Let $T:L^2(\mu)\to L^2(\mu)$ be a linear and continuous operator, where $L^2(\mu)$ is the (real) $L^2$-space to some measure space $(\Omega,\Sigma,\mu)$.

$T$ is assumed to be sign-preserving in the sense that $$ v(x) \cdot (Tv)(x) \ge0 $$ for $\mu$-almost all $x\in \Omega$ and all $v\in L^2(\mu)$.

Does this imply that $T$ is a multiplication? That is, does there exist $\phi\in L^\infty(\mu)$ such that $Tv = u\cdot v$?

I could show the following property: $$ \chi_{A^c} \cdot (L\chi_A) = 0 $$ $\mu$-almost everywhere for all characteristic functions $\chi_A$ of $A\in\Sigma$. This would prove the question for $\mathbb R^n$ or $l^2(\mathbb N)$. I was not able to prove the question in the general case.

Let $T:L^2(\mu)\to L^2(\mu)$ be a linear and continuous operator, where $L^2(\mu)$ is the (real) $L^2$-space to some $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$.

$T$ is assumed to be sign-preserving in the sense that $$ v(x) \cdot (Tv)(x) \ge0 $$ for $\mu$-almost all $x\in \Omega$ and all $v\in L^2(\mu)$.

Does this imply that $T$ is a multiplication? That is, does there exist $\phi\in L^\infty(\mu)$ such that $Tv = u\cdot v$?

I could show the following property: $$ \chi_{A^c} \cdot (T\chi_A) = 0 $$ $\mu$-almost everywhere for all characteristic functions $\chi_A$ of $A\in\Sigma$. This would prove the question for $\mathbb R^n$ or $l^2(\mathbb N)$. I was not able to prove the question in the general case.