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 $(Lf|f)=0$ for every $f$, which means that $L^*+L=0$.

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 $L^*+L=0$.

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 $1$.

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 $(Lf|f)=0$ for every $f$, which means that $L^*+L=0$.