Question

Let X be a Banach space. Suppose that dim X = n and {e₁, e₂, ..., eₙ} is a basis of X. Prove that if λ₁, λ₂, ..., λₙ are linear functionals on X given by λᵢ(eⱼ) = δᵢⱼ, then λ₁, λ₂, ..., λₙ is a basis of X* (the dual space of X) and hence dim X* = n.

Answer 1

Any linear functional in the dual space X* of a Banach space X with finite dimension n can be uniquely expressed as a linear combination of functionals λᵢ defined by λᵢ(eᵣ) = δᵢᵣ, proving that they form a basis of X* and hence dim X* = n.

Step-by-step explanation:

In mathematics, particularly in functional analysis, a set of linear functionals can serve as a basis for the dual space X* if every functional in X* can be uniquely expressed as a linear combination of these functionals. The set {λ₁, λ₂, ..., λₙ} defined by λᵢ(eᵢ) = δᵢᵣ (where δ is the Kronecker delta) satisfies this criterion in the Banach space X with finite dimension n.

For the dual basis proof, take any linear functional f in X*. Since {e₁, e₂, ..., eₙ} is a basis for X, any vector x in X can be written as x = Σ cᵢeᵢ. Therefore, we can define f(x) = Σ cᵢf(eᵢ), and since the functionals λᵢ are defined as λᵢ(eᵣ) = δᵢᵣ, the linearity of f implies f = Σ f(eᵢ)λᵢ, showing that the λᵢ’s form a basis for X*. Consequently, dim X* = n, proving the dimension of the dual space is equal to the dimension of the original Banach space X.