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5 corrected spelling

4 typo

# CoefficientsfromStoneWeirstrassversus Fourier transformonL^1:Aren'tthee^ikxfunctionsstilldense?Transform

Hey I'm a bit confused about

Usually one shows the fourier density of the functions $\sin(kx)$ in $L^2([0,1])$ using the Fourier transform. This in fact comes from the Stone-Weirstrass theorem however and then uses the density of continuous functions in $L^1([0,1])$. If I apply L^2([0,1])$. However, the Stone Weirstrass's Weirstrass theorem can be used to show, it tells me for example, that the functions$e^{ikx}$are indeed dense in the continuous functions C([0,1])$C([0,1])$and hence dense in$L^1([0,1])$. Is the difference that the coefficients are not the Fourier ones? ieL^1([0,1])$ as well. there must be other So we obtain (not-necessarily-unique) coefficients $c_k$ so such that one has $f(x) f_k(x) =c_ke^{ikx}$ converge to any given $f \sum c_k e^{ikx}$ converging in L^1([01])$. How should I think about these coefficients? How do they relate to the Fourier series of$L^1([0,1])$? Thanks!f$ (with basis $e^{ikx})$?

3 I improved the title and changed a typo in my teX notation for $e^{ikx}$.

# Fourier Transformtransformon L^1:Aren'tthee^ikxfunctionsstilldense?

Hey I'm a bit confused about the fourier transform in $L^1([0,1])$. If I apply Stone Weirstrass's theorem, it tells me that $e^ikx$ e^{ikx}$are indeed dense in the continuous functions C([0,1]) and hence dense in$L^1([0,1])$. Is the difference that the coefficients are not the Fourier ones? ie. there must be other coefficients$c_k$so that one has$f(x) = \sum c_k e^ikx$.e^{ikx}$ converging in $L^1([0,1])$?

Thanks!

2 added 23 characters in body
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