Fourier transform of $\delta'(t)$ is $iw$?

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.

• 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. – D.B. Nov 23 '18 at 16:57

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 $$\mathcal{F}u(\xi) = \langle u(x), e^{-i\xi x} \rangle.$$