When is a function a Fourier transform of an integrable function?

Question

Specifically, in the case $f(\xi)=\frac{1}{(1+\xi ^2)^\epsilon}$ where $0<\epsilon<1$.

I wish to prove this is a Fourier transform of a $L_1$ function. Any insight into the manner would be appreciated.

Answer 1

Let $$ g_R(x) = \int_{-R}^R \frac1{(1+\xi^2)^\epsilon} e^{i\xi x} \, d\xi .$$ If you integrate by parts, you get $$ g_R(x) = \frac1{ix} \left[\frac1{(1+\xi^2)^\epsilon} e^{i\xi x}\right]_{-R}^R + \frac1{ix} \int_{-R}^R \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} e^{i\xi x} \, d\xi .$$ Therefore $\check f(x) = g(x) = \lim_{R\to \infty} g_R(x)$ is defined for $x \ne 0$, and $|g(x)| \le C_1/|x|$. Integrate by parts again, and you will see that $|g(x)| \le C_2/|x|^2$. (Here $C_1$ and $C_2$ are positive constants, where for example $C_1 = \int_{-\infty}^\infty 2 \epsilon |\xi|/(1+\xi^2)^{1+\epsilon} \, d\xi$.) Hence $g(x)$ decays fast enough as $x\to \pm\infty$ to be in $L^1$.

Now we need to show that $g(x)$ blows up slowly enough as $x\to 0$ to be in $L^1$. See that $$ g(x) = \frac2x \int_0^\infty \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} \sin(\xi x) \, d\xi .$$ So use $|\sin(a)| \le \min\{|a|,1\}$ to estimate $$ \left|\int_0^\infty \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} \sin(\xi x) \, d\xi \right|$$ $$ \le \int_0^{1/x} \frac{2\epsilon\xi^2 x}{(1+\xi^2)^{1+\epsilon}} \, d\xi + \int_{1/x}^\infty \frac{2\epsilon\xi}{(1+\xi^2)^{1+\epsilon}} d\xi $$ Elementary estimates show it decays as least as fast as $x^{2\epsilon}$ as $x \to 0$.