If $E$ is finite-dimensional and $mu$ finite show that $mathcal S(X,mu,E)$ is dense in $mathcal...












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If $E$ is finite-dimensional and $mu$ finite show that $mathcal S(X,mu,E)$ is dense in $mathcal L_infty(X,mu,E)$




This is an exercise of the book Analysis III of Amann and Escher. Here $mathcal S(X,mu,E)$ is the space of $mu$-simple functions and $mathcal L_infty(X,mu,E)$ is the space with the seminorm $|{cdot}|_infty$ of functions essentially bounded, with $E$ a finite-dimensional Banach space.



I give an attempt of a proof. Can you check the correctness of the proof or comment in my mistake? Thank you.




Proof:



Set $A_{j,n}:={xin X:|f(x)|in[|f|_infty j2^{-n},|f|_infty(j+1)2^{-n})}$ for $j=0,ldots, 2^n-1$. Then clearly each $A_{j,n}$ is measurable, $A_{j,n}cap A_{k,n}=emptyset$ when $jneq k$ and $X=bigcup_{j=0}^{2^n-1}A_{j,n}$.



Note that $f(A_{j,n})$ is relatively compact because $EcongBbb K^n$ (for $Bbb Kin{Bbb R,Bbb C}$) for some $ninBbb N$. Then there is a finite open cover of $f(A_{j,n})$ of open balls of arbitrarily small radius.



Then let $B_{1,n},ldots,B_{m,n}$ a finite open cover of open balls of radius $2^{-n-1}$ of $f(A_{j,n})$. Then we set $V_{k,n}:=(B_{k,n}setminus(bigcup_{h=1}^{k-1}B_{h,n}))cap f(A_{j,n})$, so the $V_{k,n}$ are a partition of borel sets of $f(A_{j,n})$, so clearly the sets $R_{k,n}:=f^{-1}(V_{k,n})cap A_{j,n}$ are measurable and a partition of $A_{j,n}$. Finally choose some $d_kin V_{k,n}$ for each $k$ and set $g_{j,n}(x):=sum_{k=1}^mchi_{R_{k,n}}(x)d_k$.



Then, by construction, $g_{j,n}$ is $mu$-simple and also is $g_n:=sum_{j=0}^{2^n-1}g_{j,n}$. More over, it have the property that $|f(x)-g_n(x)|<2^{-n}$ for each $xin X$, so we can conclude that $mathcal S(X,mu, E)​$ is dense in $mathcal L_infty(X,mu,E)$.$Box$






general-topology proof-verification lp-spaces







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    If $E$ is finite-dimensional and $mu$ finite show that $mathcal S(X,mu,E)$ is dense in $mathcal L_infty(X,mu,E)$




    This is an exercise of the book Analysis III of Amann and Escher. Here $mathcal S(X,mu,E)$ is the space of $mu$-simple functions and $mathcal L_infty(X,mu,E)$ is the space with the seminorm $|{cdot}|_infty$ of functions essentially bounded, with $E$ a finite-dimensional Banach space.



    I give an attempt of a proof. Can you check the correctness of the proof or comment in my mistake? Thank you.




    Proof:



    Set $A_{j,n}:={xin X:|f(x)|in[|f|_infty j2^{-n},|f|_infty(j+1)2^{-n})}$ for $j=0,ldots, 2^n-1$. Then clearly each $A_{j,n}$ is measurable, $A_{j,n}cap A_{k,n}=emptyset$ when $jneq k$ and $X=bigcup_{j=0}^{2^n-1}A_{j,n}$.



    Note that $f(A_{j,n})$ is relatively compact because $EcongBbb K^n$ (for $Bbb Kin{Bbb R,Bbb C}$) for some $ninBbb N$. Then there is a finite open cover of $f(A_{j,n})$ of open balls of arbitrarily small radius.



    Then let $B_{1,n},ldots,B_{m,n}$ a finite open cover of open balls of radius $2^{-n-1}$ of $f(A_{j,n})$. Then we set $V_{k,n}:=(B_{k,n}setminus(bigcup_{h=1}^{k-1}B_{h,n}))cap f(A_{j,n})$, so the $V_{k,n}$ are a partition of borel sets of $f(A_{j,n})$, so clearly the sets $R_{k,n}:=f^{-1}(V_{k,n})cap A_{j,n}$ are measurable and a partition of $A_{j,n}$. Finally choose some $d_kin V_{k,n}$ for each $k$ and set $g_{j,n}(x):=sum_{k=1}^mchi_{R_{k,n}}(x)d_k$.



    Then, by construction, $g_{j,n}$ is $mu$-simple and also is $g_n:=sum_{j=0}^{2^n-1}g_{j,n}$. More over, it have the property that $|f(x)-g_n(x)|<2^{-n}$ for each $xin X$, so we can conclude that $mathcal S(X,mu, E)​$ is dense in $mathcal L_infty(X,mu,E)$.$Box$






    general-topology proof-verification lp-spaces







    share|cite|improve this question











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      $begingroup$



      If $E$ is finite-dimensional and $mu$ finite show that $mathcal S(X,mu,E)$ is dense in $mathcal L_infty(X,mu,E)$




      This is an exercise of the book Analysis III of Amann and Escher. Here $mathcal S(X,mu,E)$ is the space of $mu$-simple functions and $mathcal L_infty(X,mu,E)$ is the space with the seminorm $|{cdot}|_infty$ of functions essentially bounded, with $E$ a finite-dimensional Banach space.



      I give an attempt of a proof. Can you check the correctness of the proof or comment in my mistake? Thank you.




      Proof:



      Set $A_{j,n}:={xin X:|f(x)|in[|f|_infty j2^{-n},|f|_infty(j+1)2^{-n})}$ for $j=0,ldots, 2^n-1$. Then clearly each $A_{j,n}$ is measurable, $A_{j,n}cap A_{k,n}=emptyset$ when $jneq k$ and $X=bigcup_{j=0}^{2^n-1}A_{j,n}$.



      Note that $f(A_{j,n})$ is relatively compact because $EcongBbb K^n$ (for $Bbb Kin{Bbb R,Bbb C}$) for some $ninBbb N$. Then there is a finite open cover of $f(A_{j,n})$ of open balls of arbitrarily small radius.



      Then let $B_{1,n},ldots,B_{m,n}$ a finite open cover of open balls of radius $2^{-n-1}$ of $f(A_{j,n})$. Then we set $V_{k,n}:=(B_{k,n}setminus(bigcup_{h=1}^{k-1}B_{h,n}))cap f(A_{j,n})$, so the $V_{k,n}$ are a partition of borel sets of $f(A_{j,n})$, so clearly the sets $R_{k,n}:=f^{-1}(V_{k,n})cap A_{j,n}$ are measurable and a partition of $A_{j,n}$. Finally choose some $d_kin V_{k,n}$ for each $k$ and set $g_{j,n}(x):=sum_{k=1}^mchi_{R_{k,n}}(x)d_k$.



      Then, by construction, $g_{j,n}$ is $mu$-simple and also is $g_n:=sum_{j=0}^{2^n-1}g_{j,n}$. More over, it have the property that $|f(x)-g_n(x)|<2^{-n}$ for each $xin X$, so we can conclude that $mathcal S(X,mu, E)​$ is dense in $mathcal L_infty(X,mu,E)$.$Box$






      general-topology proof-verification lp-spaces







      share|cite|improve this question











      $endgroup$





      If $E$ is finite-dimensional and $mu$ finite show that $mathcal S(X,mu,E)$ is dense in $mathcal L_infty(X,mu,E)$




      This is an exercise of the book Analysis III of Amann and Escher. Here $mathcal S(X,mu,E)$ is the space of $mu$-simple functions and $mathcal L_infty(X,mu,E)$ is the space with the seminorm $|{cdot}|_infty$ of functions essentially bounded, with $E$ a finite-dimensional Banach space.



      I give an attempt of a proof. Can you check the correctness of the proof or comment in my mistake? Thank you.




      Proof:



      Set $A_{j,n}:={xin X:|f(x)|in[|f|_infty j2^{-n},|f|_infty(j+1)2^{-n})}$ for $j=0,ldots, 2^n-1$. Then clearly each $A_{j,n}$ is measurable, $A_{j,n}cap A_{k,n}=emptyset$ when $jneq k$ and $X=bigcup_{j=0}^{2^n-1}A_{j,n}$.



      Note that $f(A_{j,n})$ is relatively compact because $EcongBbb K^n$ (for $Bbb Kin{Bbb R,Bbb C}$) for some $ninBbb N$. Then there is a finite open cover of $f(A_{j,n})$ of open balls of arbitrarily small radius.



      Then let $B_{1,n},ldots,B_{m,n}$ a finite open cover of open balls of radius $2^{-n-1}$ of $f(A_{j,n})$. Then we set $V_{k,n}:=(B_{k,n}setminus(bigcup_{h=1}^{k-1}B_{h,n}))cap f(A_{j,n})$, so the $V_{k,n}$ are a partition of borel sets of $f(A_{j,n})$, so clearly the sets $R_{k,n}:=f^{-1}(V_{k,n})cap A_{j,n}$ are measurable and a partition of $A_{j,n}$. Finally choose some $d_kin V_{k,n}$ for each $k$ and set $g_{j,n}(x):=sum_{k=1}^mchi_{R_{k,n}}(x)d_k$.



      Then, by construction, $g_{j,n}$ is $mu$-simple and also is $g_n:=sum_{j=0}^{2^n-1}g_{j,n}$. More over, it have the property that $|f(x)-g_n(x)|<2^{-n}$ for each $xin X$, so we can conclude that $mathcal S(X,mu, E)​$ is dense in $mathcal L_infty(X,mu,E)$.$Box$






      general-topology proof-verification lp-spaces




      general-topology proof-verification lp-spaces






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      edited Nov 30 '18 at 15:59







      Masacroso

















      asked Oct 28 '18 at 8:58









      MasacrosoMasacroso

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          You use the fact that E has finite dimension when you say ''$f(A_{j,n}$ is relatively compact " . In a finite-dimensional space, it is true that a bounded set is relatively compact because it's closure is bounded and closed thus compact. But in a infinite-dimensional space, a Riesz's theorem claims that all the balls are not compact. For this reason if $E$ is not finite-dimensional then $f(A_{j,n})$ is bounded but not anything more and you can't find a finite open cover.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            yes, I noticed and edited at the same time you write the answer
            $endgroup$
            – Masacroso
            Oct 28 '18 at 9:37











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          $begingroup$

          You use the fact that E has finite dimension when you say ''$f(A_{j,n}$ is relatively compact " . In a finite-dimensional space, it is true that a bounded set is relatively compact because it's closure is bounded and closed thus compact. But in a infinite-dimensional space, a Riesz's theorem claims that all the balls are not compact. For this reason if $E$ is not finite-dimensional then $f(A_{j,n})$ is bounded but not anything more and you can't find a finite open cover.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            yes, I noticed and edited at the same time you write the answer
            $endgroup$
            – Masacroso
            Oct 28 '18 at 9:37
















          1












          $begingroup$

          You use the fact that E has finite dimension when you say ''$f(A_{j,n}$ is relatively compact " . In a finite-dimensional space, it is true that a bounded set is relatively compact because it's closure is bounded and closed thus compact. But in a infinite-dimensional space, a Riesz's theorem claims that all the balls are not compact. For this reason if $E$ is not finite-dimensional then $f(A_{j,n})$ is bounded but not anything more and you can't find a finite open cover.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            yes, I noticed and edited at the same time you write the answer
            $endgroup$
            – Masacroso
            Oct 28 '18 at 9:37














          1












          1








          1





          $begingroup$

          You use the fact that E has finite dimension when you say ''$f(A_{j,n}$ is relatively compact " . In a finite-dimensional space, it is true that a bounded set is relatively compact because it's closure is bounded and closed thus compact. But in a infinite-dimensional space, a Riesz's theorem claims that all the balls are not compact. For this reason if $E$ is not finite-dimensional then $f(A_{j,n})$ is bounded but not anything more and you can't find a finite open cover.






          share|cite|improve this answer









          $endgroup$



          You use the fact that E has finite dimension when you say ''$f(A_{j,n}$ is relatively compact " . In a finite-dimensional space, it is true that a bounded set is relatively compact because it's closure is bounded and closed thus compact. But in a infinite-dimensional space, a Riesz's theorem claims that all the balls are not compact. For this reason if $E$ is not finite-dimensional then $f(A_{j,n})$ is bounded but not anything more and you can't find a finite open cover.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 28 '18 at 9:36









          Swann Swann

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          • $begingroup$
            yes, I noticed and edited at the same time you write the answer
            $endgroup$
            – Masacroso
            Oct 28 '18 at 9:37


















          • $begingroup$
            yes, I noticed and edited at the same time you write the answer
            $endgroup$
            – Masacroso
            Oct 28 '18 at 9:37
















          $begingroup$
          yes, I noticed and edited at the same time you write the answer
          $endgroup$
          – Masacroso
          Oct 28 '18 at 9:37




          $begingroup$
          yes, I noticed and edited at the same time you write the answer
          $endgroup$
          – Masacroso
          Oct 28 '18 at 9:37


















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