Showing weak convergence in $sigma(L^p,L^{p'})$












1












$begingroup$


I am solving the following exercise.



enter image description here



I have followed the below solution for the first part, and it seems okay. Except for why (S2) and (S3) implies that $g_nto f$ almost everywhere. We should not get $g_nto f$ almost everywhere just because $f_nto f$ and $g_nintext{conv}{f_1,f_2,dots,f_n}$. Right? What am I missing here?



enter image description here



I am a bit unsure about how things change in the second part. I do not know where to start either. Any help is much appreciated. (The book in question is Brezis.)






functional-analysis weak-convergence







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$endgroup$












  • $begingroup$
    Is "conv" of an infinite set the closure of all the finite convex combinations or just the collection of the finite convex combinations?
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:27
















1












$begingroup$


I am solving the following exercise.



enter image description here



I have followed the below solution for the first part, and it seems okay. Except for why (S2) and (S3) implies that $g_nto f$ almost everywhere. We should not get $g_nto f$ almost everywhere just because $f_nto f$ and $g_nintext{conv}{f_1,f_2,dots,f_n}$. Right? What am I missing here?



enter image description here



I am a bit unsure about how things change in the second part. I do not know where to start either. Any help is much appreciated. (The book in question is Brezis.)






functional-analysis weak-convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is "conv" of an infinite set the closure of all the finite convex combinations or just the collection of the finite convex combinations?
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:27














1












1








1





$begingroup$


I am solving the following exercise.



enter image description here



I have followed the below solution for the first part, and it seems okay. Except for why (S2) and (S3) implies that $g_nto f$ almost everywhere. We should not get $g_nto f$ almost everywhere just because $f_nto f$ and $g_nintext{conv}{f_1,f_2,dots,f_n}$. Right? What am I missing here?



enter image description here



I am a bit unsure about how things change in the second part. I do not know where to start either. Any help is much appreciated. (The book in question is Brezis.)






functional-analysis weak-convergence







share|cite|improve this question









$endgroup$




I am solving the following exercise.



enter image description here



I have followed the below solution for the first part, and it seems okay. Except for why (S2) and (S3) implies that $g_nto f$ almost everywhere. We should not get $g_nto f$ almost everywhere just because $f_nto f$ and $g_nintext{conv}{f_1,f_2,dots,f_n}$. Right? What am I missing here?



enter image description here



I am a bit unsure about how things change in the second part. I do not know where to start either. Any help is much appreciated. (The book in question is Brezis.)






functional-analysis weak-convergence




functional-analysis weak-convergence






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asked Dec 9 '18 at 19:21









Logarithmic DerivativeLogarithmic Derivative

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  • $begingroup$
    Is "conv" of an infinite set the closure of all the finite convex combinations or just the collection of the finite convex combinations?
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:27


















  • $begingroup$
    Is "conv" of an infinite set the closure of all the finite convex combinations or just the collection of the finite convex combinations?
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:27
















$begingroup$
Is "conv" of an infinite set the closure of all the finite convex combinations or just the collection of the finite convex combinations?
$endgroup$
– Davide Giraudo
Dec 11 '18 at 9:27




$begingroup$
Is "conv" of an infinite set the closure of all the finite convex combinations or just the collection of the finite convex combinations?
$endgroup$
– Davide Giraudo
Dec 11 '18 at 9:27










1 Answer
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1












$begingroup$

Let $xin Omega$. Let $varepsilongt 0$ be fixed. Let $ngeqslant 1$: since $g_n(x)inoperatorname{conv}left{f_k(x),kgeqslant n right}$, there exists an integer $m=m(x,n)geqslant n$ and numbers $lambda_k=lambda_k(x,n)in left[0,1right]$ such that $sum_{k=n}^mlambda_k=1$ and
$$
leftlvert g_n(x)-sum_{k=n}^m lambda_kf_k(x)rightrvertlt varepsilon.
$$
Since $sum_{k=n}^mlambda_kf(x)=f(x)$, it follows that
begin{align}
leftlvert g_n(x)-f(x) rightrvert &=leftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x)+sum_{k=n}^m lambda_kf_k(x)
-lambda_kf(x) rightrvert \
&leqslantleftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x) rightrvert+leftlvert sum_{k=n}^m lambda_kf_k(x)
-lambda_kf(x) rightrvert\
&leqslantvarepsilon+ sum_{k=n}^m lambda_kleftlvert f_k(x)
- f(x) rightrvert\
&leqslant varepsilon+sup_{kgeqslant n}leftlvert f_k(x)
- f(x) rightrvert sum_{k=n}^m lambda_k.
end{align}
We got that for all $xinOmega$ and all positive $varepsilon$ and all integer $n$,
$$leftlvert g_n(x)-f(x) rightrvertleqslantvarepsilon+sup_{kgeqslant n}leftlvert f_k(x)
- f(x) rightrvert
$$
hence letting $varepsilonto 0$ gives
$$leftlvert g_n(x)-f(x) rightrvertleqslant sup_{kgeqslant n}leftlvert f_k(x)
- f(x) rightrvert,
$$
which is sufficient to derive the almost sure convergence of $(g_n)$ to $f$.






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    1 Answer
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    1 Answer
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    1












    $begingroup$

    Let $xin Omega$. Let $varepsilongt 0$ be fixed. Let $ngeqslant 1$: since $g_n(x)inoperatorname{conv}left{f_k(x),kgeqslant n right}$, there exists an integer $m=m(x,n)geqslant n$ and numbers $lambda_k=lambda_k(x,n)in left[0,1right]$ such that $sum_{k=n}^mlambda_k=1$ and
    $$
    leftlvert g_n(x)-sum_{k=n}^m lambda_kf_k(x)rightrvertlt varepsilon.
    $$
    Since $sum_{k=n}^mlambda_kf(x)=f(x)$, it follows that
    begin{align}
    leftlvert g_n(x)-f(x) rightrvert &=leftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x)+sum_{k=n}^m lambda_kf_k(x)
    -lambda_kf(x) rightrvert \
    &leqslantleftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x) rightrvert+leftlvert sum_{k=n}^m lambda_kf_k(x)
    -lambda_kf(x) rightrvert\
    &leqslantvarepsilon+ sum_{k=n}^m lambda_kleftlvert f_k(x)
    - f(x) rightrvert\
    &leqslant varepsilon+sup_{kgeqslant n}leftlvert f_k(x)
    - f(x) rightrvert sum_{k=n}^m lambda_k.
    end{align}
    We got that for all $xinOmega$ and all positive $varepsilon$ and all integer $n$,
    $$leftlvert g_n(x)-f(x) rightrvertleqslantvarepsilon+sup_{kgeqslant n}leftlvert f_k(x)
    - f(x) rightrvert
    $$
    hence letting $varepsilonto 0$ gives
    $$leftlvert g_n(x)-f(x) rightrvertleqslant sup_{kgeqslant n}leftlvert f_k(x)
    - f(x) rightrvert,
    $$
    which is sufficient to derive the almost sure convergence of $(g_n)$ to $f$.






    share|cite|improve this answer









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      1












      $begingroup$

      Let $xin Omega$. Let $varepsilongt 0$ be fixed. Let $ngeqslant 1$: since $g_n(x)inoperatorname{conv}left{f_k(x),kgeqslant n right}$, there exists an integer $m=m(x,n)geqslant n$ and numbers $lambda_k=lambda_k(x,n)in left[0,1right]$ such that $sum_{k=n}^mlambda_k=1$ and
      $$
      leftlvert g_n(x)-sum_{k=n}^m lambda_kf_k(x)rightrvertlt varepsilon.
      $$
      Since $sum_{k=n}^mlambda_kf(x)=f(x)$, it follows that
      begin{align}
      leftlvert g_n(x)-f(x) rightrvert &=leftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x)+sum_{k=n}^m lambda_kf_k(x)
      -lambda_kf(x) rightrvert \
      &leqslantleftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x) rightrvert+leftlvert sum_{k=n}^m lambda_kf_k(x)
      -lambda_kf(x) rightrvert\
      &leqslantvarepsilon+ sum_{k=n}^m lambda_kleftlvert f_k(x)
      - f(x) rightrvert\
      &leqslant varepsilon+sup_{kgeqslant n}leftlvert f_k(x)
      - f(x) rightrvert sum_{k=n}^m lambda_k.
      end{align}
      We got that for all $xinOmega$ and all positive $varepsilon$ and all integer $n$,
      $$leftlvert g_n(x)-f(x) rightrvertleqslantvarepsilon+sup_{kgeqslant n}leftlvert f_k(x)
      - f(x) rightrvert
      $$
      hence letting $varepsilonto 0$ gives
      $$leftlvert g_n(x)-f(x) rightrvertleqslant sup_{kgeqslant n}leftlvert f_k(x)
      - f(x) rightrvert,
      $$
      which is sufficient to derive the almost sure convergence of $(g_n)$ to $f$.






      share|cite|improve this answer









      $endgroup$
















        1












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        1





        $begingroup$

        Let $xin Omega$. Let $varepsilongt 0$ be fixed. Let $ngeqslant 1$: since $g_n(x)inoperatorname{conv}left{f_k(x),kgeqslant n right}$, there exists an integer $m=m(x,n)geqslant n$ and numbers $lambda_k=lambda_k(x,n)in left[0,1right]$ such that $sum_{k=n}^mlambda_k=1$ and
        $$
        leftlvert g_n(x)-sum_{k=n}^m lambda_kf_k(x)rightrvertlt varepsilon.
        $$
        Since $sum_{k=n}^mlambda_kf(x)=f(x)$, it follows that
        begin{align}
        leftlvert g_n(x)-f(x) rightrvert &=leftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x)+sum_{k=n}^m lambda_kf_k(x)
        -lambda_kf(x) rightrvert \
        &leqslantleftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x) rightrvert+leftlvert sum_{k=n}^m lambda_kf_k(x)
        -lambda_kf(x) rightrvert\
        &leqslantvarepsilon+ sum_{k=n}^m lambda_kleftlvert f_k(x)
        - f(x) rightrvert\
        &leqslant varepsilon+sup_{kgeqslant n}leftlvert f_k(x)
        - f(x) rightrvert sum_{k=n}^m lambda_k.
        end{align}
        We got that for all $xinOmega$ and all positive $varepsilon$ and all integer $n$,
        $$leftlvert g_n(x)-f(x) rightrvertleqslantvarepsilon+sup_{kgeqslant n}leftlvert f_k(x)
        - f(x) rightrvert
        $$
        hence letting $varepsilonto 0$ gives
        $$leftlvert g_n(x)-f(x) rightrvertleqslant sup_{kgeqslant n}leftlvert f_k(x)
        - f(x) rightrvert,
        $$
        which is sufficient to derive the almost sure convergence of $(g_n)$ to $f$.






        share|cite|improve this answer









        $endgroup$



        Let $xin Omega$. Let $varepsilongt 0$ be fixed. Let $ngeqslant 1$: since $g_n(x)inoperatorname{conv}left{f_k(x),kgeqslant n right}$, there exists an integer $m=m(x,n)geqslant n$ and numbers $lambda_k=lambda_k(x,n)in left[0,1right]$ such that $sum_{k=n}^mlambda_k=1$ and
        $$
        leftlvert g_n(x)-sum_{k=n}^m lambda_kf_k(x)rightrvertlt varepsilon.
        $$
        Since $sum_{k=n}^mlambda_kf(x)=f(x)$, it follows that
        begin{align}
        leftlvert g_n(x)-f(x) rightrvert &=leftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x)+sum_{k=n}^m lambda_kf_k(x)
        -lambda_kf(x) rightrvert \
        &leqslantleftlvert g_n(x)- sum_{k=n}^m lambda_kf_k(x) rightrvert+leftlvert sum_{k=n}^m lambda_kf_k(x)
        -lambda_kf(x) rightrvert\
        &leqslantvarepsilon+ sum_{k=n}^m lambda_kleftlvert f_k(x)
        - f(x) rightrvert\
        &leqslant varepsilon+sup_{kgeqslant n}leftlvert f_k(x)
        - f(x) rightrvert sum_{k=n}^m lambda_k.
        end{align}
        We got that for all $xinOmega$ and all positive $varepsilon$ and all integer $n$,
        $$leftlvert g_n(x)-f(x) rightrvertleqslantvarepsilon+sup_{kgeqslant n}leftlvert f_k(x)
        - f(x) rightrvert
        $$
        hence letting $varepsilonto 0$ gives
        $$leftlvert g_n(x)-f(x) rightrvertleqslant sup_{kgeqslant n}leftlvert f_k(x)
        - f(x) rightrvert,
        $$
        which is sufficient to derive the almost sure convergence of $(g_n)$ to $f$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 9:40









        Davide GiraudoDavide Giraudo

        127k17154268




        127k17154268






























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