I have a question concerning the commutator of translation invariant operators on $L^2(\mathbb{R})$. Recall that $S:L^2(\mathbb{R})\to L^2(\mathbb{R})$ is translation invariant if $Su_t=u_tS$ for all $t\in\mathbb{R}$ where $(u_tf)(x)=f(x+t)$. The space of such operators is well-studied due for example to the work of Hörmander. The question is the following: Let $$ \mathcal{A}:=\{S:L^2(\mathbb{R})\to L^2(\mathbb{R}):\ S\ \text{is translation-invariant}\}, $$ what is $\mathcal{A}'$ (the commutant of $\mathcal{A}$ inside $L^2(\mathbb{R})$)? Is there any explicit description of this space, and if so, do you have some literature sources for such results? Thank you very much in advance for your help.
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2$\begingroup$ Don't you mean the commutant in $B(L^2(R))$ rather than in $L^2(R)$? $\endgroup$– YCorCommented Mar 19, 2020 at 20:18
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$\begingroup$ @YCor Sorry of course I mean $B(L^2(\mathbb{R}))$. $\endgroup$– Lenard VelasquezCommented Mar 26, 2020 at 10:16
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1 Answer
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Go to the Fourier transform picture. Then these become the operators that commute with multiplication by exponentials, which means they commute with all multiplication operators, which means you have just described the operators which become multiplication operators when you conjugate by the Fourier transform.
answered Mar 19, 2020 at 13:05
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$\begingroup$ (I.e., $\mathcal{A}' = \mathcal{A}$.) $\endgroup$ Commented Mar 19, 2020 at 13:48
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$\begingroup$ Thank you for the clarification. Is there a way to see that $\mathcal{A}'=\mathcal{A}$ without using the Fourier transform? In fact one has that if I apply the Fourier transform to $\mathcal{A}$ this space becomes $L^{\infty}(\mathbb{R})$. How can one use this? $\endgroup$ Commented Mar 26, 2020 at 10:20
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$\begingroup$ "Is there a way to see ...?" --- in principle, yes, but I think you'd have to build up a lot of machinery that's immediately available for multiplication operators. "How can one use this?" --- sorry, I don't understand, use it for what? $\endgroup$ Commented Mar 26, 2020 at 13:54