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

  • $\begingroup$ That doesn't sound right, consider the case $\phi_2 \equiv 0,\phi_1 \equiv c$, $f \equiv c$, where $c$ is a constant (I guess a large constant in this case). Unless the measure space is presumed infinite so that $L^2$ doesn't contain $L^\infty$? But even then if $f \in L^2 \cap L^\infty$ then you can set $\phi$ proportional to $f$ with a large proportionality constant to break it. $\endgroup$ – Ian Feb 16 '18 at 15:49
  • $\begingroup$ Taking $f = t h$ for a fixed $h \in L^2(\mu)$ (not zero) and $t>0$ shows that the right-side is scaling like $t^4$, but the left-side like $t^2$. Thus, the inequality can not be true. Maybe you forget a square-root somewhere? $\endgroup$ – p4sch Feb 16 '18 at 15:50

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


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 $$\left|\int \phi_1|f|^2\right|^2+\left|\int \phi_2|f|^2\right|^2=|\alpha_j|^2+|\beta_j|^2.$$

  • $\begingroup$ Well sorry, but if you look up the condition for equality in Cauchy-Schwarz you see you can't have equality unless $\phi_1$ and $\phi_2$ are both constant. $\endgroup$ – David C. Ullrich Feb 16 '18 at 16:44
  • $\begingroup$ I was leaving out the "$\sup_{||f||=1}$", sorry. You cannot have equality for fixed $f$, which of course doesn't say the sups cannot be equal. $\endgroup$ – David C. Ullrich Feb 16 '18 at 22:01
  • $\begingroup$ See edit. The proof gives some insight into what's really going on in the operator-theoretic version, perhaps allowing you to prove it without the spectral theorem: In fact both sides of the equality are just $||A_1^*A_1+A_2^*A_2||$. $\endgroup$ – David C. Ullrich Feb 17 '18 at 12:38
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    $\begingroup$ I'm very sorry. For $\|TS\|=\|ST\|$, I'm misreading the proposition of the paper . I'm very sorry and thank for your example. $\endgroup$ – Schüler Mar 3 '18 at 15:58
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    $\begingroup$ @daw Huh??? What are you talking about? I didn't say anything about step functions! Simple functions are dense in $L^\infty$. $\endgroup$ – David C. Ullrich Mar 6 '18 at 15:28

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