Positive semigroups and convex operator Let $\{Z(t)\}_{t\geq 0}$ be a strongly continuous positive semigroup on a Banach lattice $V$ (endowed with ordering $\leq$). Let $\phi:V\rightarrow V$ be a convex operator. I want to prove that $$\phi(Z(t)f)\leq Z(t)(\phi f),\quad f\in V,\quad t>0.$$
Can someone out there help me out?
 A: OK, I write my comment into an answer. A linear operator is a special case of a convex one and then I believe your statement does not hold without further conditions.
What I have in mind is in $X:=C_0(\mathbb{R})$ taking the left shift group as $Z(t)$ and as $\phi$ taking mirroring along the y-axis, i.e., $\phi f(s)=f(-s)$.
If you take now a bell-shaped curve like $f(s)=e^{-s^2}$, then $f$ is positive, $\phi f = f$, and $Z(t)(\phi f)$ has maximum at $s=-t$, and it is between 0 and 1 elsewhere.
On the other hand, $\phi(Z(t)f)$ has a maximum at $s=t$ and it is immediate that we cannot compare the two functions in the usual ordering.
Wolfram Alpha gives me these images.

ADDED 01/05/2014:
You can easily modify this example to the following: Let $\Gamma\subset\mathbb{C}$ be the unit circle, $Z(t)$ the usual rotation group and $\phi$ mirroring to the $x$-axis ($\phi(e^{is}) = e^{-is}$). Taking a nonconstant continuous function $f$ on the circle you can draw the same conclusions. Then you have an example in a $C(K)$ space such that $Z(t)1 = 1$.
ADDED 01/07/2014:
Let $f(z)=\Re z+1\geq 0$, then $(Z(t)f)(e^{is}) = f(e^{i(t+s)}) = \cos(t+s)+1$, hence $\phi(Z(t)f)(e^{is}) = \cos(t-s)+1$.
On the other hand, $\phi(f)=f$, and  $Z(t)(\phi f)(e^{is}) = (Z(t)f)(e^{is})=\cos(t+s)+1$. Hence, they are not comparable and you can produce a similar picture as above to demonstrate this.
