Show that for each \(\epsilon>0\) the following function is the Fourier transform of an \(L^1(\RR^n)\) function: \[ F(\xi) \da \qty{1 \over 1 + \abs{\xi}^2}^{\epsilon} .\]
Hint: show that
\[ K_\delta(x) &\da \delta^{-n/2} e^{-\pi \abs{x}^2 \over \delta} \\ f(x) &\da \int_0^{\infty } K_{\delta}(x) e^{-\pi \delta} \delta^{\epsilon - 1} \,d \delta \\ \Gamma(s) &\da \int_0^{\infty } e^{-t} t^{s-1} \dt \\ \implies \fourier{f}(\xi) &= \int_0^{\infty } e^{- \pi \delta \abs{\xi}^2} e^{ -\pi \delta} \delta^{\epsilon - 1} = \pi^{-s} \Gamma(\epsilon) F(\xi) .\]
Solution
Setup note. Normalize the Fourier transform as \(\fourier{f}(\xi) = \int f(x) e^{-2\pi i x\cdot\xi}\,dx\). The Gaussian \(K_\delta(x) = \delta^{-n/2}e^{-\pi|x|^2/\delta}\) has Fourier transform \(e^{-\pi\delta|\xi|^2}\), and \(\int K_\delta = 1\) for every \(\delta > 0\).
1. Define \(f(x) \da \int_0^\infty K_\delta(x)\, e^{-\pi\delta}\,\delta^{\eps-1}\,d\delta\) for \(\eps > 0\).
2. \(f \in L^1(\RR^n)\).
by Tonelli and \(\int K_\delta = 1\), \[ \int_{\RR^n}|f(x)|\,dx \le \int_0^\infty e^{-\pi\delta}\,\delta^{\eps-1} \Big(\int_{\RR^n} K_\delta(x)\,dx\Big)\,d\delta = \int_0^\infty e^{-\pi\delta}\delta^{\eps-1}\,d\delta = \frac{\Gamma(\eps)}{\pi^\eps} < \infty . \]
3. Compute \(\fourier{f}\).
since \(f\) is an integral of \(L^1\) functions, the Fourier transform passes under the integral (Fubini for the absolutely convergent double integral): \[ \fourier{f}(\xi) = \int_0^\infty \fourier{K_\delta}(\xi)\, e^{-\pi\delta}\,\delta^{\eps-1}\,d\delta = \int_0^\infty e^{-\pi\delta|\xi|^2}\,e^{-\pi\delta}\,\delta^{\eps-1}\,d\delta = \int_0^\infty e^{-\pi\delta(1+|\xi|^2)}\,\delta^{\eps-1}\,d\delta . \]
4. Evaluate the last integral.
substitute \(t = \pi\delta(1+|\xi|^2)\): \[ \int_0^\infty e^{-\pi\delta(1+|\xi|^2)}\,\delta^{\eps-1}\,d\delta = \Big(\pi(1+|\xi|^2)\Big)^{-\eps} \int_0^\infty e^{-t} t^{\eps-1}\,dt = \pi^{-\eps}\,\Gamma(\eps)\,\Big(1+|\xi|^2\Big)^{-\eps} . \]
5. Conclude.
2. Hence \(F\) is the Fourier transform of an \(L^1\) function.
6. Q.E.D.