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

Schüler
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  • 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. – Ian Feb 16 '18 at 15:49
  • 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? – p4sch Feb 16 '18 at 15:50

2 Answers2

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

David C. Ullrich
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  • 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. – David C. Ullrich Feb 16 '18 at 16:44
  • 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. – David C. Ullrich Feb 16 '18 at 22:01
  • 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 $||A_1^A_1+A_2^A_2||$. – David C. Ullrich Feb 17 '18 at 12:38
  • Why? I didn't say I saw how to prove that that expression was the norm directly. The argument from the spectral theorem implies that it is the norm, because $||A_1^A_1+A_2^A_2||=|||\phi_1|^2+|\phi_2|^2||_\infty$. – David C. Ullrich Feb 17 '18 at 13:19
  • Why there exists $j$ so that $$||,|\phi_1|^2+|\phi_2|^2||_\infty=|\alpha_j|^2+|\beta_j|^2?$$ and thank you. – Schüler Mar 03 '18 at 09:23
  • @Schüler That's totally obvious. Sorry, but you're going to have to find someone else to explain this to you. You recently spent a long time insisting that $||ST||=||TS||$, after I'd shown you a simple counterexample. So the idea of trying to explain things to you seem pointless. – David C. Ullrich Mar 03 '18 at 15:20
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    I'm very sorry. For $|TS|=|ST|$, I'm misreading the proposition of the paper . I'm very sorry and thank for your example. – Schüler Mar 03 '18 at 15:58
  • @daw I think step functions are dense in $L^\infty$ (see https://math.stackexchange.com/questions/98680/are-simple-functions-dense-in-l-infty). Maybe you're thinking of continuous functions? –  Mar 06 '18 at 15:17
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    @daw Huh??? What are you talking about? I didn't say anything about step functions! Simple functions are dense in $L^\infty$. – David C. Ullrich Mar 06 '18 at 15:28
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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 $$\sup_{\|f\|_{L^2(\mu)}=1}\left(\left|\int_X \phi_1 |f|^2d\mu\right |^2 + \left|\int_X \phi_2 |f|^2d\mu\right |^2\right) \ge M-\varepsilon .$$

Student
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