$\sup_{\|\,f\|_{L^2(\mu)}=1}\big(\int_X \varphi_1^2\, f^2\,d\mu + \int_X \varphi_2^2 \,f^2\,d\mu\big)=\big\|\varphi_1^2+\varphi_2^2\big\|_\infty$ Let $(X,\mu)$ be a measure space.

Let $\varphi_1,\varphi_2\in L^\infty(\mu)$, why 
  $$\sup_{\|f\|_{L^2(\mu)}= 1}\left(\int_X |\phi_1|^2 |f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)=||\,|\phi_1|^2+|\phi_2|^2||_\infty\;?$$

 A: This part
$$
\sup_{\|\,f\|_{L^2(\mu)}\leq 1}\left(\int_X |\phi_1|^2 |\,f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)\le\|\,|\phi_1|^2+|\phi_2|^2\|_\infty=M
$$
is straight-forward. For the inverse inequality, the definition of essential supremum (or $L^\infty$-norm) implies that, for every $\varepsilon>0$, there
$$
\mu\left(\{x: |\phi_1(x)|^2+|\phi_2(x)|^2>M-\varepsilon \}\right)>0.
$$
Set 
$$U_\varepsilon=\{x: |\phi_1(x)|^2+|\phi_2(x)|^2>M-\varepsilon \}
$$ 
and pick $W_\varepsilon\subset U_\varepsilon$, so that
$$
0<\mu(W_\varepsilon)<\infty
$$
and set
$$
f_\varepsilon=\frac{1}{\sqrt{\mu(W\varepsilon)}}\chi_{W_\varepsilon}
$$
Clearly, $$\int_X |f_\varepsilon|^2\,d\mu=1$$ 
and
$$
\left(\int_X |\phi_1|^2 |\,f_\varepsilon|^2d\mu + \int_X |\phi_2|^2 |\,f_\varepsilon|^2d\mu\right)\ge 
\int_{W_\varepsilon}\big( |\phi_1|^2+|\phi_2|^2\big) |\,f_\varepsilon|^2d\mu \ge M-\varepsilon.
$$
Hence
$$
\sup_{\|\,f\|_{L^2(\mu)}\leq 1}\left(\int_X |\phi_1|^2 |\,f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)\ge M-\varepsilon,
$$
for all $\varepsilon>0$.
