User:Eric Kvaalen/Notes/Fourier transform

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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 -ωa and ωa. 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.

If a function is frequency limited in that its Fourier transform has support only between -ωa and ωa, then its values at t = nπ / ωa (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). Therefore, it is possible to specify a function f(t) arbitrarily at the infinite set of points t = nπ / ωa. 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).