A question about Riesz spaces












0












$begingroup$


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 $x in E^+$.




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 $S wedge T = 0$







functional-analysis operator-theory vector-lattices







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$endgroup$












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
















0












$begingroup$


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 $x in E^+$.




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 $S wedge T = 0$







functional-analysis operator-theory vector-lattices







share|cite|improve this question











$endgroup$












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














0












0








0





$begingroup$


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 $x in E^+$.




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 $S wedge T = 0$







functional-analysis operator-theory vector-lattices







share|cite|improve this question











$endgroup$




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 $x in E^+$.




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 $S wedge T = 0$







functional-analysis operator-theory vector-lattices




functional-analysis operator-theory vector-lattices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 19:56







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asked Dec 29 '18 at 16:38









I-love-mathI-love-math

666




666












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


















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
















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




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










1 Answer
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$begingroup$

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






share|cite|improve this answer









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    $begingroup$

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






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

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






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

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






        share|cite|improve this answer









        $endgroup$



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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 17:05









        Ben WBen W

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