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Let $\lambda>0$ be given. Define $$G_{\lambda}(\xi) = \chi_{_{\lbrace |\xi|^{2} \leq \lambda \rbrace }}. $$ and $$ E_{0}(\lambda)f = \mathcal{F}^{-1}[G_{\lambda}(|\xi|^{2})\mathcal{F}(f)], \ \ f \in L^{2}(\mathbb{R}^{n}) $$ How do I show that

$$(E_{0}(\lambda)f|f) = \|E_{0}(\lambda)f\|^{2}_{L^{2}}, \ \ f \in L^{2}(\mathbb{R}^{n})\qquad ?$$

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    $\begingroup$ The identity says that $E_0$ is a projection, and this is true because (multiplication by) $G_{\lambda}$ is a projection and the Fourier transform is unitary. $\endgroup$
    – Christian Remling
    Commented Jun 9, 2022 at 15:18
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    $\begingroup$ I think your question would have been more suitable for math.stackexchange.com . $\endgroup$
    – Christian Remling
    Commented Jun 9, 2022 at 15:19
  • $\begingroup$ @ChristianRemling I don't care which site I ask my question on $\endgroup$
    – user253963
    Commented Jun 9, 2022 at 16:37
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    $\begingroup$ AGL, but some other people do care, and @ChristianRemling's point is that you would receive more positive reception for this question on math.stackexchange $\endgroup$
    – paul garrett
    Commented Jun 10, 2022 at 0:10

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Your identity amounts to $$\frac12(E_0(\lambda)^*+E_0(\lambda))=E_0^*(\lambda) E_0(\lambda).$$ Since ${\cal F}^*=\cal F$, this is equivalent to saying that $$\frac12(G_\lambda+\overline{G}_\lambda)=|G_\lambda|^2,$$ which is true because $G_\lambda(\xi)$ equals either $0$ or $1$.

Edit. To explain the first equality, let us define the bounded linear operator $$L:=E_0(\lambda)-E_0^*(\lambda) E_0(\lambda).$$ Your assumption is that $(Lf\mid f)=0$ for every $f$, which means that $L^*+L=0$.

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  • $\begingroup$ I didn't understand the first equality, I'm sorry $\endgroup$
    – user253963
    Commented Jun 9, 2022 at 16:36
  • $\begingroup$ @AGL. I edited my answer, to explain this equality. $\endgroup$
    – Denis Serre
    Commented Jun 9, 2022 at 17:06
  • $\begingroup$ How do I show that $\lambda \to \|E_{0}(\lambda)f\|^{2}_{L^{2}}$ is absolutely continuous? $\endgroup$
    – user253963
    Commented Jun 10, 2022 at 19:16
  • $\begingroup$ mathoverflow.net/questions/424448/… $\endgroup$
    – user253963
    Commented Jun 10, 2022 at 19:17

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