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 or vector larttice 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 $$x \vee y := \sup \{x,y\} \quad , \quad x \wedge y := \inf\{x ,y \} \quad , \quad |x|=x \vee (-x).$$
A Riesz space is called 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 $F$.
An operator $T:E \to F$ between two Riesz spaces is said to be regular if it can be written as a difference of two positive operators. Of course this is equivalent to saying that there exists a posiive operator $S:E \to F$ satisfying $T \le S$.
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 $x \in E^+$.
Now I want to prove the following exercise :
Consider the continuous function $g : [0,1] \to [0,1]$ defined by $g(x)=x$ if $0 \le x \le \frac{1}{2}$ and $g(x)=\frac{1}{2}$ if $\frac{1}{2} < x \le \frac{1}{2}$. Now define the operator $T:C[0,1] \to C[0,1]$ by $[Tf](x)=f(g(x))-f(\frac{1}{2})$. Show that $T$ is a regular operator whose modulus does not exist.