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Let $\delta(t)$ be the Dirac delta function. We know $\delta'(t)$ can be also seen as an operator on test functions with compact support. Let $\phi \in C_0(\infty)$, $$ \int_{-\infty}^{\infty} \delta'(t)\phi(t)dt=-\phi'(0). $$ So is the Fourier transform of $\delta'(t)$, i.e. $$ \int_{-\infty}^{\infty} \delta'(t) e^{-iwt}dt=- [e^{-iwt}]'|_{t=0}=iw $$ But $e^{iwt}$ is not compactly supported.

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  • $\begingroup$ I'm not an expert on this. But I think the reason the distributional properties are defined in the context of test functions: $C_0^{\infty}$ is that we want to ensure that the integral over $R$ does not become unbounded as $|t| \rightarrow \infty$. The delta is special because it is supported on any subset of $R$ that includes $0$. I would just try doing integration by parts, treating $\delta'$ as a function. $\endgroup$ – D.B. Nov 23 '18 at 16:57
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Distributions with compact support, like $\delta',$ can be extended to linear functionals on $C^\infty$ by taking some $\rho \in C^\infty_c$ such that $\rho \equiv 1$ on a neighborhood of the support of the distribution and then setting $\langle u, \phi \rangle = \langle u, \rho\phi \rangle$ for $\phi \in C^\infty.$ It can be shown that this value is independent of the choice of $\rho$ and thus well-defined. Therefore $\langle \delta'(x), e^{-i\xi x} \rangle$ is okay even if $e^{-i\xi x}$ does not have compact support.

Also, the Fourier transformable distributions are linear functionals on the Schwartz space $\mathcal{S}$ instead of on $C^\infty_c.$ That space is closed under Fourier transforms, which $C^\infty_c$ is not. These distributions are called tempered and the Fourier transform is defined by $\langle \mathcal{F}u, \phi \rangle = \langle u, \mathcal{F}\phi \rangle.$ If $u$ has compact support then $\mathcal{F}u(\xi) = \langle u(x), e^{-i\xi x} \rangle.$

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