Timeline for Coefficients from Stone–Weierstrass versus Fourier transform
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2010 at 3:08 | vote | accept | Dorian | ||
| Aug 25, 2010 at 13:18 | |||||
| Aug 25, 2010 at 3:07 | vote | accept | Dorian | ||
| Aug 25, 2010 at 3:07 | |||||
| Aug 25, 2010 at 2:53 | comment | added | Zen Harper | In answer to Dorian's comment question: yes, dense span is exactly the same as saying existence of $f_n$. However, just knowing a set with dense span in general DOESN'T tell you anything about which coefficients you should actually use! But Fourier series have very special additional properties which can be exploited, as Mike Hall says. (Basically because the Hilbert space $L^2$ is densely embedded in $L^1$, so we can use Hilbert space arguments and orthogonality). | |
| Aug 25, 2010 at 2:44 | comment | added | Mike Hall | Note that it does not follow that $f$ can be written as an infinite trigonometric series in the $L^1$ sense. In the $L^2$ case you can use Hilbert space geometry to prove that the best $L^2$ approximation to $f$ by an $n$th degree trig polynomial is given by the $n$th partial Fourier series. Stone-Weierstrass implies that trig polynomials are dense in $L^2$, and hence we have $L^2$ convergence of Fourier series. $L^1$, on the other hand, isn't so nice geometrically (there's no inner product which gives the norm). | |
| Aug 25, 2010 at 2:42 | comment | added | Mike Hall | Oh yes, their span is certainly dense in $L^1$, like you said in your post. It is dense in $C([0,1])$ in the sup norm by Stone-Weierstrass, which implies that it's dense in $L^1$ norm as well, from which follows density in all of $L^1$. Since $L^1$ is a metric space, this means there is a sequence of trig polynomials converging to $f$ in $L^1$. | |
| Aug 24, 2010 at 23:31 | comment | added | Dorian | Hey Mike, What you're saying sounds reasonable but I"m a bit confused by something. Isn't saying that there is some $f_n = \sum_k c_{n,k} e^{2 \pi i k x}$ where $||f_n-f||_{L^1} \to 0$ the same as saying that the span of trigonometric polynomials is dense in $L^1$? | |
| Aug 24, 2010 at 22:02 | history | answered | Mike Hall | CC BY-SA 2.5 |