My question overlaps somewhat with: Is Plancherel's theorem true for tempered distribution?
I am trying to better understand the answer provided there, and have some additional questions. Let $S$ be a tempered distribution and $f$ be a Schwartz function. Then the following identity holds:
$$\int_{\mathbb{R}^d}\hat{S}(x)f(x)\,dx = \langle \hat{S}, f \rangle = \langle S,\hat{f} \rangle = \int_{\mathbb{R}^d}S(x)\hat{f}(x)\,dx$$
As far as I understood we do not have conjugation, since the identity $\langle \cdot, \cdot \rangle$ is not the $L^2$ inner product. Is there a specific way to indicate the inner product without conjugation?
Let $\overline{f}$ be the complex conjugate of $f$, let $\check{f}$ be the inverse Fourier transform of $f$, and let us have the identity $\overline{\hat{f}} = \check{\overline{f}}$ - which I have derived here: Fourier transform of conjugate function. Then:
$$\langle \hat{S},\hat{f} \rangle_{L^2} = \langle \hat{S},\overline{\hat{f}} \rangle = \langle \hat{S}, \check{\overline{f}} \rangle = \langle S, \overline{f} \rangle = \langle S, f \rangle_{L^2}$$
What are the assumptions for this to hold beyond $S$ being a tempered distribution and $f$ being a Schwartz function? Can the $C^{\infty}$ requirement be relaxed to allow for discontinuities over sets of measure zero? I am particularly interested if anything more can be relaxed in the case where $S$ is a Dirac delta.
Can a specific meaning be attached to the conjugation of a tempered distribution? Notably a user clarified that this can be done for the Dirac delta $\delta_y^* = \delta_{-y}$: Conjugate of Dirac delta
I am asking since, in general, I would like to be able to swap the places of $S$ and $f$ while the above identity should still hold:
$$\langle \hat{f}, \hat{S} \rangle_{L^2} = \langle f, S \rangle_{L^2}$$
Additionally, I am led to believe that if we have tempered distributions $S$ and $T$, then the following doesn't always make sense:
$$\langle \hat{S}, T \rangle = \langle S, \hat{T} \rangle$$
But in the case of $S(x) = \delta_y = \delta(x-y)$ and $T = \delta_z$ the expression seems to be well formed, specifically:
$$\langle \hat{\delta}_y,\delta_z \rangle = \int_{\mathbb{R}^d}\exp(-2\pi i \langle x, y \rangle)\delta(x-z)\,dx = \exp(-2\pi i \langle z, y \rangle) = $$ $$= \int_{\mathbb{R}^d}\delta(x-y)\exp(-2\pi i \langle z, x \rangle)\,dx =\langle \delta_y,\hat{\delta}_z \rangle$$
Additionally while $\langle \delta_y, \delta_z \rangle$ seems to make no sense, on the other hand we have:
$$\langle \hat{\delta}_y, \hat{\delta}_z \rangle = \int_{\mathbb{R}^d}\exp(-2\pi i \langle x, y+z \rangle)\,dx = \delta(y+z)$$
which seems to be defined. Is there anything that generalizes and formalizes this?