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

1. Feb 28, 2013

benf.stokes

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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

Last edited: Feb 28, 2013
2. Feb 28, 2013

micromass

Staff Emeritus
You won't be able to show uniform convergence.

Try to prove it by contradiction. Assume that $f(1)\neq 0$.

3. Feb 28, 2013

benf.stokes

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 $$f_n$$

4. Feb 28, 2013

micromass

Staff Emeritus
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?

5. Feb 28, 2013

benf.stokes

Hmm, i see it now. I was confusing norms. Thanks for the help!