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.
On
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 $H^s(\mathbb{R}^n)$.
On
Yes. For any $f\in\mathcal{S}'(\mathbb{R}^{n})$, there exists a sequence $f_{n}\in \mathcal{C}^{\infty}_{0}(\mathbb{R}^{n})$ such that $f_{n}\to f$ under the topology of $\mathcal{S}'(\mathbb{R}^{n})$. According to Plancherel's theorem, we have $\langle f_{n},g\rangle=\langle\hat{{f}_{n}},\hat{g}\rangle$. Then by tending $n$ to $\infty$ we have $\langle f,g\rangle=\langle\hat{f},\hat{g}\rangle$.
Notation: if $f_{n}\to f$ under the topology of $\mathcal{S}'(\mathbb{R}^{n})$, we have $\hat{f_{n}}\to\hat{f}$ under the topology of $\mathcal{S}'(\mathbb{R}^{n})$.
One has to be careful with complex conjugation here. The claimed identity is $$ \int f \bar g = \int \hat f \bar {\hat g} \tag1$$ Conjugating the (unitary) Fourier transform gives the inverse Fourier transform of conjugation. So,
$$ \int \hat f\bar {\hat g} = \int \hat f \check {\bar g} \tag2$$ 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: $$ \int \hat f \check {\bar g} = \int f \bar g \tag3$$