If we define the operator $K(a)=F^{−1}aF$
where $ F:L^2({\mathbb R})\to L^2({\mathbb R})$, is the fourier transform given by $$\left(Ff\right)\left(x\right)=\int_{{\mathbb R}}{f\left(t\right)e^{itx}dt,\ \ \ \ x\in {\mathbb R}}.$$ and $a$ is a function on $L^{\infty}$, which makes the operator K map the dense subset $L^2({\mathbb R})\cap L^p({\mathbb R}),$ of $L^p({\mathbb R}),$ onto itself and extends to a bounded linear operator on $L^p({\mathbb R})$.
Any idea on how to prove (or how to see) that K is a convolution operator?