I am trying to prove that $\Pi^-_\mu=\mathbb{1}(-\Delta\leq\mu)$ formally minimises $\gamma\mapsto\text{Tr}((-\Delta-\mu)\gamma)$ among the bounded operators on $L^2$ with $0\leq\gamma\leq 1$.
Attepmt at proof: Let $(\phi_n)$ be an orthonormal basis of $L^2$. Then, formally, \begin{align*} \text{Tr}((-\Delta-\mu)\gamma)=\sum_n(\phi_n,(-\Delta-\mu)\gamma\phi_n)=\sum_n\int\overline{\widehat\phi_n(\xi)}(|\xi|^2-\mu)\widehat{\gamma\phi_n}(\xi)\,\mathrm{d}\xi. \end{align*} We somehow want to use the property $0\leq\gamma\leq 1$ which implies that $$0\leq\int\overline{\widehat\psi(\xi)}\,\widehat{\gamma\psi(\xi)}\,\mathrm{d}\xi\leq\int|{\widehat\psi}(\xi)|^2\,\mathrm{d}\xi$$ for all $\psi\in L^2$, but I am not sure how to do so as we do not know that $\gamma$ commutes with $(-\Delta-\mu)$.