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

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SUMMARY

The discussion centers on proving that the set G = {f ∈ C[-1,1] | f(1) = 0} is a closed subspace of H = C[-1,1] with the L² norm. Participants explore the challenge of demonstrating uniform convergence for a sequence (f_n) under the L² norm, ultimately concluding that proving by contradiction is effective. The key insight is that if f(1) ≠ 0, then the continuity of functions leads to a contradiction, affirming that G is indeed closed.

PREREQUISITES
  • Understanding of L² inner product: <f,g> = ∫-11 f(t) &overline;g(t) dt
  • Familiarity with the concept of closed subspaces in functional analysis
  • Knowledge of uniform convergence and its implications in analysis
  • Basic principles of continuity in the context of real-valued functions
NEXT STEPS
  • Study the properties of closed subspaces in L² spaces
  • Learn about uniform convergence and its criteria in functional analysis
  • Explore the implications of continuity on function limits
  • Investigate examples of sequences in C[-1,1] and their convergence behavior
USEFUL FOR

Mathematicians, students of functional analysis, and anyone studying properties of function spaces and convergence in analysis will benefit from this discussion.

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: &lt;f,g&gt;\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||&lt; \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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