Consider the distributional identity (see Chapter 2.6 of Kanwal's Generalized Functions, for example) $$ f(x) \delta'(x-\xi) = - f'(\xi) \delta(x - \xi) + f(\xi) \delta'(x-\xi) \tag{1} $$ I can see how this identity must be true under and integral (using a test function $\phi(x)$), in that $$ \int_{-\infty}^{\infty} \mathrm{d}x\ f(x) \delta'(x-\xi) \phi(x) = \int_{-\infty}^{\infty} \mathrm{d}x\ \big[ - f'(\xi) \delta(x - \xi) + f(\xi) \delta'(x-\xi) \big] \phi(x) \ . $$
I am curious if there is a way to understand the equation (1) in Fourier space ie. is there a way to see that $$ f(x) \int_{-\infty}^{\infty} \frac{\mathrm{d}p}{2\pi} i p e^{i p (x - \xi)} = - f'(\xi) \int_{-\infty}^{\infty} \frac{\mathrm{d}p}{2\pi} e^{i p (x - \xi) } + f(\xi) \int_{-\infty}^{\infty} \frac{\mathrm{d}p}{2\pi} i p e^{i p (x - \xi)} $$ without putting the LHS and RHS underneath an integral with a test function?
P.S. Although this argument is probably in bad taste, I can for example make some progress on the identity $x \delta(x) = 0$ in Fourier space: $$ x\delta(x) = \int_{-\infty}^{\infty} \frac{\mathrm{d}p}{2\pi} x e^{i p x} = \int_{-\infty}^{\infty} \frac{\mathrm{d}p}{2\pi} \frac{\mathrm{d}}{\mathrm{d} p} \left\{ - i e^{i p x} \right\} \to 0 $$ which is zero if you can ignore the contribution from the endpoints. So I am looking for something analogous here.