Two implications of an operator that preserves positivity on L2 Suppose $(X,\mathcal{A},m)$ is a probability space. Let A be an operator on function spaces. I am currently taking a course in measure theory, and in the lecture slides the following two remarks are made.

*

*If $A: L_2(X) \to L_2(X)$ preserves positivity and $A1=1$, then $A:L_{\infty}\to L_{\infty}$ is a contraction.


*If $A:L_2(X) \to L_2(X)$ preserves positivity and also preserves an integral, then it extends to a contraction $A: L_1(X) \to L_1(X)$.
My question is why are the above statements true? In particular, the condition about preserving positivity seems difficult to use for me because what we want to show in both cases is a statement that can be expressed purely in terms of the norms of functions (in which case it doesn't matter whether the function is non-negative or not). The fact that we are trying to prove some kind of inequality involving norms also suggests some use of Holder's inequality, but I am struggling to fill in the details.
 A: *

*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.


*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.
