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Let ($X$,$A$,$\mu$) be a $\sigma$-finite measure space and $g$ $\in$ $L^{\infty}(\mu)$. Define $M_g$: $L^2(\mu)$ $\longrightarrow$ $L^2(\mu)$ by setting $M_g(f)$ = $fg$. Show that if $M_g$ a positive definite operator then $g$ $\geq$ $0$ a.e. on $X$. I have already shown that $M_g$ is self-adjoint if $g(x)$ $\in$ $\mathbb{R}$ for a.e $x$ $\in$ $X$. I have got upto $\int_X g|f|^2d\mu $ $\geq$ $0$ for each $f$ $\in$ $L^2(\mu)$. How do I conclude that $g$ $\geq$ $0$ a.e from here? Thanks for any help.

Ester
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Suppose that $g < 0$ on some set of non-zero measure $A$. Since $\mu$ is $\sigma$-finite, we can assume $\mu(A)$ to be finite. We have for the indicator function that $\chi_A \in L^2(\mu)$ and we also have $$ \int_X g \chi_A d\mu < 0, $$ which is a contradiction. So $g$ is positive almost everywhere.

Demophilus
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