# Dense subset

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• Jan 10th 2011, 04:04 PM
slevvio
Well my argument was was that the $\theta$ i've got is a continuous linear functional, and $\sin n$ is in the kernel for all $n$, except when $n = 1$. However $\theta(\cos 2x) \not= 0$ hence is not in the kernel. The kernel is closed by continuity, hence any sequence in it, in particular, a sequence of the sin functions, cannot converge to an element outside the closed set. The only problematic term is $\sin$ which is not in the kernel, but that sequence is just $\{\sin, \sin, \sin, ... \}$ which converges to $\sin$.

I was just wondering where I have gone wrong here? Did you mean I should use Arzelą to find a fallacy in the above argument?
• Jan 10th 2011, 11:55 PM
Opalg
Quote:

Originally Posted by Rebesques

The last part of the argument is that if $\|s_n(g)-g\|_1 = \int_{-\pi}^\pi |s_n(g)(x)-g(x)|\,dx \to0$ then $\|s_n(f)-f\|_1 = \int_0^\pi |s_n(f)(x)-f(x)|\,dx \to0$, because $f(x)=g(x)$ when $x\geqslant0$. The integral of a positive function over part of an interval must be less than the integral over the whole interval.
• Jan 11th 2011, 06:49 AM
slevvio
Let g be an odd function

Then $g(0) = -g(0) \implies g(0) = 0$.

Let $f:[0, \pi] \rightarrow \mathbb{C},$given by $f(x) = 1$.

Then $f(0) \not= 0$, hence cannot be extended to an odd function. So your method only works when f(0) = 0 and so it reasonable to believe the span is not dense:)

But I don't know much of a difference this makes on the argument, because let's say we proceed with this constant function 1. then there is a discontinuity, does this stop us being able to talk about inner products, integrals etc?
• Jan 11th 2011, 07:38 AM
Opalg
Quote:

Originally Posted by slevvio
Let g be an odd function

Then $g(0) = -g(0) \implies g(0) = 0$.

Let $f:[0, \pi] \rightarrow \mathbb{C},$ given by $f(x) = 1$.

Then $f(0) \not= 0$, hence cannot be extended to an odd function. So your method only works when f(0) = 0 and so it reasonable to believe the span is not dense :) I claim that my method does work. (Tongueout)

But I don't know much of a difference this makes on the argument, because let's say we proceed with this constant function 1. then there is a discontinuity, does this stop us being able to talk about inner products, integrals etc? That's exactly correct. It doesn't matter if g has a discontinuity at 0. It still has a Fourier series, which converges to g in the L^2 and L^1 norms.

Admittedly, I was being a bit sloppy in calling g an odd function. It becomes odd if we redefine g(0) to be 0. Changing the value a function at one point doesn't affect its Fourier series, and it doesn't change the $L^1$-norm.
• Jan 12th 2011, 03:41 AM
slevvio
Opalg, I wasn't too sure about what you were saying since i only learned about the space of continuous functions, but I read some more and I see what you say is correct! So the span is dense. I spoke to my lecturer and we concluded that his original idea that the span wasn't dense is wrong.

Here is why I was wrong. I claimed that the span could not contain any sequences that tend to cos2x since all the $\sin n$ 's (except $\sin 1$) were in a closed subset and cos2x isnt. However throwing sin in gives a sequence which does converge i.e. the partial sums of the fourier series

Thanks everyone for this very long and informative thread
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