# Is a sign-preserving operator on $L^2$ a multiplication?

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.

• So you are able to prove it for $\sigma$-finite spaces right? – Mizar Jul 29 at 8:49
• @Mizar I am able to prove this property regarding characteristic functions. I do not see, hot this implies that $T$ is a multiplication operator. – daw Jul 29 at 9:12
• OK, I added the proof in that case to my answer! – Mizar Jul 29 at 10:04

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.

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.

• Thanks for the nice answer! I was missing the trick $Tu = u\cdot T1$. – daw Jul 29 at 10:49
• How would one define $v$ in the non-finite case? Then $1\not\in L^2$. – daw Jul 29 at 10:59
• Interestingly, the assumption on $T$ already implies $T1\in L^\infty(\mu)$. – daw Jul 29 at 11:00
• Ah yes, $T1$ is not bounded a priori, but you discover it along the proof. I will edit my answer slightly. To deduce the $\sigma$-finite case, say $\Omega=\bigsqcup E_i$ (disjoint union) with $\mu(E_i)<\infty$ and declare $v$ to be $T1_{E_i}$ on $E_i$ (tell me if you have trouble seeing how to conclude from here). – Mizar Jul 29 at 11:08