Prove [itex]C[a,b][/itex] a closed linear subspace of [itex]L^{\infty}[a,b][/itex]

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jgens
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Homework Statement



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

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The Attempt at a Solution



I have already showed that each equivalence class contains exactly one continuous function. To prove that [itex]C[a,b][/itex] is a closed subspace, it is enough to notice that on [itex]C[a,b][/itex] we have [itex]||\cdot||_{\infty} = ||\cdot||_{\mathrm{max}}[/itex], 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 [itex]C[a,b][/itex] is closed if and only if for each [itex]f \in L^{\infty}[a,b] \setminus C[a,b][/itex] there exists a continuous linear functional [itex]\psi[/itex] which vanishes on [itex]C[a,b][/itex] but does not vanish at [itex]f[/itex]. Any help with this?
 

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jbunniii
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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.
 

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