Question about masa

Question

For finite measure $\mu$ we know $L^{\infty}(X,\mu)$ is masa in $B(L^2(X,\mu))$, for if the measure is replaced by infinite measure is the same result will be valid, what about $\sigma$-finite case. Please clarify.

Answer 1

Yes, when $\mu$ is $\sigma$-finite, $L^\infty(X,\mu)$ is a masa. The proof is basically an extension of the one for the finite case.

In the non $\sigma$-finite case, the result is not true. For any space $(X,\mu)$ that is not localizable (see this question and answer ), there exists $L\in L^1(\mu)^*$ that is not given by a function in $L^\infty$. For each $f\in L^2(\mu)$, consider the map $\alpha_f:L^2(\mu)\to \mathbb C$ given by $\alpha_f(g)=L(g\bar f)$. This is linear and bounded, so by the Riesz Representation Theorem there exists $T_f\in L^2(\mu)$ with $\alpha_f(g)=\int_X g\,\overline {T_f}\,d\mu$. Define $T:L^2(\mu)\to L^2(\mu)$ by $Tf=T_f$. This is clearly linear and bounded (since $\|T_f\|=\|\alpha_f\|\leq\|L\|\,\|f\|$).

For any $h\in L^\infty(\mu)$, $g\in L^2(\mu)$, $$ TM_hg=T(hg), $$ and $$ \int_X k\,\overline{T_{hg}}\,d\mu=\alpha_{hg}(k)=L(k\overline{hg}) =L((k\bar h)g)=\int_Xk\bar h\,\overline{T_g}\,d\mu. $$ Thus $T(hg)=h\,Tg$, so $$ TM_hg=T(hg)=h\,Tg=M_hTg. $$ Then $TM_h=M_hT$, and $T\in L^\infty(\mu)'$. But $T$ is not a multiplication operator; if it were, say $T=M_\phi$, then $Tf=\phi f$, and $$ L(g\bar f)=\alpha_f(g)=\int_X g\,\overline{\phi f }\,d\mu=\int_X(g\bar f)\,\bar\phi\,d\mu, $$ contradicting that $L$ is not given by a function in $L^\infty$.