Non-unique Hahn Banach extension for a functional defined on a closed subspace of $l^1$












3












$begingroup$


I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems







share|cite|improve this question









$endgroup$












  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43


















3












$begingroup$


I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems







share|cite|improve this question









$endgroup$












  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43
















3












3








3





$begingroup$


I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems







share|cite|improve this question









$endgroup$




I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems




sequences-and-series functional-analysis duality-theorems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 16:27









VoBVoB

734413




734413












  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43




















  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43


















$begingroup$
Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:15






$begingroup$
Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:15














$begingroup$
I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
$endgroup$
– VoB
Dec 19 '18 at 17:43




$begingroup$
I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
$endgroup$
– VoB
Dec 19 '18 at 17:43












$begingroup$
So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
$endgroup$
– VoB
Dec 19 '18 at 17:45






$begingroup$
So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
$endgroup$
– VoB
Dec 19 '18 at 17:45














$begingroup$
You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
$endgroup$
– SmileyCraft
Dec 19 '18 at 18:22




$begingroup$
You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
$endgroup$
– SmileyCraft
Dec 19 '18 at 18:22












$begingroup$
Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
$endgroup$
– VoB
Dec 19 '18 at 18:43






$begingroup$
Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
$endgroup$
– VoB
Dec 19 '18 at 18:43












1 Answer
1






active

oldest

votes


















1












$begingroup$

Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
$$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
$$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



Now we have
$$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



so $|Phi e_{2015}| le |T|$.



We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
    $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
    so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
    $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
    so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



    Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



    Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



    Now we have
    $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



    so $|Phi e_{2015}| le |T|$.



    We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
      $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
      so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
      $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
      so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



      Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



      Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



      Now we have
      $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



      so $|Phi e_{2015}| le |T|$.



      We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
        $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
        so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
        $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
        so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



        Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



        Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



        Now we have
        $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



        so $|Phi e_{2015}| le |T|$.



        We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






        share|cite|improve this answer









        $endgroup$



        Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
        $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
        so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
        $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
        so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



        Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



        Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



        Now we have
        $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



        so $|Phi e_{2015}| le |T|$.



        We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 20:53









        mechanodroidmechanodroid

        27.6k62447




        27.6k62447






























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