Comparing two summable conditions











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Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:



(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;



(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;



Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).



That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!










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  • this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
    – Masacroso
    Nov 14 at 12:24










  • Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
    – Junnan
    2 days ago















up vote
3
down vote

favorite












Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:



(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;



(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;



Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).



That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!










share|cite|improve this question
























  • this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
    – Masacroso
    Nov 14 at 12:24










  • Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
    – Junnan
    2 days ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:



(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;



(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;



Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).



That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!










share|cite|improve this question















Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:



(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;



(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;



Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).



That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!







functional-analysis analysis banach-spaces examples-counterexamples






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edited 2 days ago

























asked Nov 14 at 11:34









Greywhite

655




655












  • this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
    – Masacroso
    Nov 14 at 12:24










  • Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
    – Junnan
    2 days ago


















  • this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
    – Masacroso
    Nov 14 at 12:24










  • Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
    – Junnan
    2 days ago
















this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24




this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24












Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago




Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago










2 Answers
2






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I thought of an example, hopefully is correct.



Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:



(1) $sum_{kgeq1}|a_k^n|<infty$;



(2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
$$

(Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);



(3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
$$

(Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
Now, write
$$
a_k^n=frac{1}{n^2+k^2}
$$



Point (1) is trivially true.
For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
$$
sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
$$

one has
$$
left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
$$

The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
$$

which is finite since $sum_{ngeq1}a_k^n<infty$.



Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
$$
sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
$$

I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.



I hope it works!



P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
$$
a(n,k)=frac{1}{n^2+k^2}
$$

whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.






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    up vote
    1
    down vote













    I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.



    I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
    $$
    $$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$



    That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$



    Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$



    Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$



    by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.



    Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$



    You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$



    and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.






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      2 Answers
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      2 Answers
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      up vote
      1
      down vote













      I thought of an example, hopefully is correct.



      Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:



      (1) $sum_{kgeq1}|a_k^n|<infty$;



      (2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
      $$
      sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
      $$

      (Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);



      (3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
      $$
      sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
      $$

      (Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
      Now, write
      $$
      a_k^n=frac{1}{n^2+k^2}
      $$



      Point (1) is trivially true.
      For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
      $$
      sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
      $$

      one has
      $$
      left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
      leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
      $$

      The second term is surely convergent. For the first, using Fubini's theorem, it equals
      $$
      sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
      $$

      which is finite since $sum_{ngeq1}a_k^n<infty$.



      Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
      $$
      sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
      $$

      I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.



      I hope it works!



      P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
      $$
      a(n,k)=frac{1}{n^2+k^2}
      $$

      whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.






      share|cite|improve this answer



























        up vote
        1
        down vote













        I thought of an example, hopefully is correct.



        Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:



        (1) $sum_{kgeq1}|a_k^n|<infty$;



        (2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
        $$
        sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
        $$

        (Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);



        (3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
        $$
        sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
        $$

        (Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
        Now, write
        $$
        a_k^n=frac{1}{n^2+k^2}
        $$



        Point (1) is trivially true.
        For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
        $$
        sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
        $$

        one has
        $$
        left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
        leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
        $$

        The second term is surely convergent. For the first, using Fubini's theorem, it equals
        $$
        sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
        $$

        which is finite since $sum_{ngeq1}a_k^n<infty$.



        Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
        $$
        sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
        $$

        I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.



        I hope it works!



        P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
        $$
        a(n,k)=frac{1}{n^2+k^2}
        $$

        whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.






        share|cite|improve this answer

























          up vote
          1
          down vote










          up vote
          1
          down vote









          I thought of an example, hopefully is correct.



          Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:



          (1) $sum_{kgeq1}|a_k^n|<infty$;



          (2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
          $$
          sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
          $$

          (Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);



          (3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
          $$
          sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
          $$

          (Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
          Now, write
          $$
          a_k^n=frac{1}{n^2+k^2}
          $$



          Point (1) is trivially true.
          For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
          $$
          sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
          $$

          one has
          $$
          left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
          leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
          $$

          The second term is surely convergent. For the first, using Fubini's theorem, it equals
          $$
          sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
          $$

          which is finite since $sum_{ngeq1}a_k^n<infty$.



          Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
          $$
          sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
          $$

          I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.



          I hope it works!



          P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
          $$
          a(n,k)=frac{1}{n^2+k^2}
          $$

          whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.






          share|cite|improve this answer














          I thought of an example, hopefully is correct.



          Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:



          (1) $sum_{kgeq1}|a_k^n|<infty$;



          (2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
          $$
          sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
          $$

          (Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);



          (3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
          $$
          sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
          $$

          (Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
          Now, write
          $$
          a_k^n=frac{1}{n^2+k^2}
          $$



          Point (1) is trivially true.
          For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
          $$
          sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
          $$

          one has
          $$
          left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
          leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
          $$

          The second term is surely convergent. For the first, using Fubini's theorem, it equals
          $$
          sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
          $$

          which is finite since $sum_{ngeq1}a_k^n<infty$.



          Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
          $$
          sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
          $$

          I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.



          I hope it works!



          P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
          $$
          a(n,k)=frac{1}{n^2+k^2}
          $$

          whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.







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          edited 2 days ago

























          answered 2 days ago









          Marco

          1909




          1909






















              up vote
              1
              down vote













              I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.



              I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
              $$
              $$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$



              That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$



              Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$



              Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$



              by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.



              Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$



              You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$



              and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.






              share|cite|improve this answer

























                up vote
                1
                down vote













                I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.



                I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
                $$
                $$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$



                That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$



                Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$



                Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$



                by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.



                Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$



                You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$



                and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.



                  I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
                  $$
                  $$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$



                  That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$



                  Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$



                  Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$



                  by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.



                  Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$



                  You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$



                  and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.






                  share|cite|improve this answer












                  I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.



                  I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
                  $$
                  $$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$



                  That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$



                  Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$



                  Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$



                  by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.



                  Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$



                  You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$



                  and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.







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                  answered yesterday









                  Greywhite

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