$ \sup_{\|\,f\|_{L^2(\mu)}\leq 1}\left(\int_X |\phi_1|^2 |\,f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)\le\|\,|\phi_1|^2+|\phi_2|^2\|_\infty?$

Question

Let $(X,\mu)$ be a measure space.

Let $\varphi_1,\varphi_2\in L^\infty(\mu)$, why it is straight-forward that $$ \sup_{\|\,f\|_{L^2(\mu)}\leq 1}\left(\int_X |\phi_1|^2 |\,f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)\le\|\,|\phi_1|^2+|\phi_2|^2\|_\infty:=M\;? $$

Answer 1

Let $f\in L^2(\mu)$ with $\|f\|_2\leq 1$. Then $\int_X|f|^2d\mu\leq 1$. Since $|\phi_1|^2+|\phi_2|^2\leq\||\phi_1|^2+\|\phi_2|^2\|_\infty$ $\mu$-a.e., $\int_X(|\phi_1|^2+|\phi_2|^2)|f|^2\leq\||\phi_1|^2+|\phi_2|^2\|_\infty$ Take $\sup$ for those functions $f$ and you're done.