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

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Homework Help Overview

The problem involves proving that a specific subspace G of the space C[-1,1] with the L^2 norm is closed. The subspace G consists of functions f in H such that f(1) = 0.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to establish a connection between the L^2 norm and uniform convergence to show that if a sequence of functions converges in the L^2 norm, it must also converge uniformly. Others suggest proving by contradiction, questioning the implications of assuming f(1) ≠ 0.

Discussion Status

Participants are exploring various approaches to the problem, including the implications of continuity and the behavior of sequences of functions under the L^2 norm. Some guidance has been offered regarding the nature of convergence and the properties of the integral norm.

Contextual Notes

There is a discussion about the limitations of uniform convergence in this context and the potential for discontinuities in the limit of the sequence of functions. The original poster expresses confusion regarding the norms involved.

benf.stokes
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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: <f,g>\to \int_{-1}^{1}f(t)\overline{g(t)} dt

The Attempt at a Solution



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

Try to prove it by contradiction. Assume that f(1)\neq 0.
 
Hmm, if f(1)\neq 0 then f(t)-f_n(t) \neq 0 in a open interval around one.
But shouldn't it be possible to have a f_n 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 f_n
 
benf.stokes said:
But shouldn't it be possible to have a f_n 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 f_n

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

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

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

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