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Let $(E,\mathcal E,\mu)$ be a probability space and $A$ be a self-adjoint bounded linear operator on $L^2(\mu)$. Assume $Af\ge0$ for all $f\in\mathcal L^2(\mu)$ with $f\ge0$. Are we able to show the Cauchy-Schwarz-like inequality $$\langle Af,f\rangle_{L^2(\mu)}\le\left\|f\right\|_{L^2(\mu)}^2\sup_{\substack{g\in\mathcal L^2(\mu)\\\left\|g\right\|_{L^2(\mu)}\le1\\g\ge0}}\left\|Ag\right\|_{L^2(\mu)}\tag1$$ for all $f\in\mathcal L^2(\mu)$ with $f\ge0$? Since the standard prove doesn't work, we need a different argument.

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First we assume that $f\in\mathcal L^2(\mu)$ with $f\ge0$ and $\left\|f\right\|_{L^2(\mu)}=1$. Then we have by the standard Cauchy-Schwartz that $$ \langle Af,f\rangle_{L^2(\mu)}\le \| Af\|_{L^2(\mu)}\| f\|_{L^2(\mu)}=\| Af\|_{L^2(\mu)} \le \sup_{\substack{g\in\mathcal L^2(\mu)\\\left\|g\right\|_{L^2(\mu)}\le1\\g\ge0}}\left\|Ag\right\|_{L^2(\mu)}\tag1. $$ Now for arbitrary $f\in\mathcal L^2(\mu)$ with $f\ge0$ and $\left\|f\right\|_{L^2(\mu)}>0$, we apply the result to the scaled function $\tilde{f}=f/\left\|f\right\|_{L^2(\mu)}$ and deduce that $$ \langle A\frac{f}{\left\|f\right\|_{L^2(\mu)}},\frac{f}{\left\|f\right\|_{L^2(\mu)}}\rangle_{L^2(\mu)} =\langle A\tilde{f},\tilde{f}\rangle_{L^2(\mu)} \le \sup_{\substack{g\in\mathcal L^2(\mu)\\\left\|g\right\|_{L^2(\mu)}\le1\\g\ge0}}\left\|Ag\right\|_{L^2(\mu)}\tag1. , $$ which together with the linearity of $A$ gives us the desired inequality.

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  • $\begingroup$ Thank you for your answer. I'm sorry. This is trivial. I've actually tried to solve a slightly different problem: math.stackexchange.com/q/3467325/47771. Maybe you can take a look. $\endgroup$ – 0xbadf00d Dec 7 '19 at 20:30

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