# A question about Riesz spaces

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 $$S \wedge T = 0$$

• You have a typo in your theorem. It should be $(S\wedge T)(x)=\inf\{S(y)+T(z):y,z\in E^+,y+z=x\}$. 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 $$(S\wedge T)(f)=0$$.