Does conjugation of a Dirac delta function makes sense: $\delta^*(x-y)$? Namely is $\delta^*_y = \delta_y$? I am asking specifically in the context of this identity (which is the same as Plancherel's identity, however $\delta$ is not in $L_1$):
$$\int_{\mathbb{R}^d}f(x)\delta^*(x-y)\,dx = \langle f, \delta_y \rangle = f(y) = \mathcal{F}^{-1}[\hat{f}](y) = \langle\hat{f},\hat{\delta}_y \rangle = \int_{\mathbb{R}^d}\hat{f}(u)\exp(2\pi i \langle u,y\rangle)\,du$$
Note that:
$$\mathcal{F}[\delta_y](u) = \int_{\mathbb{R}}\delta(x-y)\exp(-2\pi i \langle u, x \rangle)\,dx = \exp(-2\pi i \langle u, y \rangle)$$
And then:
$$\hat{\delta}^*_y = \exp(2\pi i \langle u, y \rangle)$$
Is this valid?