# $\sup_{\|\,f\|_{L^2(\mu)}=1}\big(\int_X \varphi_1^2\, f^2\,d\mu + \int_X \varphi_2^2 \,f^2\,d\mu\big)=\big\|\varphi_1^2+\varphi_2^2\big\|_\infty$

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

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

• Unless the $\phi$'s vanish the left hand side is clearly infinite. Did you intend to write something else? And do you have any thoughts on how to proceed? – Umberto P. Feb 28 '18 at 14:21
• Clearly $$\sup_{\|f\|_{L^2(\mu)}}\left(\int_X |\phi_1|^2 |f|^2d\mu + \int_X |\phi_2|^2 |f|^2d\mu\right)\leq||\,|\phi_1|^2+|\phi_2|^2||_\infty.$$ But I'm facing difficulties to show the reverse inequality – Schüler Feb 28 '18 at 14:25
• I made an edit that hopefully corrected the question to what it should be… – Dirk Feb 28 '18 at 14:31

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 $\varepsilon>0$.
• Thank you for your answer, why you don't set $$f_\varepsilon=\frac{1}{\sqrt{\mu(U\varepsilon)}}\chi_{U_\varepsilon}$$? – Schüler Feb 28 '18 at 16:05
• @Schüler Because, it is possible that $\mu(U_\varepsilon)=\infty$. That's why we got in all the trouble to define $W_\varepsilon$. – Yiorgos S. Smyrlis Feb 28 '18 at 16:38
• If $(X,\mu)$ is not $\sigma$-finite measure space. Why we can pick $W_\varepsilon\subset U_\varepsilon$, such that $$0<\mu(W_\varepsilon)<\infty ??$$ Thank you. – Schüler Mar 6 '18 at 6:04
• If $\mu=0$ this identity is still true, since the essential supremum is zero! If $\mu(E)=\infty$, for all $E\ne\varnothing$, then $L^2(\mu)=\{0\}$. – Yiorgos S. Smyrlis Mar 6 '18 at 6:45