无穷维向量空间的双对偶空间

hbghlyj

到雙对偶空間内的單射
存在一個由$V$到其雙对偶$V^{**}$的自然映射$\psi$，定義為
\[\left ( \psi(v) \right )(\varphi)=\varphi(v)\forall v\in V,\varphi \in V^*\]
$\psi$是單射；当且仅当$V$的維數有限時，$\psi$是個同構。

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Richard Kaye, Robert Wilson - Linear algebra-Oxford University Press (1998) page 93
Exercise 5.14 Let $V=\ell^2$ be the set of all real sequences $\left(a_n\right)$ such that $\sum_{n=1}^{\infty} a_n^2$ converges.
(a) Prove that $V$ is a vector space under componentwise addition and scalar multiplication.
(b) Define a suitable inner product on $V$, and prove that it is an inner product.
(c) Deduce that if $\left(a_n\right)$ and $\left(b_n\right)$ are in $V$ then
\[
\left(\sum_{n=1}^{\infty}\left(a_n+b_n\right)^2\right)^{\frac{1}{2}} \leqslant\left(\sum_{n=1}^{\infty}\left(a_n\right)^2\right)^{\frac{1}{2}}+\left(\sum_{n=1}^{\infty}\left(b_n\right)^2\right)^{\frac{1}{2}} .
\]
(d) Let $U$ be the subspace of all finite sequences (i.e. those which are 0 from some point on). Show that $U^{\perp}=\{0\}$, and deduce that $\left(U^{\perp}\right)^{\perp}=V$ and $\left(U^{\perp}\right)^{\perp} \neq U$.

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Math Counterexamples - A vector space not isomorphic to its double dual

A counter-example in the infinite-dimensional case

Let \(E=\mathbb{F}^{(\mathbb{N})}\) denote the vector space of sequences of elements of \(\mathbb{F}\) that are eventually \(0\) and \((e_i)_{n \in \mathbb{N}}\) its canonical basis. Also let \(V=\mathbb{F}^{\mathbb{N}}\) denote the vector space of sequences of elements of \(\mathbb{F}\). For \(v=(v_n) \in V\), let define \(\varphi_v \in E^*\) by:
\[\varphi_v(a)= \sum_{n=0}^\infty a_n v_n\] for \(a=(a_n) \in E\) (the sum exists as \(a=(a_n)\) is eventually \(0\)).
The map:
\[\Theta : \left|
\begin{array}{ll} V & \longrightarrow E^*\\
v & \longmapsto \varphi_v \end{array} \right. \] is an injective linear map. On the other hand, let’s take \(\varphi \in E^*\) and note \(u_n=\varphi(e_n)\) for \(n \in \mathbb{N}\). We have \(\varphi = \varphi_u\) with \(u=(u_n)\). Which proves that \(\Theta\) is an isomorphism.
Assuming axiom of choice, and taking the field of real numbers as \(\mathbb{F}\), \(V\) has a basis and has uncountable dimension, whereas \(E\) has countable dimension.
So \(\dim E < \dim E^* = \dim V \leq \dim E^{**}\) and \(E\) cannot be isomorphic to its double-dual. In fact, assuming axiom of choice an infinite-dimension vector space is never isomorph to its double dual.

abababa

只要V是自反空间，典范映射就一定是同构映射吧，不要求V是有限维。

hbghlyj

似乎混淆了“连续线性映射”和“线性映射”

hbghlyj

abababa 发表于 2022-11-10 11:13
不要求V是有限维。

math6140spring22
Theorem. Let $V$ be an infinite dimensional $F$-vector space. Then $\dim_F V < \dim_F V^∗$.

hbghlyj

bidual of an infinite dimensional vector space is bigger than the vector space itself
@Daniel McLaury
8 upvotes

Well, the dual of an infinite-dimensional vector space is already larger that the original space, so of course the dual of the dual will be even larger. (The English term for an espace vectoriel is a “vector space,” rather than a “vectorial space.”)
To see that the dual is larger, take some basis $\beta$ for your vector space $V$. Elements of $V$ consist of finite sums of the form $\lambda_1 e_1 + \cdots + \lambda_n e_n$, where the $e_i \in \beta$. On the other hand, $V^*$ contains everything of the form $\sum_{e \in \beta} c_e e^\ast$ — evaluating such a gadget at an element of $V$ will still produce a finite sum, since only finitely many $e_i$ appear in any given element of $V$.
When the basis is finite, of course, there’s no difference between a finite sum and an arbitrary one. When it’s infinite, though, the sets have different cardinalities.

@Alexandre Borovik
2 upvotes

It is a very deep question indeed. I have to admit that I cannot give any intuitive explanation which is simpler than a reduction to some basic set theory.

To make the question as close tothe set theory as possible, let us restrict our attention to the case of a vector space $V$ over the field $ \mathbb{F}_2$ of two elements. If my memory does not betray me, it is an old result by Paul Eklof that existence of a basis in an arbitrary vector space over $\mathbb{F}_2$ is equivalent to the Axiom of Choice (it is easy in one direction: the Axiom of Choice, in the form of the Zorn Lemma, impliesthe existence of a basis. So, let us accept it, and let $B$ be a basis in $V, $ which means that every $v\in V$ is defined by its support in $B$, t hat is, by the (finite) set of elements in $B$ which sum up to $v$. Therefore $V$ has the same cardinality as the set of finite subsets of $B$; if $B$ is infinite, than it is easy to prove that the set of all finite subsets of $B$ has the same cardinality as $B, $hence $V$ has the same cardinality as $B$.

Now let us look at the dual space $V ^ *$, that is, the set of all linear functionals $\alpha: V \to \mathbb{F}_2$. Each $\alpha$ is uniquely determined by its support in $B$, that is, by the set $\{\, b \in B \mid \alpha(b) = 1\,\}$. Therefore $V^*$ is in one-to-one correspondence with the set 2^B of all subsets in $B$. But it is a classical result by Cantor that $2^B$ has larger cardinality than $B$ and hence $V^*$ has larger cardinality than $V$. Of course, the cardinality of the bidual $V^{**}$ is even larger.

The case of an arbitrary field $F$ can be handled in a similar way, but we will need to deal with the cardinality of the set $F^B$.

Is this intuitive? Well, it is intuiteve for me because it was the first thought that crossed my mind. But I am not a set theorists and I have no idea whether the same can be proven without the Axiom of Choice. Also, the term “infinite dimensional vector space is ambigous: does it mean “ a vector space without a finite basis” or “a vector space with an infinite basis”? And can anything that intimately involves the Axiom of choice be called intuitive?