Why is the Fourier transform of $1/|x|^\alpha$ a function?

I would like to prove that the Fourier transform of $f(x) = 1/|x|^\alpha$ is a function, where $x\in \mathbb{R}^n$ and $0<\alpha <n$. It's clear for me that it is a tempered distribution. I need this information to complete the proof that the Fourier transform of $1/|x|^\alpha$ is $k/|x|^{n-\alpha}$ for some $k\in \mathbb C$. The calculation of the Fourier transform of $f$, knowing that $\widehat f$ is a function, is here.

• you have to consider $\langle \varphi, f \rangle = \langle \hat{\varphi}, \hat{f} \rangle$ for well chosen test functions $\varphi$ to prove that $\langle \delta_a, \hat{f} \rangle$ is finite for $a \ne 0$ so that your constant $k$ exists – reuns Jan 20 '16 at 16:16
• Can you help me? Why $f$ is a tempered distribution? – Senna Feb 22 at 19:59
• Look at the decomposition $f=u+v$ as in the answer below. The function $u$ is in $L^1$ and $v$ is in $L^\infty$. So, you just need to use that $L^p \subset S'$ for all $p$, which is proved here. – Hugocito Feb 22 at 20:42

Let $\alpha>n/2$. Then, $f$ can be split as the sum of two pieces: let $B_a$ be the ball of radius $a$ centred at the origin, then \begin{align} f&\equiv u+v\\ &= \chi_{B_a}(x)\frac{1}{|x|^{\alpha}}+\left(1-\chi_{B_a}(x)\right)\frac{1}{|x|^{\alpha}}; \end{align} $u$ lies in $L^1(\mathbb R^n)$, for $0<\alpha<n$, and $v$ lies in $L^2(\mathbb R^n)$, precisely for $\alpha>n/2$. This ensures $\widehat u \in L^\infty(\mathbb R^n)$ and $\widehat v\in L^2(\mathbb R^n)$ and that $\widehat f$ needs indeed to be a function in $L^1_{\text{loc}}(\mathbb R^n)$ from an abstract point of view. So $f,\widehat f\in L^1_{\text{loc}}(\mathbb R^d)$.
In general, $f$ lies in $\mathscr S'(\mathbb R^d)$, which is stable under Fourier transform and, moreover, the Fourier is a bijection of $\mathscr S'$ onto itself.
If $a<n/2$, start instead from g=$1/|\xi|^{n-\alpha}$. Running the above argument, we find that $\widehat g$ must lie in $L^1_{\text{loc}}(\mathbb R^n)$. This argument in turn proves that $\widehat g$ is $f$ up to constants, so that in fact $\widehat f=g$ up to constants.
This leaves out the case $\alpha=n/2$ which is gained by continuity of the Fourier transform on $\mathscr S'$: since $$\langle |x|^{-n/2}-|x|^{\varepsilon-n/2},\psi(x)\rangle\to0$$ as $\varepsilon\to0$, for every $\psi\in\mathscr S$, then also $\widehat{|x|^{-n/2}}=|\xi|^{-n/2}$ up to constants.