Why $\sup_{\|f\|_2=1}(\left|\int_X \phi_1 |f|^2d\mu\right |^2 + \left|\int_X \phi_2 |f|^2d\mu\right |^2 )\ge ||\,|\phi_1|^2+|\phi_2|^2||_\infty\;?$

Question

Let $(X,\mu)$ be a measure space (not necessary $\sigma$-finite) and $\varphi_1,\varphi_2\in L^\infty(\mu)$.

I want to show that $$\sup_{\|f\|_{L^2(\mu)}=1}\left(\left|\int_X \phi_1 |f|^2d\mu\right |^2 + \left|\int_X \phi_2 |f|^2d\mu\right |^2\right) \ge \sup_{\|f\|_{L^2(\mu)}=1}\left(\int_X |\phi_1|^2 |f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right).$$

By this answer (1), we have

$$\sup_{\|f\|_{L^2(\mu)}= 1}\left(\int_X |\phi_1|^2 |f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)=||\,|\phi_1|^2+|\phi_2|^2||_\infty:=M.$$

How we can prove that $$\sup_{\|f\|_{L^2(\mu)}=1}\left(\left|\int_X \phi_1 |f|^2d\mu\right |^2 + \left|\int_X \phi_2 |f|^2d\mu\right |^2\right) \ge M\;?$$

Answer 1

Note that by the definition of essential supremum, we may two find discs $D_1$ and $D_2$ such that $$\Omega=\{x\in X;\;\phi_1(x)\in D_1,\,\phi_2(x)\in D_2\},$$ has a measure $>0$. Also $|\theta_1|^2+|\theta_2|^2>M-\varepsilon$ for every $\theta_i\in D_i,\;i=1,2$. Choose $f_\varepsilon$ with $\|f_\varepsilon\|_2=1$ concentrated in $\Omega$. Then $\int \phi_i |f|^2d\mu\in D_i,\;i=1,2$. It implies that $$\sup_{\|f\|_{L^2(\mu)}=1}\left(\left|\int_X \phi_1 |f|^2d\mu\right |^2 + \left|\int_X \phi_2 |f|^2d\mu\right |^2\right) \ge M-\varepsilon .$$

Answer 2

In fact the two supremums are equal, at least for a $\sigma$-finite measure space (or any measure space such that $||\phi||_\infty=||\phi||_{(L^1)^*}$ for $\phi\in L^\infty$.)

First, if $\int|f|^2=1$ then Cauchy-Schwarz shows that$$\left|\int\phi|f|^2\right| \le\int(|\phi|\,|f|)|f|\le\left(\int|\phi|^2|f|^2\right)^{1/2}||f||_2 =\left(\int|\phi|^2|f|^2\right)^{1/2}.$$

Hence $$\left|\int\phi_1|f|^2\right|^2+\left|\int\phi_2|f|^2\right|^2 \le\int|\phi_1|^2|f|^2+\int|\phi_2|^2|f|^2.$$

For the other direction, it's convenient to note that both sides are continuous in $\phi_1$ and $\phi_2$ (in the norm topology) so we may assume that $\phi_1$ and $\phi_2$ are simple functions.

In any case it's clear that $$\sup_{||f||_2=1}\int(|\phi_1|^2+|\phi_2|^2)|f|^2=||\,|\phi_1|^2+|\phi_2|^2||_\infty.$$Now, muttering the words "common refinement", suppose that $$\phi_1=\sum\alpha_j\chi_{E_j},\phi_2=\sum\beta_j\chi_{E_j},$$where the $E_j$ are disjoint and have positive measure. There exists $j$ so that $$||\,|\phi_1|^2+|\phi_2|^2||_\infty=|\alpha_j|^2+|\beta_j|^2;$$now if $||f||_2=1$ and $f$ iis supported on $E_j$ we have $$\left|\int \phi_1|f|^2\right|^2+\left|\int \phi_2|f|^2\right|^2=|\alpha_j|^2+|\beta_j|^2.$$