# Cauchy-Schwarz-like inequality for a self-adjoint nonnegativity-preserving operator on $L^2$

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.

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.