Proving C[a,b] is a Closed Subspace of L^{\infty}[a,b]

[jgens]

Homework Statement

Let [a,b] be a closed, bounded interval of real numbers and consider L^{\infty}[a,b]. Let X be the subspace of L^{\infty}[a,b] comprising those equivalence classes that contain a continuous function. Show that such an equivalence class contains exactly one continuous function; and thus, X is linearly isomorphic to C[a,b]. Show that C[a,b] is a closed subspace of the Banach space L^{\infty}[a,b].

Homework Equations

N/A

The Attempt at a Solution

I have already showed that each equivalence class contains exactly one continuous function. To prove that C[a,b] is a closed subspace, it is enough to notice that on C[a,b] we have ||\cdot||_{\infty} = ||\cdot||_{\mathrm{max}}, and that the uniform limit of a uniformly convergent sequence of continuous functions is continuous. So there does not seem to be much to this problem.

My text introduces this problem in the context of the Hahn-Banach Theorem along with other results about linear functionals. In particular, I know that C[a,b] is closed if and only if for each f \in L^{\infty}[a,b] \setminus C[a,b] there exists a continuous linear functional \psi which vanishes on C[a,b] but does not vanish at f. Any help with this?

 

[jbunniii]

The question seems a bit strange, as it takes pains to distinguish between X, which is a subspace, and C[a,b], which is only isomorphic to X.

The elements of C[a,b] are functions, not equivalence classes of functions, so C[a,b] is not a subspace (closed or otherwise) of L^infinity.

I wonder if the author didn't intend to ask you to show that X is a closed subspace of L^infinity. Not that this is necessarily any harder to prove, unless I'm missing a subtlety.