Question

Let X be a Banach space. Suppose X has infinite dimension, and {e₁, e₂, ..., eₙ} is a linearly independent set of vectors whose linear span is dense in X.

Suppose λᵢ ∈ X* for all i ≥ 1 with λᵢ(eⱼ) = δᵢⱼ. Prove or disprove that the closure of span(λᵢ) (the linear span of λᵢ for all i) equals X*.

Note: X* denotes the space of linear functionals.

Answer 1

To prove or disprove that the closure of the linear span of a set of linear functionals λᵢ equals X*, we need to show that X* is contained in the closure of the linear span and vice versa.

Step-by-step explanation:

To prove or disprove that the closure of the linear span of a set of linear functionals λᵢ equals X*, we need to show that X* is contained in the closure of the linear span and vice versa.

Let f be an element in X*. Since the linear span of {e₁, e₂, ..., eₙ} is dense in X, we can find a sequence {xₙ} in the linear span that converges to x in X. Then, we can define a sequence of linear functionals {λˢ} as λˢ(xₙ) = f(xₙ) for every n. It can be shown that λˢ is a Cauchy sequence in X*, and since X* is Banach, λˢ converges to some linear functional g in X*. Thus, f is in the closure of the linear span of λˢ.

Conversely, if g is in the closure of the linear span of λˢ, we can choose xₙ such that |λˢ(xₙ) - g(xₙ)| < 1/n for every n. Then, we can define a linear functional f in X* as f(x) = lim(λˢ(xₙ)) as n approaches infinity. This f(x) is well-defined and satisfies f(x) = g(x) for every x in X. Therefore, the closure of the linear span of λᵢ equals X*.