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.