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 $x \in E^+$.

Now By this theorem I want to prove the following exercise from the first section of this book:

Consider the positive operators $S,T : L_1[0,1] \to L_1[0,1]$ defind by $$S(f)=f \quad , \quad T(f)=[\int_0^1 f(x) dx].1$$ Then show that $S \wedge T = 0$

  • $\begingroup$ You have a typo in your theorem. It should be $(S\wedge T)(x)=\inf\{S(y)+T(z):y,z\in E^+,y+z=x\}$. $\endgroup$
    – Ben W
    Dec 29 '18 at 17:06

Fix $f\in L_1[0,1]$. For each $\epsilon>0$ find $\delta\in(0,1)$ so that if $A\subset[0,1]$ has measure $<\delta$ then $\|f\boldsymbol{1}_A\|_{L_1[0,1]}<\epsilon$. For each such $A$, write $$f=f\boldsymbol{1}_{[0,1]\setminus A}+f\boldsymbol{1}_A.$$ Observe that $$(S\wedge T)(f)\leq f\boldsymbol{1}_{[0,1]\setminus A}+\epsilon\boldsymbol{1}.$$ In particular, $(S\wedge T)(f)\leq\epsilon$ on $A$. Since $A$ is an arbitrary set of measure $<\delta$, we have $(S\wedge T)(f)\leq\epsilon$ on $[0,1]$. But $\epsilon>0$ was arbitrary too, so $(S\wedge T)(f)=0$.


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