that $B = lambda$ does not even contain a positive element.
answered Jul 31 at 18:56
AGF
204
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
add a comment |Â
up vote
1
down vote
In general, it seems like no. Consider $A = mathbbC$. For any $lambda in mathbbC setminus 0$, $langle lambda rangle = mathbbC$. However, $lambda$ can be chosen so that $B = lambda$ does not even contain a positive element.
answered Jul 31 at 18:56
AGF
204
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In general, it seems like no. Consider $A = mathbbC$. For any $lambda in mathbbC setminus 0$, $langle lambda rangle = mathbbC$. However, $lambda$ can be chosen so that $B = lambda$ does not even contain a positive element.
answered Jul 31 at 18:56
AGF
204
In general, it seems like no. Consider $A = mathbbC$. For any $lambda in mathbbC setminus 0$, $langle lambda rangle = mathbbC$. However, $lambda$ can be chosen so that $B = lambda$ does not even contain a positive element.
answered Jul 31 at 18:56
AGF
204
answered Jul 31 at 18:56
AGF
204
answered Jul 31 at 18:56
AGF
204
answered Jul 31 at 18:56
AGF
204
204
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
add a comment |Â
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
True. I have added in the post that $B$ is closed under the $*$ operation, linear combinations, and products
– Lucas
Jul 31 at 19:17
add a comment |Â
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