Proof that a given subspace of C[−1,1] with L2 norm is closed

In summary, to prove that G is a closed subspace of H, we can use a contradiction argument. Assume that f(1) is not equal to 0, and that f_n converges to f. Then, using continuity and the L2-norm, we can find a point x_n that is close to 1 and far from 0. This contradicts the fact that f_n(1) = 0, showing that f(1) must equal 0 and therefore, f belongs to G.
  • #1
benf.stokes
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0

Homework Statement



Let H= C[-1,1] with L^2 norm and consider G={f belongs to H| f(1) = 0}. Show that G is a closed subspace of H.


Homework Equations



L^2 inner product: [tex]<f,g>\to \int_{-1}^{1}f(t)\overline{g(t)} dt[/tex]

The Attempt at a Solution



I've been trying to prove this for a while but i can't establish that given [tex]||f_{n}-f||< \epsilon[/tex] (where the norm is the L^2 one) we have uniform convergence for the sequence [tex](f_{n})[/tex]If I could prove this the result would follow easily given that G is contained in its closure and if [tex](f_{n})[/tex] converged uniformly we would have [tex]f(1)=\lim_{x \to 1} \lim_{n \to \infty} f_n(x)=\lim_{n \to \infty} \lim_{x \to 1} f_n=0[/tex] and thus that f belongs G
 
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  • #2
You won't be able to show uniform convergence.

Try to prove it by contradiction. Assume that [itex]f(1)\neq 0[/itex].
 
  • #3
Hmm, if [tex]f(1)\neq 0[/tex] then [tex]f(t)-f_n(t) \neq 0[/tex] in a open interval around one.
But shouldn't it be possible to have a [tex]f_n[/tex] as close as possible to f(t) until 1-1/n say and then decrease linearly to zero? I know this function in particular is discontinuous in the n to infinity limit but can you give me a hint of how i prove it must me discontinuous for all possible [tex]f_n[/tex]
 
  • #4
benf.stokes said:
But shouldn't it be possible to have a [tex]f_n[/tex] as close as possible to f(t) until 1-1/n say and then decrease linearly to zero? I know this function in particular is discontinuous in the n to infinity limit but can you give me a hint of how i prove it must me discontinuous for all possible [tex]f_n[/tex]

This function [itex]f_n[/itex] that you mention converges to [itex]f[/itex]. Remember that you're dealing with the integral norm. The integral norm behaves very strangely.

But let's proceed naively. We know [itex]f_n\rightarrow f[/itex]. We assumed [itex]f(1)\neq 0[/itex]. And we know that [itex]f_n(1)=0[/itex].

Can you use continuity to find a point [itex]x_n[/itex] such that [itex]f(x_n)[/itex] is "far away" from 0. But [itex]f[/itex] and [itex]f_n[/itex] are closed in the L2-norm. Can you use this to say anything about the [itex]x_n[/itex]?? It seems reasonably that [itex]x_n[/itex] is rather close to 1. Can you give a formula for how close?
 
  • #5
Hmm, i see it now. I was confusing norms. Thanks for the help!
 

1. What is a subspace in the context of C[-1,1] with L2 norm?

A subspace in the context of C[-1,1] with L2 norm is a subset of the vector space C[-1,1] that satisfies the following conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication.

2. How is the L2 norm defined in this context?

The L2 norm in this context refers to the standard Euclidean norm, which is defined as the square root of the sum of squared values of a vector's components. In the case of C[-1,1], the vector's components are the function values at different points within the interval [-1,1].

3. What does it mean for a subspace to be closed?

A subspace is considered closed if it contains all of its limit points. In other words, if a sequence of vectors within the subspace converges to a vector outside of the subspace, then that vector must also be included in the subspace for it to be considered closed.

4. How can one prove that a given subspace is closed in C[-1,1] with L2 norm?

To prove that a given subspace is closed in C[-1,1] with L2 norm, one can use the definition of a closed subspace and show that it satisfies the necessary conditions. This can involve demonstrating that the subspace contains the zero vector, is closed under vector addition, and is closed under scalar multiplication.

5. Why is it important to prove that a subspace is closed in C[-1,1] with L2 norm?

Proving that a subspace is closed in C[-1,1] with L2 norm is important because it ensures that the subspace is a complete vector space. This is necessary for many mathematical and scientific applications, such as signal processing, where the completeness of a vector space is essential for accurate and efficient calculations.

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