# Vector spaces in duality

It is not necessary to assume the concept of a vector space for simply and directly defining the notion of a pair of vector spaces in duality, in the following way:

A pair of dual vector spaces, is a particular case of duality system.
It is a pair of sets (E,E'), where the elements of E are called vectors, those of E' are called covectors (or linear forms) together with an operation of scalar product from E×E' to ℝ : for each u∈E and x∈E', we have u⋅x∈ℝ, and that is subject to the following axioms:

### Separation axioms

They are formally written as follows:

∀u,v∈E, (∀x∈E', u⋅x = v⋅x) ⇒ u=v
∀x,y∈E', (∀u∈E, u⋅x = u⋅y) ⇒ x=y

In more intuitive words : Each element of E is completely specified by its scalar products by the elements of E' ; and the same exchanging E and E'.

In more sophisticated words : both curried forms of the scalar product : a map from E to ℝE', and a map from E ' to ℝE, are injective; thus E plays the role of (is identifiable with) a subset of ℝE ', and E' plays the role of a subset of ℝE 'as identified to its image in ℝE.

### More axioms

We have a zero element in E, whose scalar product with any element of E' is zero; and a zero element of E', whose scalar product by any element of E is zero. We shall abusively denote the zeros of all sets by the same symbol 0, letting their precise identity be given by the operations where they are used, as this won't cause any ambiguity:

• ∀x∈E', 0⋅x = 0
• ∀u∈E', u⋅0 = 0

We define an operation of addition in E  as the addition of the scalar products with any element of E' ; and the same in E'. Formally:

• ∀u,v∈E, ∀x∈E', (u+v)⋅x = u⋅x + v⋅x
• ∀x,y∈E', ∀u∈E, u⋅(x+y) = u⋅x + u⋅y

In other words : for all u,v∈E, the function from E' to R defined by (x ↦ u⋅x + v⋅x) "belongs to" (or : is represented through the scalar product by) an element of E that is denoted u+v.

We define an operation of multiplication of elements of E by any real number, as multiplying by the same number, the scalar product with any element of E'. And the same in E'. Formally :

• ∀a∈ℝ, ∀u∈E, ∀x∈E', (au)⋅x = a(u⋅x) = u⋅(ax)

Remark : the zero element of E can as well be obtained from any element u of E by the multiplication by 0∈ℝ: 0 = 0u. But declaring the constant 0∈E further says that E is nonempty.
The above definitions of the operations in E can also be equivalently expressed in the following two ways

• E represents (though the scalar product) a vector subspace of the space of functions from E' to ℝ. (stable by addition and multiplication by a scalar).
• E is a vector space, and the elements of E' represent linear functions from E to ℝ.

(the below will be reworked later)

A subspace is a subalgebra (i.e. stable) for addition and multiplication by a number.
Note that (as for any duality system) from any pair (E,E') with an operation from E×E' to ℝ we can produce a pair of spaces in duality by replacing E by the vector space generated by its image in ℝE', and the same for E'. This can be done in any order without affecting the result and indeed gives a pair of dual vector spaces by doing this replacement just once on each side, because this replacement for E does not affect the linear relations in E' (relations in E' that are true in ℝE remain true when replacing E by the vector space generated by its image in ℝE'), thanks to the algebraic relations between addition and multiplication (commutativity, associativity, distributivity).

If (E,E') is a pair of dual spaces and E is finite-dimensional then E' is the space of all linear forms on E and has the same dimension.
There are counter-examples in the infinite-dimensional case. For example take E = E' = the set of continuous maps from [-1,1] to ℝ, and the operation of integral of the product of these functions.
Then the Dirac mass in 0 (or in any other point of [0,1], which maps any f in E to f(0), is outside E'. It may still be understood as a limit of a sequence of elements of E'.