Well my argument was was that the i've got is a continuous linear functional, and is in the kernel for all , except when . However 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 which is not in the kernel, but that sequence is just which converges to .
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?
Let g be an odd function
Then .
Let given by .
Then , 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?
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 's (except ) 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