 Solution

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus