structural:scholar_edition, information:classification, stylistic:tone, structural:summarization
Decay of Fourier transform

The above result tells us about the decay of the Fourier transform, since it follows that if f {\displaystyle f} and f (k) {\displaystyle f^{(k)}} are integrable then

| F f (ξ) | ≤ I (f) 1 + | 2 π ξ | k, where I (f) = ∫ − ∞ ∞ (| f (y) | + | f (k) (y) |) d y. {\displaystyle \vert {\mathcal {F}}f(\xi)\vert \leq {\frac {I(f)}{1+\vert 2\pi \xi \vert ^{k}}},{\text{ where }}I(f)=\int _{-\infty }^{\infty }{\Bigl (}\vert f(y)\vert +\vert f^{(k)}(y)\vert {\Bigr)}\,dy.}

In other words, if f {\displaystyle f} satisfies these conditions then its Fourier transform decays at infinity at least as quickly as 1/| ξ |. In particular, if k ≥ 2 {\displaystyle k\geq 2} then the Fourier transform is integrable.

The proof uses the fact, which is immediate from the definition of the Fourier transform, that

| F f (ξ) | ≤ ∫ − ∞ ∞ | f (y) | d y. {\displaystyle \vert {\mathcal {F}}f(\xi)\vert \leq \int _{-\infty }^{\infty }\vert f(y)\vert \,dy.}

Using the same idea on the equality stated at the start of this subsection gives

| (2 π i ξ) k F f (ξ) | ≤ ∫ − ∞ ∞ | f (k) (y) | d y. {\displaystyle \vert (2\pi i\xi)^{k}{\mathcal {F}}f(\xi)\vert \leq \int _{-\infty }^{\infty }\vert f^{(k)}(y)\vert \,dy.}

Summing these two inequalities and then dividing by 1 + |2 π ξ | gives the stated inequality.