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

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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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