Infinite-Dimensional Space Isomorphism: V vs V*

[Bacle]

Hi, Analysts:

I was just looking for a nice proof that when V is an infinite-dimensional normed space,
then V and V* are not isomorphic ( I think there is an exception if V is a Hilbert Space,
by using Riesz Representation ).

Also: while V is not always isomorphic to V* in the inf.-dim. (non-Hilbert) case:
Is V isomorphic to its (strictly smaller than the total dual) _continuous_ dual ?

Thanks for any references, help.

 

[micromass]

Hi, Analysts:

I was just looking for a nice proof that when V is an infinite-dimensional normed space,
then V and V* are not isomorphic ( I think there is an exception if V is a Hilbert Space,
by using Riesz Representation ).

No, even for Hilbert spaces, there are more vectors in V* than V. Here's a nice proof of this fact:

Let V be an F-vector space.V can be identified with the following subset of F^A:

[tex]V\cong \{f\in F^A~\vert~f(\alpha)=0~\text{except for finitely many}~\alpha\}[/tex]

Indeed, with every function f, we associate the vector

[tex]\sum_{\alpha\in A} f(\alpha)\alpha}[/tex].

However, the algebraic dual V* can be described as [tex]V^\ast\cong F^A[/tex] (since every element in F^A extends to an element in V*).

So we see that V and V* coincide for finite sets A. However, if A is not finite, then V* has much more elements than V.

Also: while V is not always isomorphic to V* in the inf.-dim. (non-Hilbert) case:
Is V isomorphic to its (strictly smaller than the total dual) _continuous_ dual ?

This is an interesting question. The answer is that V is somethimes very similar to it's continuous dual. For example, finite-dimensional normed spaces are isomorphic to it's continuous dual. And the Riesz-representation theorem yields that the Hilbert space is anti-isomorphic to it's continuous dual.

More interesting is the question whether V is isomorphic to it's double dual. The spaces which satisfy this are called reflexive spaces and are quite interesting in functional analysis...

 

[Bacle]

Sorry, it seems the quote function has been disabled. Thanks for your answer.

I see, so if V is infinite-dimensional, it is isomorphic to the direct sum of (cardinality-many)
copies of F, i.e., all cardinality-ples with entries of F, with compact support?

And, re F^A , as the set of all maps into F, or as the direct product, or
equivalently, the collection of all cardinality-many-ples with entries in F, but support
is not necessarily finite.

So, is your argument that we can get a functional by doing entry-by-entry multiplication
of v* in V* with some v in V, i.e, if v* in F^A , and v in V:

v*(v) := v*1.v1 +v*2.v2+...+v*k.vk

which is well-defined since it has finite support,

where v*i is the i-th entry of an element in F<sup>A>/sup> ?

 

[micromass]

Yes, that's exactly how I identify V* with F^A! Thus you see that V* is much larger than V.

 

[Landau]

For people who are wondering what micromass means by A: it is a basis of V (or equivalently just any set whose cardinality equals the dimension of V).