If a linear operator preserves positive functions, then it leaves some linear functional invariant I have the following question which I cannot seem to make any progress on:
Suppose that $T:C[0,1]\to C[0,1]$ is a linear operator satisfying that $Tf\geq 0$ whenever $f\geq 0$, and $T1=1$. Show that there exists a linear functional $\ell:C[0,1]$ satisfying $\ell(1)=1, \ell(f)\geq 0$ if $f\geq 0$ and $\ell(Tf)=\ell(f)$.
My biggest issue is with the last condition, I just don't see how we know enough about $T$ to construct something that satisfies this. An obvious candidate would be $T^{-1}$ but how do we know that even exists or is well defined?

Your positive operator $T$ is given by a Markov kernel, i.e., $Tf(x)=\int f(y)\,k(x,dy)$ where $k$ is a kernel of probability measures on $[0,1]$. The positive linear functional that you want is given by an invariant probability measure $\pi$, that is, a measure that satisfies $k(x,dy)\pi(dx)=\pi(dy).$ In this case, for $\ell(f):=\int f \,d\pi$ we have $\ell(Tf)=\ell(f)$.
The existence of such a $\pi$ is guaranteed by the Markov-Kakutani fixed point theorem, see What is a stationary measure? by Alex Furman.
