0
$\begingroup$

Theorem:

Let $V$ be an vector space. Then the dual space of $V$'s dual space is canonically isomorphic to $V$.

I am able to prove that $V$ is a subspace of $V^{**}$, the map $\mathbf{v}(\mathbf{v}^*) = \mathbf{v}^*(\mathbf{v}), \mathbf{v} \in V, \mathbf{v}^* \in V^*$ is injective, and $\mathrm{dim}(V) = \mathrm{dim}(V^*) = \mathrm{dim}(V^{**})$. I don't know how to continue from this point.

$\endgroup$
0
$\begingroup$

This only follows when $V$ is finite dimensional. The proof then follows from the rank-nullity theorem. So we have $T:V\to V^{**}$ by $T(v)=ev_v$, the evaluation map, i.e. $ev_v(f)=f(v)$ for $f\in V^*$. Then $T$ is injective: Suppose $T(v)=ev_0=ev_v=0$, or evaluation at $0$. If $v\not=0$, then we can find an $f\in V^*$ such that $f(v)\not=0$ (complete $\{v\}$ to a basis $\{v,v_1,\dots, v_{n-1}\}$, then define $v^*\in V^*$ by $v^*(v)=1$, $v^*(v_j)=0$ for all $j$). Then $ev_v(f)=f(v)\not=0=f(0)=ev_0$. Thus $T(v)=0\iff v=0$ so $T$ is injective. Then by Rank-Nullity

$$\dim V=\dim \text{Im}(T)+\dim\text{ker}(T)=\dim V^*=\dim V^{**}$$

By the fact that, for finite dimensional vector spaces, $\dim V=\dim V^{*}$.

There is a subtlety here: the isomorphism $V\to V^{*}$ and $V^{\ast}\to V^{\ast\ast}$ are not canonical (depend on basis choice), as does the proof I provided that for $v\in V, v\not=0$, there is $f\in V^*$ such that $f(v)\not=0$. However, combining everything doesn't require the choice of a basis, just that one exists.

An counterexample is provided here in the case that $V$ is not finite dimensional.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Tp Moya : in your 3rd line you have a term $v_v$ . i think this is a typo. To OP : A real Hilbert space is canonically its own dual and second dual. In general neither of these is true for infinite dimensional spaces. $\endgroup$ – DanielWainfleet Aug 27 '15 at 16:30
0
$\begingroup$

Let $l_0$ be the set of real sequences $(r_n)_{n \in N}$ such that $\lim_{n \to \infty}r_n=0$ with norm $||(a_n)_{n \in N} ||= \sup \{ |a_n \}_{n \in N}$.Then $l_0^*=l_1$ is the set of real sequences $q= (q_n)_{n \in N}$ satisfying $||q||= \sum_{n \in N} |a_n|< \infty $... [ For $ r=(r_n)_n \in l_0$ and $q=(q_n)_n \in l_1$,we have $q(r)=\sum q_np_n ]$... And $l_0^{**}=l_1^*=l_\infty $ is the set of all bounded real sequences $(s_n)_{n \in N}$ with $||(s_n)_n||= \sup_n |s_n|$. The canonical embedding of $l_0$ into $l_0^{**}$ is obviously not a surjection, so it's not an isomorphism.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.