Proof that the following map $Phi:ell^1to(ell^infty)'$ is not surjective












2












$begingroup$


I am working on the dual spaces of sequence spaces, and I want to show that the map $$
Phi:ell^1to(ell^infty)',qquad(Phi y)(x)=sum_{iinmathbb{N}}y_ix_i
$$



is not surjective. I have already shown it is a linear isometry. Can I use the Hahn-Banach theorem to find $y'in (ell^infty)'$ such that there is no $yinell^1$ with $Phi y=y'$.



I have written my own answer the following way (corollary 4.14 is a corollary in my lecture notes stating that $X$ is seperable if $X'$ is. I have proven earlier that $ell^infty$ is inseperable)



Proposition 2: The map $$Phi_infty:ell^1to(ell^infty)',qquad(Phi_infty y)(x)=sum_{iinmathbb{N}}x^iy^i$$ is not surjective.



The proof is based on the following lemma, which consists of two parts.
Lemma 1. Let $X$ and $Y$ be normed spaces. Then the following claims are true:




  1. If $f:Xto Y$ is a surjective linear isometry, then $f$ is a homeomorphism.


  2. Let $f: Xto Y$ be a homeomorphism. If $X$ is a seperable space, then $Y$ is a seperable space.



Proof of 1: We remark that every isometry is automatically injective. Since $f$ is also surjective, $f$ is a bijection. Now, we calculate
$|f|_{op}=sup{||f(x)||_Y,big|,||x||_Xleq 1}=sup{||x||_X,big|,||x||_Xleq 1 }=1<infty$, hence $f$ is bounded. Since $f$ is linear, $f$ is continuous. We will now show that $f^{-1}$ is continuous. We remark that $f^{-1}$ is linear. We also remark that for all $yin Y$, $||y||_Y=||f(f^{-1}(y))||_Y=||f^{-1}(y)||_X$, hence $f^{-1}$ is an isometry, too. Then, $|f^{-1}|_{op}=sup{||f^{-1}(y)||_X,big|,||y||_Yleq 1}=sup{||y||_Y,big|,||y||_Yleq 1}=1<infty$, hence $f^{-1}$ is bounded. It follows that $f^{-1}$ is continuous, hence $f$ is a homeomorphism.

Proof of 2: Let $X$ be a seperable normed space and let $f:Xto Y$ be a homeomorphism. Since $X$ is seperable, there exists a countable dense subset $Asubseteq X$. Then, $f(A)$ is a countable subset of $Y$. We will show that $f(A)$ is dense in $Y$.

Let $V$ be an open set in $Y$. By continuity of $f$, $f^{-1}(V)$ is open in $X$, so $Acap f^{-1}(V)neqemptyset$. By bijectivity, we see that
$emptysetneq f(Acap f^{-1}(V))=f(A)cap f(f^{-1}(V))=f(A)cap V$. This holds for all open sets in $Y$, hence $f(A)$ is dense in $Y$. It follows that $Y$ is seperable.



We can now prove proposition 2.



By theorem 4.6, $Phi_infty$ is a well-defined linear isometry. We remark that $ell^1$ is seperable. It follows by corollary 4.14 that $(ell^infty)'$ is inseperable since $ell^infty$ is inseperable.

We will give a proof by contradiction. Suppose that $Phi_infty$ is surjective. Then, $Phi_infty$ is a surjective linear isometry and by lemma 1 part textit{i}, a homeomorphism. Since $ell^{1}$ is seperable, it follows by lemma 1 part textit{ii} that $Phi_infty(ell^{1})=(ell^infty)'$ is seperable, but this is a contradiction since we know by corollary 4.14 that $(ell^infty)'$ is inseperable. Therefore, our assumption that $Phi_infty$ was surjective is false, hence $Phi_infty$ is not surjective










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The standard approach here is to prove that $ell^1$ is separable, whereas $ell^infty$ and $(ell^infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $ell^infty$ coming from $ell^1$ are just evaluations at a point, but there must be more functionals because $mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology.
    $endgroup$
    – Ashwin Trisal
    Jan 14 at 7:48












  • $begingroup$
    Related: Dual of $l^infty$ is not $l^1$
    $endgroup$
    – Martin Sleziak
    Jan 19 at 1:34
















2












$begingroup$


I am working on the dual spaces of sequence spaces, and I want to show that the map $$
Phi:ell^1to(ell^infty)',qquad(Phi y)(x)=sum_{iinmathbb{N}}y_ix_i
$$



is not surjective. I have already shown it is a linear isometry. Can I use the Hahn-Banach theorem to find $y'in (ell^infty)'$ such that there is no $yinell^1$ with $Phi y=y'$.



I have written my own answer the following way (corollary 4.14 is a corollary in my lecture notes stating that $X$ is seperable if $X'$ is. I have proven earlier that $ell^infty$ is inseperable)



Proposition 2: The map $$Phi_infty:ell^1to(ell^infty)',qquad(Phi_infty y)(x)=sum_{iinmathbb{N}}x^iy^i$$ is not surjective.



The proof is based on the following lemma, which consists of two parts.
Lemma 1. Let $X$ and $Y$ be normed spaces. Then the following claims are true:




  1. If $f:Xto Y$ is a surjective linear isometry, then $f$ is a homeomorphism.


  2. Let $f: Xto Y$ be a homeomorphism. If $X$ is a seperable space, then $Y$ is a seperable space.



Proof of 1: We remark that every isometry is automatically injective. Since $f$ is also surjective, $f$ is a bijection. Now, we calculate
$|f|_{op}=sup{||f(x)||_Y,big|,||x||_Xleq 1}=sup{||x||_X,big|,||x||_Xleq 1 }=1<infty$, hence $f$ is bounded. Since $f$ is linear, $f$ is continuous. We will now show that $f^{-1}$ is continuous. We remark that $f^{-1}$ is linear. We also remark that for all $yin Y$, $||y||_Y=||f(f^{-1}(y))||_Y=||f^{-1}(y)||_X$, hence $f^{-1}$ is an isometry, too. Then, $|f^{-1}|_{op}=sup{||f^{-1}(y)||_X,big|,||y||_Yleq 1}=sup{||y||_Y,big|,||y||_Yleq 1}=1<infty$, hence $f^{-1}$ is bounded. It follows that $f^{-1}$ is continuous, hence $f$ is a homeomorphism.

Proof of 2: Let $X$ be a seperable normed space and let $f:Xto Y$ be a homeomorphism. Since $X$ is seperable, there exists a countable dense subset $Asubseteq X$. Then, $f(A)$ is a countable subset of $Y$. We will show that $f(A)$ is dense in $Y$.

Let $V$ be an open set in $Y$. By continuity of $f$, $f^{-1}(V)$ is open in $X$, so $Acap f^{-1}(V)neqemptyset$. By bijectivity, we see that
$emptysetneq f(Acap f^{-1}(V))=f(A)cap f(f^{-1}(V))=f(A)cap V$. This holds for all open sets in $Y$, hence $f(A)$ is dense in $Y$. It follows that $Y$ is seperable.



We can now prove proposition 2.



By theorem 4.6, $Phi_infty$ is a well-defined linear isometry. We remark that $ell^1$ is seperable. It follows by corollary 4.14 that $(ell^infty)'$ is inseperable since $ell^infty$ is inseperable.

We will give a proof by contradiction. Suppose that $Phi_infty$ is surjective. Then, $Phi_infty$ is a surjective linear isometry and by lemma 1 part textit{i}, a homeomorphism. Since $ell^{1}$ is seperable, it follows by lemma 1 part textit{ii} that $Phi_infty(ell^{1})=(ell^infty)'$ is seperable, but this is a contradiction since we know by corollary 4.14 that $(ell^infty)'$ is inseperable. Therefore, our assumption that $Phi_infty$ was surjective is false, hence $Phi_infty$ is not surjective










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The standard approach here is to prove that $ell^1$ is separable, whereas $ell^infty$ and $(ell^infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $ell^infty$ coming from $ell^1$ are just evaluations at a point, but there must be more functionals because $mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology.
    $endgroup$
    – Ashwin Trisal
    Jan 14 at 7:48












  • $begingroup$
    Related: Dual of $l^infty$ is not $l^1$
    $endgroup$
    – Martin Sleziak
    Jan 19 at 1:34














2












2








2





$begingroup$


I am working on the dual spaces of sequence spaces, and I want to show that the map $$
Phi:ell^1to(ell^infty)',qquad(Phi y)(x)=sum_{iinmathbb{N}}y_ix_i
$$



is not surjective. I have already shown it is a linear isometry. Can I use the Hahn-Banach theorem to find $y'in (ell^infty)'$ such that there is no $yinell^1$ with $Phi y=y'$.



I have written my own answer the following way (corollary 4.14 is a corollary in my lecture notes stating that $X$ is seperable if $X'$ is. I have proven earlier that $ell^infty$ is inseperable)



Proposition 2: The map $$Phi_infty:ell^1to(ell^infty)',qquad(Phi_infty y)(x)=sum_{iinmathbb{N}}x^iy^i$$ is not surjective.



The proof is based on the following lemma, which consists of two parts.
Lemma 1. Let $X$ and $Y$ be normed spaces. Then the following claims are true:




  1. If $f:Xto Y$ is a surjective linear isometry, then $f$ is a homeomorphism.


  2. Let $f: Xto Y$ be a homeomorphism. If $X$ is a seperable space, then $Y$ is a seperable space.



Proof of 1: We remark that every isometry is automatically injective. Since $f$ is also surjective, $f$ is a bijection. Now, we calculate
$|f|_{op}=sup{||f(x)||_Y,big|,||x||_Xleq 1}=sup{||x||_X,big|,||x||_Xleq 1 }=1<infty$, hence $f$ is bounded. Since $f$ is linear, $f$ is continuous. We will now show that $f^{-1}$ is continuous. We remark that $f^{-1}$ is linear. We also remark that for all $yin Y$, $||y||_Y=||f(f^{-1}(y))||_Y=||f^{-1}(y)||_X$, hence $f^{-1}$ is an isometry, too. Then, $|f^{-1}|_{op}=sup{||f^{-1}(y)||_X,big|,||y||_Yleq 1}=sup{||y||_Y,big|,||y||_Yleq 1}=1<infty$, hence $f^{-1}$ is bounded. It follows that $f^{-1}$ is continuous, hence $f$ is a homeomorphism.

Proof of 2: Let $X$ be a seperable normed space and let $f:Xto Y$ be a homeomorphism. Since $X$ is seperable, there exists a countable dense subset $Asubseteq X$. Then, $f(A)$ is a countable subset of $Y$. We will show that $f(A)$ is dense in $Y$.

Let $V$ be an open set in $Y$. By continuity of $f$, $f^{-1}(V)$ is open in $X$, so $Acap f^{-1}(V)neqemptyset$. By bijectivity, we see that
$emptysetneq f(Acap f^{-1}(V))=f(A)cap f(f^{-1}(V))=f(A)cap V$. This holds for all open sets in $Y$, hence $f(A)$ is dense in $Y$. It follows that $Y$ is seperable.



We can now prove proposition 2.



By theorem 4.6, $Phi_infty$ is a well-defined linear isometry. We remark that $ell^1$ is seperable. It follows by corollary 4.14 that $(ell^infty)'$ is inseperable since $ell^infty$ is inseperable.

We will give a proof by contradiction. Suppose that $Phi_infty$ is surjective. Then, $Phi_infty$ is a surjective linear isometry and by lemma 1 part textit{i}, a homeomorphism. Since $ell^{1}$ is seperable, it follows by lemma 1 part textit{ii} that $Phi_infty(ell^{1})=(ell^infty)'$ is seperable, but this is a contradiction since we know by corollary 4.14 that $(ell^infty)'$ is inseperable. Therefore, our assumption that $Phi_infty$ was surjective is false, hence $Phi_infty$ is not surjective










share|cite|improve this question











$endgroup$




I am working on the dual spaces of sequence spaces, and I want to show that the map $$
Phi:ell^1to(ell^infty)',qquad(Phi y)(x)=sum_{iinmathbb{N}}y_ix_i
$$



is not surjective. I have already shown it is a linear isometry. Can I use the Hahn-Banach theorem to find $y'in (ell^infty)'$ such that there is no $yinell^1$ with $Phi y=y'$.



I have written my own answer the following way (corollary 4.14 is a corollary in my lecture notes stating that $X$ is seperable if $X'$ is. I have proven earlier that $ell^infty$ is inseperable)



Proposition 2: The map $$Phi_infty:ell^1to(ell^infty)',qquad(Phi_infty y)(x)=sum_{iinmathbb{N}}x^iy^i$$ is not surjective.



The proof is based on the following lemma, which consists of two parts.
Lemma 1. Let $X$ and $Y$ be normed spaces. Then the following claims are true:




  1. If $f:Xto Y$ is a surjective linear isometry, then $f$ is a homeomorphism.


  2. Let $f: Xto Y$ be a homeomorphism. If $X$ is a seperable space, then $Y$ is a seperable space.



Proof of 1: We remark that every isometry is automatically injective. Since $f$ is also surjective, $f$ is a bijection. Now, we calculate
$|f|_{op}=sup{||f(x)||_Y,big|,||x||_Xleq 1}=sup{||x||_X,big|,||x||_Xleq 1 }=1<infty$, hence $f$ is bounded. Since $f$ is linear, $f$ is continuous. We will now show that $f^{-1}$ is continuous. We remark that $f^{-1}$ is linear. We also remark that for all $yin Y$, $||y||_Y=||f(f^{-1}(y))||_Y=||f^{-1}(y)||_X$, hence $f^{-1}$ is an isometry, too. Then, $|f^{-1}|_{op}=sup{||f^{-1}(y)||_X,big|,||y||_Yleq 1}=sup{||y||_Y,big|,||y||_Yleq 1}=1<infty$, hence $f^{-1}$ is bounded. It follows that $f^{-1}$ is continuous, hence $f$ is a homeomorphism.

Proof of 2: Let $X$ be a seperable normed space and let $f:Xto Y$ be a homeomorphism. Since $X$ is seperable, there exists a countable dense subset $Asubseteq X$. Then, $f(A)$ is a countable subset of $Y$. We will show that $f(A)$ is dense in $Y$.

Let $V$ be an open set in $Y$. By continuity of $f$, $f^{-1}(V)$ is open in $X$, so $Acap f^{-1}(V)neqemptyset$. By bijectivity, we see that
$emptysetneq f(Acap f^{-1}(V))=f(A)cap f(f^{-1}(V))=f(A)cap V$. This holds for all open sets in $Y$, hence $f(A)$ is dense in $Y$. It follows that $Y$ is seperable.



We can now prove proposition 2.



By theorem 4.6, $Phi_infty$ is a well-defined linear isometry. We remark that $ell^1$ is seperable. It follows by corollary 4.14 that $(ell^infty)'$ is inseperable since $ell^infty$ is inseperable.

We will give a proof by contradiction. Suppose that $Phi_infty$ is surjective. Then, $Phi_infty$ is a surjective linear isometry and by lemma 1 part textit{i}, a homeomorphism. Since $ell^{1}$ is seperable, it follows by lemma 1 part textit{ii} that $Phi_infty(ell^{1})=(ell^infty)'$ is seperable, but this is a contradiction since we know by corollary 4.14 that $(ell^infty)'$ is inseperable. Therefore, our assumption that $Phi_infty$ was surjective is false, hence $Phi_infty$ is not surjective







sequences-and-series functional-analysis linear-transformations dual-spaces separable-spaces






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edited Jan 19 at 1:36









Martin Sleziak

44.9k10122275




44.9k10122275










asked Jan 14 at 7:44









JamesJames

956318




956318








  • 1




    $begingroup$
    The standard approach here is to prove that $ell^1$ is separable, whereas $ell^infty$ and $(ell^infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $ell^infty$ coming from $ell^1$ are just evaluations at a point, but there must be more functionals because $mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology.
    $endgroup$
    – Ashwin Trisal
    Jan 14 at 7:48












  • $begingroup$
    Related: Dual of $l^infty$ is not $l^1$
    $endgroup$
    – Martin Sleziak
    Jan 19 at 1:34














  • 1




    $begingroup$
    The standard approach here is to prove that $ell^1$ is separable, whereas $ell^infty$ and $(ell^infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $ell^infty$ coming from $ell^1$ are just evaluations at a point, but there must be more functionals because $mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology.
    $endgroup$
    – Ashwin Trisal
    Jan 14 at 7:48












  • $begingroup$
    Related: Dual of $l^infty$ is not $l^1$
    $endgroup$
    – Martin Sleziak
    Jan 19 at 1:34








1




1




$begingroup$
The standard approach here is to prove that $ell^1$ is separable, whereas $ell^infty$ and $(ell^infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $ell^infty$ coming from $ell^1$ are just evaluations at a point, but there must be more functionals because $mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology.
$endgroup$
– Ashwin Trisal
Jan 14 at 7:48






$begingroup$
The standard approach here is to prove that $ell^1$ is separable, whereas $ell^infty$ and $(ell^infty)^*$ are not. There is a very over-the-top argument where you show that the multiplicative linear functionals on $ell^infty$ coming from $ell^1$ are just evaluations at a point, but there must be more functionals because $mathbb N$ is not compact, but the pure states of a von Neumann algebra must be compact in the weak star topology.
$endgroup$
– Ashwin Trisal
Jan 14 at 7:48














$begingroup$
Related: Dual of $l^infty$ is not $l^1$
$endgroup$
– Martin Sleziak
Jan 19 at 1:34




$begingroup$
Related: Dual of $l^infty$ is not $l^1$
$endgroup$
– Martin Sleziak
Jan 19 at 1:34










3 Answers
3






active

oldest

votes


















3












$begingroup$

One way of showing this is to use the fact that If $X^{*}$ is separable then so is $X$. Take $X=ell^{infty}$. If the given map is surjective then $X^{*}$ is isometrically isomorphic to $ell^{1}$ which makes it separable. But $X=ell^{infty}$ is not separable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
    $endgroup$
    – James
    Jan 14 at 7:54












  • $begingroup$
    Is an isometric isomorphism naturally a homeomorphism?
    $endgroup$
    – James
    Jan 14 at 8:23






  • 1




    $begingroup$
    @James Of course, it is.
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 8:24



















5












$begingroup$

Here is a different approach. Consider the subspace
$$c := { x in ell^infty mid lim_{ntoinfty}x_n text{ exists}}.$$



Can you imagine a functional on $c$ which (if extended to all of $ell^infty$ via Hahn-Banach) is not of the form $Phi y$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
    $endgroup$
    – GEdgar
    Jan 14 at 11:58










  • $begingroup$
    Which functional is it?
    $endgroup$
    – James
    Jan 14 at 12:12










  • $begingroup$
    The functional is the one assumed to exist in the definition of $c$.
    $endgroup$
    – GEdgar
    Jan 15 at 15:37



















1












$begingroup$

Similarly to gerw's answer. Just use that, as a $C^ast$-algebra $ell^infty(mathbb N)$ is isomorphic to $C(beta mathbb{N})$, the continuous function over the compact space given by $beta mathbb{N}$, the Stone-Cech compactification of $mathbb{N}$. By Riesz theorem $(ell^infty)^ast = M(beta mathbb{N})$, the finite Radon measures over $beta mathbb{N}$. The points in $beta mathbb{N} setminus mathbb{N}$ are proper ultrafiters and evaluating on them gives functionals that are not in the image of $ell^1$.






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    3 Answers
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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    One way of showing this is to use the fact that If $X^{*}$ is separable then so is $X$. Take $X=ell^{infty}$. If the given map is surjective then $X^{*}$ is isometrically isomorphic to $ell^{1}$ which makes it separable. But $X=ell^{infty}$ is not separable.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
      $endgroup$
      – James
      Jan 14 at 7:54












    • $begingroup$
      Is an isometric isomorphism naturally a homeomorphism?
      $endgroup$
      – James
      Jan 14 at 8:23






    • 1




      $begingroup$
      @James Of course, it is.
      $endgroup$
      – Kavi Rama Murthy
      Jan 14 at 8:24
















    3












    $begingroup$

    One way of showing this is to use the fact that If $X^{*}$ is separable then so is $X$. Take $X=ell^{infty}$. If the given map is surjective then $X^{*}$ is isometrically isomorphic to $ell^{1}$ which makes it separable. But $X=ell^{infty}$ is not separable.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
      $endgroup$
      – James
      Jan 14 at 7:54












    • $begingroup$
      Is an isometric isomorphism naturally a homeomorphism?
      $endgroup$
      – James
      Jan 14 at 8:23






    • 1




      $begingroup$
      @James Of course, it is.
      $endgroup$
      – Kavi Rama Murthy
      Jan 14 at 8:24














    3












    3








    3





    $begingroup$

    One way of showing this is to use the fact that If $X^{*}$ is separable then so is $X$. Take $X=ell^{infty}$. If the given map is surjective then $X^{*}$ is isometrically isomorphic to $ell^{1}$ which makes it separable. But $X=ell^{infty}$ is not separable.






    share|cite|improve this answer









    $endgroup$



    One way of showing this is to use the fact that If $X^{*}$ is separable then so is $X$. Take $X=ell^{infty}$. If the given map is surjective then $X^{*}$ is isometrically isomorphic to $ell^{1}$ which makes it separable. But $X=ell^{infty}$ is not separable.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 14 at 7:50









    Kavi Rama MurthyKavi Rama Murthy

    69.1k53169




    69.1k53169












    • $begingroup$
      I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
      $endgroup$
      – James
      Jan 14 at 7:54












    • $begingroup$
      Is an isometric isomorphism naturally a homeomorphism?
      $endgroup$
      – James
      Jan 14 at 8:23






    • 1




      $begingroup$
      @James Of course, it is.
      $endgroup$
      – Kavi Rama Murthy
      Jan 14 at 8:24


















    • $begingroup$
      I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
      $endgroup$
      – James
      Jan 14 at 7:54












    • $begingroup$
      Is an isometric isomorphism naturally a homeomorphism?
      $endgroup$
      – James
      Jan 14 at 8:23






    • 1




      $begingroup$
      @James Of course, it is.
      $endgroup$
      – Kavi Rama Murthy
      Jan 14 at 8:24
















    $begingroup$
    I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
    $endgroup$
    – James
    Jan 14 at 7:54






    $begingroup$
    I have already proven that $X$ is seperable if $X'$ is seperable, sosince $ell^infty$ is not seperable $(ell^infty)^*$ is not seperable.
    $endgroup$
    – James
    Jan 14 at 7:54














    $begingroup$
    Is an isometric isomorphism naturally a homeomorphism?
    $endgroup$
    – James
    Jan 14 at 8:23




    $begingroup$
    Is an isometric isomorphism naturally a homeomorphism?
    $endgroup$
    – James
    Jan 14 at 8:23




    1




    1




    $begingroup$
    @James Of course, it is.
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 8:24




    $begingroup$
    @James Of course, it is.
    $endgroup$
    – Kavi Rama Murthy
    Jan 14 at 8:24











    5












    $begingroup$

    Here is a different approach. Consider the subspace
    $$c := { x in ell^infty mid lim_{ntoinfty}x_n text{ exists}}.$$



    Can you imagine a functional on $c$ which (if extended to all of $ell^infty$ via Hahn-Banach) is not of the form $Phi y$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
      $endgroup$
      – GEdgar
      Jan 14 at 11:58










    • $begingroup$
      Which functional is it?
      $endgroup$
      – James
      Jan 14 at 12:12










    • $begingroup$
      The functional is the one assumed to exist in the definition of $c$.
      $endgroup$
      – GEdgar
      Jan 15 at 15:37
















    5












    $begingroup$

    Here is a different approach. Consider the subspace
    $$c := { x in ell^infty mid lim_{ntoinfty}x_n text{ exists}}.$$



    Can you imagine a functional on $c$ which (if extended to all of $ell^infty$ via Hahn-Banach) is not of the form $Phi y$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
      $endgroup$
      – GEdgar
      Jan 14 at 11:58










    • $begingroup$
      Which functional is it?
      $endgroup$
      – James
      Jan 14 at 12:12










    • $begingroup$
      The functional is the one assumed to exist in the definition of $c$.
      $endgroup$
      – GEdgar
      Jan 15 at 15:37














    5












    5








    5





    $begingroup$

    Here is a different approach. Consider the subspace
    $$c := { x in ell^infty mid lim_{ntoinfty}x_n text{ exists}}.$$



    Can you imagine a functional on $c$ which (if extended to all of $ell^infty$ via Hahn-Banach) is not of the form $Phi y$?






    share|cite|improve this answer









    $endgroup$



    Here is a different approach. Consider the subspace
    $$c := { x in ell^infty mid lim_{ntoinfty}x_n text{ exists}}.$$



    Can you imagine a functional on $c$ which (if extended to all of $ell^infty$ via Hahn-Banach) is not of the form $Phi y$?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 14 at 10:48









    gerwgerw

    19.7k11334




    19.7k11334












    • $begingroup$
      I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
      $endgroup$
      – GEdgar
      Jan 14 at 11:58










    • $begingroup$
      Which functional is it?
      $endgroup$
      – James
      Jan 14 at 12:12










    • $begingroup$
      The functional is the one assumed to exist in the definition of $c$.
      $endgroup$
      – GEdgar
      Jan 15 at 15:37


















    • $begingroup$
      I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
      $endgroup$
      – GEdgar
      Jan 14 at 11:58










    • $begingroup$
      Which functional is it?
      $endgroup$
      – James
      Jan 14 at 12:12










    • $begingroup$
      The functional is the one assumed to exist in the definition of $c$.
      $endgroup$
      – GEdgar
      Jan 15 at 15:37
















    $begingroup$
    I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
    $endgroup$
    – GEdgar
    Jan 14 at 11:58




    $begingroup$
    I think this is the best approach. It does not require knowledge of $C^*$ algebras. And of course showing $(l^infty)^*$ is not separable is much more than showing it is not equal to $l^1$.
    $endgroup$
    – GEdgar
    Jan 14 at 11:58












    $begingroup$
    Which functional is it?
    $endgroup$
    – James
    Jan 14 at 12:12




    $begingroup$
    Which functional is it?
    $endgroup$
    – James
    Jan 14 at 12:12












    $begingroup$
    The functional is the one assumed to exist in the definition of $c$.
    $endgroup$
    – GEdgar
    Jan 15 at 15:37




    $begingroup$
    The functional is the one assumed to exist in the definition of $c$.
    $endgroup$
    – GEdgar
    Jan 15 at 15:37











    1












    $begingroup$

    Similarly to gerw's answer. Just use that, as a $C^ast$-algebra $ell^infty(mathbb N)$ is isomorphic to $C(beta mathbb{N})$, the continuous function over the compact space given by $beta mathbb{N}$, the Stone-Cech compactification of $mathbb{N}$. By Riesz theorem $(ell^infty)^ast = M(beta mathbb{N})$, the finite Radon measures over $beta mathbb{N}$. The points in $beta mathbb{N} setminus mathbb{N}$ are proper ultrafiters and evaluating on them gives functionals that are not in the image of $ell^1$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Similarly to gerw's answer. Just use that, as a $C^ast$-algebra $ell^infty(mathbb N)$ is isomorphic to $C(beta mathbb{N})$, the continuous function over the compact space given by $beta mathbb{N}$, the Stone-Cech compactification of $mathbb{N}$. By Riesz theorem $(ell^infty)^ast = M(beta mathbb{N})$, the finite Radon measures over $beta mathbb{N}$. The points in $beta mathbb{N} setminus mathbb{N}$ are proper ultrafiters and evaluating on them gives functionals that are not in the image of $ell^1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Similarly to gerw's answer. Just use that, as a $C^ast$-algebra $ell^infty(mathbb N)$ is isomorphic to $C(beta mathbb{N})$, the continuous function over the compact space given by $beta mathbb{N}$, the Stone-Cech compactification of $mathbb{N}$. By Riesz theorem $(ell^infty)^ast = M(beta mathbb{N})$, the finite Radon measures over $beta mathbb{N}$. The points in $beta mathbb{N} setminus mathbb{N}$ are proper ultrafiters and evaluating on them gives functionals that are not in the image of $ell^1$.






        share|cite|improve this answer









        $endgroup$



        Similarly to gerw's answer. Just use that, as a $C^ast$-algebra $ell^infty(mathbb N)$ is isomorphic to $C(beta mathbb{N})$, the continuous function over the compact space given by $beta mathbb{N}$, the Stone-Cech compactification of $mathbb{N}$. By Riesz theorem $(ell^infty)^ast = M(beta mathbb{N})$, the finite Radon measures over $beta mathbb{N}$. The points in $beta mathbb{N} setminus mathbb{N}$ are proper ultrafiters and evaluating on them gives functionals that are not in the image of $ell^1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 11:41









        Adrián González-PérezAdrián González-Pérez

        1,206139




        1,206139






























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