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.



Is a sign-preserving operator on $L^2$ a multiplication?
First, if $uv=0$ then $uTv=0$ a.e.: indeed, applying the hypothesis with the function $\epsilon u+\epsilon^{-1}v$ and evaluating on $\{u\neq 0\}$, we get
$$0\le \epsilon^2uTu+\epsilon^{-2}vTv+uTv+vTu=\epsilon^2 uTu+uTv$$
a.e. and deduce the claim (sending $\epsilon\to 0$). This gives the property $(\*)$ that you showed.


If the space is $\sigma$-finite, using $(\*)$ we can easily reduce to the case that the measure is finite. Now take $v:=T1$. We claim that $Tu=uv$ (a.e.) when $u$ is simple: since $T$ is linear, it suffices to show this when $u=1\_A$ is a characteristic function. But
$$v=T1=T1\_A+T1\_{A^c}$$
and $T1\_{A^c}$ vanishes (a.e.) on $A$, hence $T1\_A=v$ (a.e.) on $A$. Since $T1\_A$ vanishes (a.e.) on $A^c$, the claim follows. Taking $A:=\{|v|>\|T\|\}$, if $\mu(A)>0$ we see that $\|T1\_A\|\_{L^2}>\|T\|\|1\_A\|\_{L^2}$, contradiction. So $v\in L^\infty$. The statement now follows since simple functions are dense.




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It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).


Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form
$$\bigcup\Big(\{s\_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t\_j\}\Big),$$
where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum\_i a\_i 1\_{\{s\_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.


Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.


If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form
$$\bigcup(\{s\_i\}\times E\_i')\cup\bigcup(E\_j\times\{t\_j\})\cup (E\times E')\cup N,$$
where $E\_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s\_i\}$, $E\_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t\_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s\_i\}\Big)\times\Big(\bigcup\{t\_j\}\Big)$.
Once you have this, you can argue as before.


