Two implications of an operator that preserves positivity on L2

Solution 1:

1. Let $f\in L^\infty$ with $\|f\|_\infty=1$. Since $\operatorname{Re} f\leq1$, we have $$ 0\leq A(1-\operatorname{Re} f)=1-A(\operatorname{Re} f). $$ Similarly, $$ A(\operatorname{Im} f)\le 1. $$ This shows that $Af\in L^\infty$. At this stage I'm not sure if there is an elementary argument. The one I know is that, begin positive on an abelian C$^*$-algebra, $A$ is completely positive. And completely positive maps satisfy $\|A\|=\|A1\|$. So $\|A\|=1$ and it is a contraction.

2. Because $A$ preserves positivity, we can do the following. The adjoint $A^*$ of $A$ also maps $L^2(X)$ to $L^2(X)$. Now, since $A$ preserves the integral, $$ \int_X (A^*1)f=\langle A^*1,f\rangle=\langle 1,Af\rangle=\int_XAf=\int_Xf. $$ As this can be done for any $f$, it follows that $A^*1=1$. Also, since $A$ preserves positivity, so does $A^*$: if $g\geq0$ and $f\geq0$, then $$ \int_X (A^*g)\,f=\int_Xg\,Af\geq0. $$ Again as $f$ can be any non-negative function we get that $A^*g\geq0$. So $A^*$ is in the conditions of item 1, and this means that $A^*$ is an $L^\infty$ contraction.

If we write $|Af|=g\,Af$ with $|g|=1$, then $$ \int_X|Af|=\int_X(Af)\,g=\int_Xf\,Ag\leq\int_X|f|\,|Ag|\leq\int_X|f|, $$ and so $A$ is an $L^1$-contraction.