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Cross-post from math.sx.

Let $\eta:\mathbb R^n\to[0,1]$ be a smooth and compactly supported function and assume $f:\mathbb R^n \to\mathbb R$ is measurable. I want to bound the commutator of the operators $\eta(-i\nabla)$, which is a well-defined bounded and selfadjoint operator by the functional calculus, and $M_f$, which is the multiplication operator given by $f$, acting on $L^2(\mathbb R^n)$.

If $f$ is Lipschitz-continuous, I can prove the commutator is bounded by using properties of the Fourier transform. However, I would like to use weaker assumptions (for example locally Hölder continuous) on $f$, as my assumption on $\eta$ is rather strong. References, proofs and counterexamples are all warmly welcome.

Edit: I should stress that I especially do not have any integrability or boundedness conditions on $f$. Examples, I am thinking of, are usually diverging at infinity, such as $f(x)=|x|$ (which is an example of the Lipschitz case) or, say, $f(x)=x^2$.

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    $\begingroup$ I do not think you can go far beyond a linear growth. Let $\hat g=\eta$ so that $g$ is in the Schwartz class and $\eta(-i\nabla) h=g*h$. Then the commutator is given by $$Ch(x)=\int_{\bf R^n}g(y) (f(x-y)-f(x))h(x-y)\, dy.$$ If $|f(x-y)-f(x)| \leq A(y)$ and $gA$ is in $L^1$, then $C$ is bounded by Young's inequality for convolutions. However this imples a linear growth for $f$, since the estimate is independent of $x$. If $f(x)=x^2$ (1d), the above difference is $2xy+y^2$. The $y^2$ term is treated as above, but the extra $x$ can make $C$ unbounded. I did not work the details, however. $\endgroup$
    – Giorgio Metafune
    Commented Aug 17, 2021 at 7:32
  • $\begingroup$ Thank you for the comment and for taking time to think about the problem, although it is of course not the answer I hoped for. ;) $\endgroup$
    – Benjamin
    Commented Aug 17, 2021 at 11:13

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