Let $V$ be a complex vector space of finite dimension and let

$\overline{V}=\{h: V\to \mathbb{C}: h\text{ antilinear }\}$

Prove that $V$ can be identified with the set of antilinear complex maps $\phi: \overline{V} \to \mathbb{C}$ in other words $V \cong \overline{\overline{V}}$.


I know how to do this if we consider the dual space (for finite dimension). In this case we define the evaluation and define a morphism that it's easily seen to be injective. In this case the evaluation is not an element of $\overline{V}$ (let's call this space the anti dual space of $V$).

I know that there exist a natural isomorphism because the book that I'm reading says a lot about some identifications under this natural isomorphism.


It's easy to prove that if $\dim V < \infty$ then $\dim_{\mathbb{C}} V = \dim_{\mathbb{C}} \overline{V}$. Is the same proof for the dualspace $V^*$. Let $v_1 ... v_n$ be a basis for $V$ for each $i \in \{1...n \}$ define $\delta_i$ as the unique function $\delta_i: V\to \mathbb{C}$ that is antilineal and such that $\delta_i(v_j)=\delta_{ij}$ (Kronecker Delta). It's easy to prove that this functions exist and are uniquely determined by it's values on the basis. It's also easy to prove that $\delta_i$ is a basis for $\overline{V}$.

Define the function $\overline{ev}:V \to \overline{\overline{V}}$ (the composition of the conjugation with the evaluation function). Prove that this function is injective and by dimensions it's an isomorphism.


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