Return to Revisions

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 $1$.