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**slevvio** Let g be an odd function

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

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

Then $\displaystyle f(0) \not= 0$, 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.