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Let $\mathcal{L}$ be a linear and continuous operator from the space of tempered distributions $\mathcal{S}'(\mathbb{R})$ to itself. The Fourier transform of a tempered distribution $f$ is denoted by $\mathcal{F}f$ and the inverse Fourier transform by $\mathcal{F}^{-1}f$. I am wondering if the following conjecture is true, which would provide a characterization of convolution (i.e. shift-equivariant) operators.

Conjecture: The operator $\mathcal{L}$ is shift-equivariant if and only if, for any $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R})$ with disjoint support (i.e., such that $\varphi_1 \times \varphi_2 = 0$), we have that \begin{equation} \langle \mathcal{L} \{ \mathcal{F}\varphi_1 \} , \mathcal{L} \{ \mathcal{F}\varphi_2 \} \rangle = 0. \end{equation}

One side of the conjecture is clear. If $\mathcal{L}$ is shift equivariant, then it can be written as $\mathcal{L} f = h * f$ for some $h$ in the space $\mathcal{O}_{c}'(\mathbb{R})$ of rapidly decaying distributions. Therefore, using the properties of the Fourier transform, we have \begin{equation} \langle \mathcal{L} \{ \mathcal{F}\varphi_1 \} , \mathcal{L} \{ \mathcal{F}\varphi_2 \} \rangle = \langle h * \mathcal{F}\varphi_1 ,h * \mathcal{F}\varphi_2 \rangle = \langle (\mathcal{F}^{-1} f ) \cdot \varphi_1 ,(\mathcal{F}^{-1} f ) \cdot \varphi_2 \rangle = \langle (\mathcal{F}^{-1} f ) ,(\mathcal{F}^{-1} f ) \cdot \varphi_1 \cdot \varphi_2 \rangle = 0. \end{equation} I am wondering if the converse of this result is true, leading to the proposed conjecture.

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    $\begingroup$ Shift-invariant means what I could call shift-equivariant, right, in the sense that the transform of the shift is the shift of the transform? $\endgroup$
    – LSpice
    Commented Oct 10, 2020 at 21:33
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    $\begingroup$ This is correct, it is more precise to talk about equivariance. I updated. $\endgroup$
    – Goulifet
    Commented Oct 11, 2020 at 2:11
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    $\begingroup$ What about, for instance, $({\cal L}f)(x)=f(-x)$? $\endgroup$
    – Michael Renardy
    Commented Oct 11, 2020 at 2:25
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    $\begingroup$ @MichaelRenardy Your counter-example can be solved by changing the statement to $\langle \mathcal{L} \{ \mathcal{F}\varphi_1 \} , \mathcal{F}\varphi_2 \rangle = 0 $. Next counter-example is $\cal{ Lf = F^{-1}( F(f)')}$. $\endgroup$
    – reuns
    Commented Oct 11, 2020 at 10:07
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    $\begingroup$ @reuns This works indeed. And actually anything of the form $\mathcal{F}^{-1} \{ h * \mathcal{F}\{f\} \}$ would, with $h$ a distribution whose support is a singleton (yours is $h = \delta'$). Therefore the conjecture is false. Thanks for your help. $\endgroup$
    – Goulifet
    Commented Oct 11, 2020 at 14:53

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