On $L^2(\mathbb{R}^d)$, we have $T_m$ defined $\widehat{T_m f} = m \widehat{f}$ is a bounded operator on $L^2$ if and only if $m \in L^\infty$.

What can be said about the same problem for more general measures? For example, I am interested in weighted $L^2$ spaces where $$ \mathrm{d}\mu(x) = (1 + |x|^{2})^{s/2} \mathrm{d}x, $$ when is $T_m$ a bounded operator on $L^2(\mu)$?

  • $\begingroup$ How did you show the first statement? If I use that $\mathcal F$ is unitary (Plancherel) in $(L^2, \|.\|_2)$, and then use Cauchy-Schwartz, I get that the operator norm is $\|m\|_{2}$. $\endgroup$
    – DominikS
    Commented Oct 4, 2023 at 7:11
  • 1
    $\begingroup$ We have $\| T_mf \|_2 = \big( \int m(\xi)^2 \widehat{f}(\xi)^2 \mathrm{d}\xi \big)^{1/2} \leq \|m\|_\infty \| \widehat{f} \|_2 = \|m\|_{\infty} \|f\|_2$ $\endgroup$
    – Jack
    Commented Oct 6, 2023 at 9:26

1 Answer 1


Let's assume $s>0$. Then $\mathcal{F} (L^2(\mu))$ is the Sobolev space $H^{2,s}$ (functions with $s$ derivatives in unweighted $L_2$), and your weighted space norm is giving the Sobolev norm of $\widehat{f}$.

To be bounded, $T$ can't destroy the smoothness of $\widehat{f}$, so you would want $m$ to be in some space like the $L_\infty$ Sobolev space of the same smoothness.

If you want a more precise statement, you might check Multiplication in Sobolev spaces for details


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