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.


1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – Eric
    Commented Jun 17, 2023 at 10:29
  • 1
    $\begingroup$ @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. $\endgroup$
    – jd27
    Commented Jun 17, 2023 at 17:23

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