Is Plancherel's theorem true for tempered distribution?

Let $f, g\in L^{2},$ by Plancherel’s theorem, we have

$$\langle f, g \rangle= \langle \hat{f}, \hat{g} \rangle.$$

My Question is:
Is it true that:
$$\langle f, g \rangle= \langle \hat{f}, \hat{g} \rangle$$

for $f\in \mathcal{S’}(\mathbb R^d)$ (tempered distribution) and $g\in \mathcal{S}(\mathbb R^d)$(Schwartz space)? If yes, how to justify.

Solutions Collecting From Web of "Is Plancherel's theorem true for tempered distribution?"

One has to be careful with complex conjugation here. The claimed identity is
\int f \bar g = \int \hat f \bar {\hat g}
Conjugating the (unitary) Fourier transform gives the inverse Fourier transform of conjugation. So,
\int \hat f\bar {\hat g} = \int \hat f \check {\bar g}
The right hand side is the value of distribution $\hat f$ on test function $\check {\bar g} $. By the definition of $\hat f$, this is computed by passing the hat to the test function, which cancels out the inverse hat:
\int \hat f \check {\bar g} = \int f \bar g

I think the answer lies in this characterization:

Explicit characterization of dual of $H^1$

knowing, in particular, that $\mathcal{F} : \mathcal{S}'(\mathbb{R}^n) \longrightarrow \mathcal{S}'(\mathbb{R}^n)$ is an isomorphism, extension of $\mathcal{F}:L^2(\mathbb{R}^n) \longrightarrow L^2(\mathbb{R}^n)$.

TrialAndError did you notice that there is no scalar product in the space of tempered distributions. But you can understand this, in the sense of the characterization of dual space $H^s({\mathbb{R}^n})^*$, or simply of $H^s(\mathbb{R}^n)$.