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


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



Spectral family associated with the Laplacian operator in $L^{2}(\mathbb{R}^{n})$
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 $L^\*+L=0$.


