I have the following inner product: $$\langle \delta_y, f \rangle = \int_{\mathbb{R}^d} \delta(x-y)f^*(x)\,dx = f^*(y)$$
For it a property similar to Plancherel's theorem can be shown with respect to the Fourier transform even though the Dirac delta is not in $L_1$:
$$f^*(y) = \int_{\mathbb{R}^d} \delta(x-y)f^*(x)\,dx = \langle \delta_y, f \rangle$$
$$ f^*(y) = \overline{\mathcal{F}^{-1}[\hat{f}](y)} = \overline{\int_{\mathbb{R}^d} \exp(2\pi i \langle u, y \rangle)\hat{f}(u)\,du} = $$ $$ = \int_{\mathbb{R}^d}\exp(-2\pi i \langle u,y\rangle)\hat{f}^*(u)\,du = \langle \hat{\delta_y}, \hat{f} \rangle$$
Where $\hat{\cdot}$ and $\mathcal{F}[\cdot]$ represents the Fourier transform.
Is there some formalization of this?
Now if I consider the identity $\langle\delta_y,\hat{f}\rangle = \langle\hat{\delta}_y,f\rangle$ instead, then it also holds:
$$\hat{f}^*(y) = \int_{\mathbb{R}^d}\delta(u-y)\hat{f}^*(u)\,du = \int_{\mathbb{R}^d}\exp(-2\pi i \langle x, y\rangle)f^*(x)\,dx = \hat{f}^*(y)$$
In that case even if $f=\delta_z$, then:
$$\langle \delta_y, \hat{\delta}_z \rangle = \int_{\mathbb{R}^d}\delta(x-y)\exp(-2\pi i \langle x, z \rangle)\,dx = \exp(-2\pi i \langle y, z \rangle)$$
References on the subject are welcome.
Edit: I had a mistake with conjugation.