The following is mentioned on Wikipedia
https://en.wikipedia.org/wiki/Multiplier_(Fourier_analysis)#Multiplier_operators_on_common_groups
"In the sense of distributions, there is no difference between multiplier operators and convolution operators; every multiplier T can also be expressed in the form $Tf = f*K$ for some distribution $K$, known as the convolution kernel of T. In this view, translation by an amount $x_0$ is convolution with a Dirac delta function $δ(· − x_0)$, differentiation is convolution with $δ'.$ "
Does anyone know of a text where this is outlined?
This amounts to the duality between convolution and multiplication under the Fourier transform (https://en.wikipedia.org/wiki/Convolution_theorem). It states that $$ \widehat{f *g} = \hat{f}\cdot\hat{g}. $$ (This holds for the inverse Fourier transform also; $\mathcal{F}^{-1} f (x)$ is just $\mathcal{F}f(-x)$.) A Fourier multiplier $m(D)$ acts on a (Schwartz, say) function $f$ by $$ (\widehat{m(D)f})(\xi) = m(\xi)\hat{f}(\xi). $$ Inverting the Fourier transform and using the convolution theorem gives $$ m(D)f(x) = \mathcal{F}^{-1}[m\cdot\hat{f}](x) = \mathcal{F}^{-1}(m) * \mathcal{F}^{-1}(\hat{f})(x) = \mathcal{F}^{-1}(m) * f(x). $$ So, a Fourier multiplier is just a convolution with the inverse Fourier transform of the multiplier $m$.