We can write, since $f\in\mathcal S(\mathbb R^n)$
\begin{align*}
\lVert x^{\alpha}\partial^{\beta}f\rVert_1&\leqslant
\int_{\mathbb R^n}|x|^{\alpha}|\partial^{\beta}f(x)|dx\\\
&=\int_{\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|\frac 1{(1+|x|)^{n+1}}dx\\\
&\leqslant C'\sup_{x\in\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|
\int_{\mathbb R^n}\frac{dx}{(1+|x|)^{n+1}}\\\
&=C'\sup_{x\in\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|
s_n\int_0^{+\infty}\frac{r^{n-1}}{(1+r)^{n+1}}dr,
\end{align*}
where $s_n$ is the area of the unit sphere in $\mathbb R^n$. The last integral
is convergent, and we get the expected result putting $C:=C's_n\int_0^{+\infty}\frac{r^{n-1}}{(1+r)^{n+1}}dr$.
For the second fact, note that $\partial^{\beta}\widehat f(x)=\int_{\mathbb R^n}
i^{\beta}t^{\beta}e^{it\cdot x}f(t)dt$, hence for $x\in\mathbb R^n$:
\begin{align*}
(1+|x|)^N|\partial^{\beta}\widehat f(x)|&=
(1+|x|)^N\left|\int_{\mathbb R^n}e^{it\cdot x}t^{\beta}f(t)dt\right|\\\
&=\sum_{k=0}^N\binom Nk|x|^k\left|\int_{\mathbb R^n}e^{it\cdot x}
t^{\beta}f(t)dt\right|\\\
&=\sum_{k=0}^N\binom Nk\sum_{|\gamma |=k}\left|\int_{\mathbb R^n}
x^{\gamma}e^{it\cdot x}
t^{\beta}f(t)dt\right|\\\
&=\sum_{|\gamma|\leqslant N}\binom Nk\left|\int_{\mathbb R^n}
x^{\gamma}e^{it\cdot x}t^{\beta}f(t)dt\right|.
\end{align*}
Let $\displaystyle I_{\gamma}(x):=\int_{\mathbb R^n}
x^{\gamma}e^{it\cdot x}t^{\beta}f(t)dt$. Integrating by parts and using
Leibniz formula, we have
\begin{align*}
|I_{\gamma}(x)|&=\left|\int_{\mathbb R^n}e^{it\cdot x}\sum_{\alpha\leqslant
\gamma}\binom{\gamma}{\alpha}\partial^{\alpha}f(t)t^{\beta-\alpha}\frac{\beta !}{(\beta-\alpha)!}dt\right|\\\
&\leqslant \beta !\sum_{\alpha\leq \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}\int_{\mathbb R^n}\left|\partial^{\alpha}f(t)t^{\beta-\alpha}\right|dt,
\end{align*}
and using the first point
\begin{align*}
|I_{\gamma}(x)|&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n}
(1+|x|)^{n+1}|x|^{\beta-\alpha}|\partial^{\alpha}f(x)|\\\
&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n}
(1+|x|)^{n+1}(1+|x|)^{\beta-\alpha}|\partial^{\alpha}f(x)|\\\
&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n}
(1+|x|)^{n+1+\beta}|\partial^{\alpha}f(x)|\\\
&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\lVert f\rVert_{(n+1+\beta,\alpha)}.
\end{align*}
Putting $A_{\gamma,\beta}=\beta\max_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}$. Then
$|I_{\gamma}(x)|\leqslant A_{\gamma,\beta}\sum_{\alpha\leq\gamma}\lVert f\rVert_{(n+1+\beta,\alpha)}$. Now, put $\displaystyle B_{N,\beta}:=\max_{|\gamma|\leqslant
N}A_{\gamma,\beta}\binom N{|\gamma|}$. We get
\begin{align*}
\lVert \widehat f\rVert_{(N,\beta)}&\leqslant B_{N,\beta}\sum_{|\gamma|\leqslant N}\:
\sum_{\alpha\leq \gamma} \lVert f\rVert_{(n+1+\beta,\alpha)}\\\
&\leqslant B_{N,\beta}\sum_{|\gamma '|\leqslant N} D(\gamma')\lVert f\rVert_{(n+1+\beta,\gamma')},
\end{align*}
where $D(\gamma')$ denote the number of times on which $\gamma'$ is
obtained in the double sum. Finally, we get
$$\lVert \widehat f\rVert_{(N,\beta)}\leqslant C_{N,\beta}\sum_{|\gamma |\leqslant N} \lVert f\rVert_{(n+1+\beta,\gamma)}$$
putting $\displaystyle C_{N,\beta}:=B_{N,\beta}\max_{|\gamma'|\leqslant N}D(\gamma')$.
Best Answer
Denoting by $\widehat{\cdot}$ the Fourier transform in $\mathbb R^d$ and letting $$ f_\lambda(x)\equiv f(\lambda x), $$ then, by definition of the Fourier transform, \begin{align} \widehat {f_\lambda}(k)&=\int_{\mathbb R^n}e^{-ik\cdot x}f(\lambda x)dx\\ &=\int_{\mathbb R^n}e^{-ik\cdot x/\lambda}f(x)\lambda^{-d}dx\\ &=\lambda^{-d}\widehat f(k/\lambda). \end{align} Let $f(x)=F(|x|)$ be a radial function; if $M\in SO(d)$, then $f_M(x)\equiv f(Mx)=f(x)$ and hence, since $\widehat{ f_M}(k)=\widehat f(Mk)$, we have $\widehat {f_M}= \widehat f$, which in turn implies that also $\widehat f$ is radial. Now, suppose $f(x)=|x|^{-\alpha}$, for $0<\alpha<d$, then $$ f_\lambda(x) = \lambda^{-\alpha}f(x) $$ and, by the above remark, $$ \widehat f (k/\lambda)= \lambda^{d}\widehat {f_\lambda}(k) = \lambda^{d-\alpha}\hat f(k). $$ The only radial $\widehat f$ homogeneous of degree $-d+\alpha$ is $$ \widehat f (k) = |k|^{\alpha-d}, $$ up to constants.
Note that $f$ 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 $L^1(\mathbb R^d)$, for $0<\alpha<d$, and $v$ lies in $L^2(\mathbb R^d)$, for $\alpha>d/2$. This ensures $\widehat u \in L^\infty(\mathbb R^d)$ and $\widehat v\in L^2(\mathbb R^d)$ and that $\widehat f$ needs indeed to be a function in $L^1_{\text{loc}}(\mathbb R^d)$ from an abstract point of view. So $f,\widehat f\in L^1_{\text{loc}}(\mathbb R^d)$.
More generally, $f$ lies in $\mathscr S'(\mathbb R^d)$, which is stable under Fourier transform.