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Examples of elements in $(X^*)^*$ that are not evaluation maps
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I just learned about the weak-* topology on $X^*$, and in this context was introduced to the 'dual of the dual of a space', and the functional $J: X rightarrow (X^*)^*$ where $J(x)[psi] = psi(x)$, i.e. $J$ takes a point $x in X$ to the functional that takes a map from $X$ to the reals, and evaluates it at the point $x$. From this, $J(X) subset (X^*)^*$, and the weak-* topology on $X^*$ is the topology induced by this $J(X)$.
It originally took me a while to parse this definition in the right way, but it has become more intuitive after I started thinking of the elements of $(X^*)^*$ as evaluation maps $J_x: X^* rightarrow mathbbR$ i.e. $J_x(psi) = psi(x)$. Formally, however, for a general space we only know $J(X) (subset (X^*)^*)$ is a set of evaluation maps. $J(X)$ might be a proper subset.
My question: am I losing intuition about the space $(X^*)^*$ if I'm thinking about its members as evaluation maps? What do members of $(X^*)^* setminus J(X)$ (i.e. those that are not evaluation maps) look like?
functional-analysis linear-transformations normed-spaces dual-spaces weak-topology
asked Jul 31 at 21:02
ertl
427110
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up vote
2
down vote
favorite
I just learned about the weak-* topology on $X^*$, and in this context was introduced to the 'dual of the dual of a space', and the functional $J: X rightarrow (X^*)^*$ where $J(x)[psi] = psi(x)$, i.e. $J$ takes a point $x in X$ to the functional that takes a map from $X$ to the reals, and evaluates it at the point $x$. From this, $J(X) subset (X^*)^*$, and the weak-* topology on $X^*$ is the topology induced by this $J(X)$.
It originally took me a while to parse this definition in the right way, but it has become more intuitive after I started thinking of the elements of $(X^*)^*$ as evaluation maps $J_x: X^* rightarrow mathbbR$ i.e. $J_x(psi) = psi(x)$. Formally, however, for a general space we only know $J(X) (subset (X^*)^*)$ is a set of evaluation maps. $J(X)$ might be a proper subset.
My question: am I losing intuition about the space $(X^*)^*$ if I'm thinking about its members as evaluation maps? What do members of $(X^*)^* setminus J(X)$ (i.e. those that are not evaluation maps) look like?
functional-analysis linear-transformations normed-spaces dual-spaces weak-topology
asked Jul 31 at 21:02
ertl
427110
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I just learned about the weak-* topology on $X^*$, and in this context was introduced to the 'dual of the dual of a space', and the functional $J: X rightarrow (X^*)^*$ where $J(x)[psi] = psi(x)$, i.e. $J$ takes a point $x in X$ to the functional that takes a map from $X$ to the reals, and evaluates it at the point $x$. From this, $J(X) subset (X^*)^*$, and the weak-* topology on $X^*$ is the topology induced by this $J(X)$.
It originally took me a while to parse this definition in the right way, but it has become more intuitive after I started thinking of the elements of $(X^*)^*$ as evaluation maps $J_x: X^* rightarrow mathbbR$ i.e. $J_x(psi) = psi(x)$. Formally, however, for a general space we only know $J(X) (subset (X^*)^*)$ is a set of evaluation maps. $J(X)$ might be a proper subset.
My question: am I losing intuition about the space $(X^*)^*$ if I'm thinking about its members as evaluation maps? What do members of $(X^*)^* setminus J(X)$ (i.e. those that are not evaluation maps) look like?
functional-analysis linear-transformations normed-spaces dual-spaces weak-topology
asked Jul 31 at 21:02
ertl
427110
I just learned about the weak-* topology on $X^*$, and in this context was introduced to the 'dual of the dual of a space', and the functional $J: X rightarrow (X^*)^*$ where $J(x)[psi] = psi(x)$, i.e. $J$ takes a point $x in X$ to the functional that takes a map from $X$ to the reals, and evaluates it at the point $x$. From this, $J(X) subset (X^*)^*$, and the weak-* topology on $X^*$ is the topology induced by this $J(X)$.
It originally took me a while to parse this definition in the right way, but it has become more intuitive after I started thinking of the elements of $(X^*)^*$ as evaluation maps $J_x: X^* rightarrow mathbbR$ i.e. $J_x(psi) = psi(x)$. Formally, however, for a general space we only know $J(X) (subset (X^*)^*)$ is a set of evaluation maps. $J(X)$ might be a proper subset.
My question: am I losing intuition about the space $(X^*)^*$ if I'm thinking about its members as evaluation maps? What do members of $(X^*)^* setminus J(X)$ (i.e. those that are not evaluation maps) look like?
functional-analysis linear-transformations normed-spaces dual-spaces weak-topology
asked Jul 31 at 21:02
ertl
427110
asked Jul 31 at 21:02
ertl
427110
asked Jul 31 at 21:02
ertl
427110
asked Jul 31 at 21:02
ertl
427110
427110
add a comment |Â
add a comment |Â
1 Answer
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There is a concept for when all elements of $X^*$ are those coming from $X$, it is called reflexive space.
The typical example of a non-reflexive space is $c_0$, the space of sequences that tend to zero, with the supremum norm.
Then $c_0^*$ is isomorphic to $ell^1$, the space of absolutely convergent series with norm the sum of the sequence of absolute values.
The double dual $(c_0)^**=(ell^1)^*$ is isomorphic to $ell^infty$, the space of bounded sequences, with the supremum norm. You have that $J:c_0to ell^infty$ is the inclusion, but you can see that it misses many elements of the co-domain. For example, the sequence $(1,1,1,...)$.
answered Jul 31 at 21:10
JessicaMcRae
1264
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There is a concept for when all elements of $X^*$ are those coming from $X$, it is called reflexive space.
The typical example of a non-reflexive space is $c_0$, the space of sequences that tend to zero, with the supremum norm.
Then $c_0^*$ is isomorphic to $ell^1$, the space of absolutely convergent series with norm the sum of the sequence of absolute values.
The double dual $(c_0)^**=(ell^1)^*$ is isomorphic to $ell^infty$, the space of bounded sequences, with the supremum norm. You have that $J:c_0to ell^infty$ is the inclusion, but you can see that it misses many elements of the co-domain. For example, the sequence $(1,1,1,...)$.
answered Jul 31 at 21:10
JessicaMcRae
1264
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
add a comment |Â
up vote
4
down vote
There is a concept for when all elements of $X^*$ are those coming from $X$, it is called reflexive space.
The typical example of a non-reflexive space is $c_0$, the space of sequences that tend to zero, with the supremum norm.
Then $c_0^*$ is isomorphic to $ell^1$, the space of absolutely convergent series with norm the sum of the sequence of absolute values.
The double dual $(c_0)^**=(ell^1)^*$ is isomorphic to $ell^infty$, the space of bounded sequences, with the supremum norm. You have that $J:c_0to ell^infty$ is the inclusion, but you can see that it misses many elements of the co-domain. For example, the sequence $(1,1,1,...)$.
answered Jul 31 at 21:10
JessicaMcRae
1264
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There is a concept for when all elements of $X^*$ are those coming from $X$, it is called reflexive space.
The typical example of a non-reflexive space is $c_0$, the space of sequences that tend to zero, with the supremum norm.
Then $c_0^*$ is isomorphic to $ell^1$, the space of absolutely convergent series with norm the sum of the sequence of absolute values.
The double dual $(c_0)^**=(ell^1)^*$ is isomorphic to $ell^infty$, the space of bounded sequences, with the supremum norm. You have that $J:c_0to ell^infty$ is the inclusion, but you can see that it misses many elements of the co-domain. For example, the sequence $(1,1,1,...)$.
answered Jul 31 at 21:10
JessicaMcRae
1264
There is a concept for when all elements of $X^*$ are those coming from $X$, it is called reflexive space.
The typical example of a non-reflexive space is $c_0$, the space of sequences that tend to zero, with the supremum norm.
Then $c_0^*$ is isomorphic to $ell^1$, the space of absolutely convergent series with norm the sum of the sequence of absolute values.
The double dual $(c_0)^**=(ell^1)^*$ is isomorphic to $ell^infty$, the space of bounded sequences, with the supremum norm. You have that $J:c_0to ell^infty$ is the inclusion, but you can see that it misses many elements of the co-domain. For example, the sequence $(1,1,1,...)$.
answered Jul 31 at 21:10
JessicaMcRae
1264
edited Jul 31 at 21:22
answered Jul 31 at 21:10
JessicaMcRae
1264
answered Jul 31 at 21:10
JessicaMcRae
1264
answered Jul 31 at 21:10
JessicaMcRae
1264
1264
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
add a comment |Â
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
$ell^1$ is the space of absolutely summable series, isn't it?
– Berci
Jul 31 at 21:13
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
Could you elaborate on "$c_0^*$ is isomorphic to $ell^1$"?
– ertl
Aug 2 at 21:23
add a comment |Â
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