# 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.

• Yes, because $\hat f$ is defined by passing the hat to the test function. There are some complex conjugates to sort out in the scalar product, that's all.
– user147263
Commented Mar 20, 2016 at 16:25
• Thanks a lot; would you please explain bit more? Commented Mar 21, 2016 at 4:04
• Plancherel's theorem in $L^2$ involves inner products, whereas the distributional pairing does not. You have to account for that. Commented Mar 22, 2016 at 13:20

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: $$\int \hat f \check {\bar g} = \int f \bar g \tag3$$

• but you should not enter the space $H^s(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$ and its dual? Commented Mar 21, 2016 at 12:03

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 $H^s(\mathbb{R}^n)$.

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 $$\mathcal{S}'(\mathbb{R}^{n})$$.