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

In summary, the conversation discusses a problem involving closed, bounded intervals of real numbers and a subspace of L^{\infty}[a,b] containing continuous functions. It is shown that each equivalence class contains exactly one continuous function, and that C[a,b] is a closed subspace of L^{\infty}[a,b]. The context of the Hahn-Banach Theorem is also mentioned, and the question of whether X is a closed subspace of L^{\infty} is raised.
  • #1
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].

Homework Equations



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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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  • #2


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.
 

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

1. What is the definition of a closed subspace?

A closed subspace is a subset of a vector space that is closed under vector addition and scalar multiplication. This means that the sum of any two vectors in the subspace is also in the subspace, and any scalar multiple of a vector in the subspace is also in the subspace.

2. How do you prove that C[a,b] is a closed subspace of L^{\infty}[a,b]?

To prove that C[a,b] is a closed subspace of L^{\infty}[a,b], we need to show that it satisfies the definition of a closed subspace. This means we need to show that the sum of any two continuous functions is also continuous, and any scalar multiple of a continuous function is also continuous. We also need to show that the functions in C[a,b] are bounded, which is a requirement for L^{\infty}[a,b].

3. Can you provide an example of a continuous function in C[a,b] that is not in L^{\infty}[a,b]?

Yes, an example of a continuous function in C[a,b] that is not in L^{\infty}[a,b] is the function f(x) = 1/x on the interval [1,2]. This function is continuous on [1,2], but it is not bounded, so it does not belong in L^{\infty}[1,2].

4. Why is it important to prove that C[a,b] is a closed subspace of L^{\infty}[a,b]?

Proving that C[a,b] is a closed subspace of L^{\infty}[a,b] is important because it shows that C[a,b] is a subset of L^{\infty}[a,b] that satisfies all the properties of a vector space. This allows us to use the tools and techniques of linear algebra to study and solve problems in the space of continuous functions on the interval [a,b].

5. Are there any other ways to prove that C[a,b] is a closed subspace of L^{\infty}[a,b]?

There are a few different ways to prove that C[a,b] is a closed subspace of L^{\infty}[a,b]. One way is to use the definition of a closed subspace and show that C[a,b] satisfies all the requirements. Another way is to use the fact that continuous functions are the limit of a sequence of piecewise linear functions, and prove that these piecewise linear functions are in C[a,b]. Finally, we can also use the fact that C[a,b] is a subset of L^{\infty}[a,b] and show that it also satisfies the definition of a closed subspace.

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