For $m \in L^\infty$, we can define the multiplier operator $T_m \in L(L^2,L^2)$ implicitly by
$\mathcal F (T_m f)(\xi) = m(\xi) \cdot (\mathcal F T_m)(\xi)$
where $\mathcal F$ is the Fourier transform. It is obvious from the defintion that $T_m$ commutes with translations.
How can you show the converse, i.e. every translation invariant $T \in L(L^2,L^2)$ is induced by a multiplier $m_T$? I have no idea how this might work.