# Understanding of Fourier transform of the convolution of two distribution

In Yosida's Functional Analysis, he covers a topic on the Fourier transform of the convolution given by distributions (at the end of Section 3, Chapter 6). He begins by pointing out the Fourier transform of a distribution $$T$$ on $$\mathbb{R}^{n}$$with compact support ($$T\in \mathcal{E}'(\mathbb{R}^{n})$$) is given by a function $$\hat{T} (\xi) =(2\pi)^{-\frac{n}{2}}T_{[x]}(e^{-i\langle x,\xi \rangle}),$$ where the subscript $$[x]$$ is marked to indicate $$T$$ acts on functions of the variable $$x$$. My interpretation for this result is that $$\hat{T}$$, as a tempered distribution, can be written in terms of integration: $$\hat{T}(\varphi) = \int_{\mathbb{R}^{n}} \hat{T}(\xi) \varphi(\xi) \ d\xi = \int_{\mathbb{R}^{n}} (2\pi)^{-\frac{n}{2}}T_{[x]}(e^{-i\langle x,\xi \rangle}) \varphi(\xi) \ d\xi$$ for each Schwartz function $$\varphi\in \mathcal{S}(\mathbb{R}^{n})$$.

Question 1. Is my interpretation correct? I can't see it clearly when I read the proof, so I would like to ask for clarification.

I' m confused when he gives the Fourier transform of the convolution quoted as follows:

Theorem. If $$T\in \mathcal{S}'(\mathbb{R}^{n})$$ and $$\varphi\in \mathcal{S}(\mathbb{R}^{n})$$, then $$\widehat{T\ast \varphi} = (2\pi)^{\frac{n}{2}}\hat{\varphi}\hat{T} \tag{1}$$ If $$T_{1}\in \mathcal{S}'(\mathbb{R}^{n})$$ and $$T_{2}\in \mathcal{E}'(\mathbb{R}^{n})$$, then $$\widehat{T_{1}\ast T_{2}} = (2\pi)^{\frac{n}{2}}\hat{T_{2}}\cdot \hat{T_{1}} \tag{2},$$ which has a sense sense $$\hat{T_{2}}$$ is given by a function.

My interpretation for Equation (1) is that this equation holds as both sides of the objects are seen as tempered distributions since $$\hat{T}$$ is nothing but a tempered distribution. However, I am confused by his explanation on $$\hat{T_{2}}$$ in this Theorem and that in the proof of Equation (2). Truly, $$\hat{T_{2}}$$ should be viewed as a function since it makes no sense to speak of multiplication of two distributions (at least at this point), but his proof of Equation (2) seems to indicate $$\hat{T_{2}}$$ should be a tempered distribution. Let me quote the proof as follows:

Proof of Equation (2). Let $$\psi_{\varepsilon}$$ be the regularization $$T_{2}\ast \psi_{\varepsilon}$$. Then the Fourier transform of $$T_{1}\ast \psi_{\varepsilon} = T_{1}\ast (T_{2}\ast \varphi_{\varepsilon}) = (T_{1}\ast T_{2})\ast \varphi_{\varepsilon}$$ is by Equation (1), equal to $$(2\pi)^{\frac{n}{2}} \hat{T_{1}} \cdot \hat{\psi_{\varepsilon}} = (2\pi)^{\frac{n}{2}} \hat{T_{1}} \cdot (2\pi)^{\frac{n}{2}}\hat{T_{2}}\cdot \hat{\varphi_{\varepsilon}} = (2\pi)^{\frac{n}{2}} \widehat{T_{1}\ast T_{2}} \cdot \hat{\varphi_{\varepsilon}}. \tag{3}$$ Hence we obtain Equation (2) by letting $$\varepsilon \to 0$$ and using $$\lim_{\varepsilon\to 0} \hat{\varphi_{\varepsilon}}(x)=1$$.

You can see in Equation (3), the middle term is given by Equation (1) and because of that, $$\hat{T_{2}}$$ should be viewed as a tempered distribution.

Question 2. What does Equation (2) truly mean? Do I misunderstand the whole thing?

Thank you first for kindly reading this long post, and I will appreciate any comment and answer.

## To question 1:

Yes it means that there is a function (the one given) so that $$\hat{T}$$ is given by integration against that function.

## To question 2:

Equation 1 also holds in the sense of functions if $$T \in \mathcal{E}^\prime$$ and we identify $$\hat T$$ with the $$C^\infty$$ function corresponding to it. It even follows that $$T * \varphi \in \mathcal{S}$$ and so the whole equation can be interpreted in the sense of functions. It is also important to keep the compatibility of the Fourier transform and the convolution in mind.

• Thank you for this answer and this clears up my confusion, but would you mind explaining what you mean by the compatibility of the Fourier transform and the convolution? Thanks!
– Eric
Commented Jun 17, 2023 at 10:29
• @Eric i mean that the (functional) Fourier transform of a function and the distributional Fourier transform of the distribution associated to that function can be identified with each other in natural way under certain conditions. Similarly the convolution of a distribution/function with a function (seen as a function) and the distributional convolution of a distribution with the distribution associated to a function can be identified under certain circumstances.
– jd27
Commented Jun 17, 2023 at 17:23