A real vector space $E$ is said to be an ordered vector space whenever it is equipped with an order relation $\ge$ that is compatible with the algebraic structure of $E$.
A Riesz space is an ordered vector space $E$ which for each pair of vectors $x,y \in E$, the supremum and the infimum of the st $\{x,y\}$ both exist in $E$. Following the classical notation, we shall write $$x \vee y := \sup \{x,y\} \quad , \quad x \wedge y := \inf\{x ,y \} .$$ An example of Riesz space is function space $E$ of real valued functions on a set $\Omega$ such that for each pair $f , g \in E$ the functions $$[f \vee g](w) := \max \{f(w),g(w)\} \quad, \quad [f \wedge g](w) := \min\{f(w) ,g(w) \} $$ both belong to $E$.
A Riesz space is caled Dedekind complete whenever every nonempty bounded above subset has a supremum .
Here $\mathcal{L}_b(E,F)$ is the vector space of all order bounded operators from $E$ to $F$.
By "postive operator" book of "Charalambos D.Aliprantis and Owen Burkinshow" we have the following theorem
Theorem(F.Riesz-Kantorovich) . If $E$ and $F$ are Riesz spaces with $F$ Dedekind complete, thenthe ordered vector space $\mathcal{L}_b(E,F)$ is a Dedekind complete Riesz space with the lattice operations $$|T| = \sup\{|Ty| : |y|\le x \},$$ $$ [S \vee T](x)=\sup\{S(y)+T(z) : y,z \in E^+ , y+z=x\} ,$$ $$ [S \wedge T](x)=\sup\{S(y)+T(z) : y,z \in E^+ , y+z=x\}$$ for all $S,T \in \mathcal{L}_b(E,F)$ and $x \in E^+$.
I want to prove the following exercise
Let $E=C[0,1], F=\mathbb{R}$, and let $S,T : E \to F$ be defined by $S(f)=f(0)$ and $T(f)=f(1)$. Then $S \wedge T = 0$.
I think since $F= \mathbb{R}$ we have $(S \wedge T)(f) = S(f) \wedge T(f) = f(0) \wedge f(1) $. Then can I conclude that $(S \wedge T)(f) = f(0) \wedge f(1) = f(0) = 0 $ ?