Let $\lambda>0$ be given. Define $$G_{\lambda}(\xi) = \chi_{_{\lbrace |\xi|^{2} \leq \lambda \rbrace }}. $$ and $$ E_{0}(\lambda)f = \mathcal{F}^{-1}[G_{\lambda}(|\xi|^{2})\mathcal{F}(f)], \ \ f \in L^{2}(\mathbb{R}^{n}) $$ I check that $$(E_{0}(\lambda)f|f) = \|E_{0}(\lambda)f\|^{2}_{L^{2}}, \ \ f \in L^{2}(\mathbb{R}^{n})\qquad $$
How do I show that $\lambda \to \|E_{0}(\lambda)f\|^{2}_{L^{2}}$ is absolutely continuous?
Since $\mathcal{F}^{-1}$ is - up to a constant - an isometry of $L^2(\mathbb{R}^n)$, we need only to prove that the map $$\lambda \mapsto ||1_{B_{\sqrt{\lambda}}}\mathcal{F}f||_2^2$$ is absolutely continuous.
Call $\sigma$ the uniform law on the (euclidian) unit sphere $\mathbb{S}_{n-1}$ and call $b_n$ the volume of the (euclidian) unit ball in $\mathbb{R}^n$. Since the map $|\mathcal{F}f|^2$ is integrable on $\mathbb{R}^n$, integration in polar coordinates yields for every $\lambda \ge 0$ $$||1_{B_{\sqrt{\lambda}}}\mathcal{F}f||_2^2 = \int_0^{\sqrt{\lambda}} \Big( \int_{\mathbb{S}_{n-1}} |(\mathcal{F}f)(ru)|^2 \mathrm{d}\sigma(u)\Big) nb_nr^{n-1} \mathrm{d}r.$$ By the change of variable $s=r^2$, one gets the integral on $[0,\lambda]$ of some integrable function from $\mathbb{R}_+$ to $\mathbb{R}$. Absolute continuity follows.