About the infimum of two operators in vector lattices
$begingroup$
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 $ ?
vector-lattices
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
$endgroup$
add a comment |
$begingroup$
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 $ ?
vector-lattices
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
$endgroup$
add a comment |
$begingroup$
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 $ ?
vector-lattices
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
$endgroup$
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 $ ?
vector-lattices
vector-lattices
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
asked Dec 30 '18 at 21:05
I-love-mathI-love-math
666
666
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