User:Eric Kvaalen/Notes/Fourier transform

If a function f(t) is totally flat in a certain range (say around 0), but not over the whole real line, then its Fourier transform must contain arbitrarily high frequencies. To see this, assume the contrary--that the Fourier transform is supported only between -&omega;a and &omega;a. The inverse transform evaluated for t=0 tells us that


 * $$ f(0) = 0 = \frac{1}{\sqrt{2\pi}} \int_{-\omega_a}^{\omega_a} F(\omega) e^{ i\omega t}\,d\omega $$

Furthermore, the derivative at zero is zero, giving:


 * $$ f'(0) = 0 = \frac{1}{\sqrt{2\pi}} \int_{-\omega_a}^{\omega_a} i \omega F(\omega) e^{ i\omega t}\,d\omega $$

Continuing to take derivatives, we find that


 * $$ 0 = \frac{1}{\sqrt{2\pi}} \int_{-\omega_a}^{\omega_a} \omega^n F(\omega) e^{ i\omega t}\,d\omega $$

for any n. This shows that the Fourier transform between -ωa and ωa is orthogonal to a basis set of polynomials in ω, and therefore F(ω) must be zero almost everywhere, both in the range -ωa to ωa and, by assumption, outside this range. But that would imply that f(t) is zero everywhere, which is contrary to the assumption that it is not flat everywhere. Demonstrandum demonstratum est.


 * The above is a consequence of the Paley-Wiener theorem. If the support of the transform had bounded support, then the function would be in Paley-Wiener space, which means it would be an entire function. An entire function that is flat on a portion of the real line is a constant. Furthermore it would have to be zero because it must be square-integrable on the real line. Eric Kvaalen (talk) 14:05, 14 August 2015 (UTC)

If a function is frequency limited in that its Fourier transform has support only between -ωa and ωa, then its values at $$t=n \pi/\omega_a$$ (for integer n) correspond to the coefficients of a Fourier series for F(ω) between -ωa and ωa (this is due to the fact that the transform of the transform is the original function [correction: it's the reverse of the original function]). Therefore, it is possible to specify a function f(t) arbitrarily at the infinite set of points $$t = n \pi/\omega_a$$. But if one does this, one cannot specify it at other points, because the Fourier series for F(ω) between -ωa and ωa defines F(ω).
 * This follows from Theorem 22 in Hilbert Spaces of Entire Functions by Louis de Branges, which says that an orthogonal basis for a space like a Paley-Wiener space is made up of reproducing kernel functions that give the values at the points $$n\pi/\omega_a$$. The coefficients in an expansion in this basis are given by the scalar products of the function with each of these reproducing kernel functions, which gives simply the values of the function at those points. Eric Kvaalen (talk) 14:05, 14 August 2015 (UTC)