Adjoint of bounded linear map is isometric isomorphism implies original map is isometric isomorphism?












2












$begingroup$


Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.



A straightforward calculation shows:



Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.



I'm trying to prove the converse.



But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).



Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.



I cannot seem to strengthen the conclusion to $T$ is surjective.



If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.



But what happens if $X$ is not complete or $T(X)$ is not closed?



In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:



$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)



There are also some Hilbert space examples in the following links, but they don't address what I am asking about:



$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?



Example: operator injective, then the adjoint is NOT surjective






functional-analysis analysis operator-theory







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.



    A straightforward calculation shows:



    Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.



    I'm trying to prove the converse.



    But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).



    Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.



    I cannot seem to strengthen the conclusion to $T$ is surjective.



    If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.



    But what happens if $X$ is not complete or $T(X)$ is not closed?



    In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:



    $T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)



    There are also some Hilbert space examples in the following links, but they don't address what I am asking about:



    $T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?



    Example: operator injective, then the adjoint is NOT surjective






    functional-analysis analysis operator-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.



      A straightforward calculation shows:



      Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.



      I'm trying to prove the converse.



      But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).



      Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.



      I cannot seem to strengthen the conclusion to $T$ is surjective.



      If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.



      But what happens if $X$ is not complete or $T(X)$ is not closed?



      In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:



      $T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)



      There are also some Hilbert space examples in the following links, but they don't address what I am asking about:



      $T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?



      Example: operator injective, then the adjoint is NOT surjective






      functional-analysis analysis operator-theory







      share|cite|improve this question











      $endgroup$




      Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.



      A straightforward calculation shows:



      Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.



      I'm trying to prove the converse.



      But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).



      Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.



      I cannot seem to strengthen the conclusion to $T$ is surjective.



      If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.



      But what happens if $X$ is not complete or $T(X)$ is not closed?



      In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:



      $T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)



      There are also some Hilbert space examples in the following links, but they don't address what I am asking about:



      $T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?



      Example: operator injective, then the adjoint is NOT surjective






      functional-analysis analysis operator-theory




      functional-analysis analysis operator-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 20:12







      MichaelGaudreau

















      asked Dec 17 '18 at 1:50









      MichaelGaudreauMichaelGaudreau

      1488




      1488






















          1 Answer
          1






          active

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          2












          $begingroup$

          The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.



          Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:25










          • $begingroup$
            So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:35






          • 1




            $begingroup$
            I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
            $endgroup$
            – Aweygan
            Dec 18 '18 at 23:52










          • $begingroup$
            Yes, I've come to the same understanding. Thanks for your excellent example.
            $endgroup$
            – MichaelGaudreau
            Dec 19 '18 at 1:33










          • $begingroup$
            You're welcome. Glad to help!
            $endgroup$
            – Aweygan
            Dec 19 '18 at 1:35











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          $begingroup$

          The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.



          Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:25










          • $begingroup$
            So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:35






          • 1




            $begingroup$
            I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
            $endgroup$
            – Aweygan
            Dec 18 '18 at 23:52










          • $begingroup$
            Yes, I've come to the same understanding. Thanks for your excellent example.
            $endgroup$
            – MichaelGaudreau
            Dec 19 '18 at 1:33










          • $begingroup$
            You're welcome. Glad to help!
            $endgroup$
            – Aweygan
            Dec 19 '18 at 1:35
















          2












          $begingroup$

          The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.



          Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:25










          • $begingroup$
            So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:35






          • 1




            $begingroup$
            I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
            $endgroup$
            – Aweygan
            Dec 18 '18 at 23:52










          • $begingroup$
            Yes, I've come to the same understanding. Thanks for your excellent example.
            $endgroup$
            – MichaelGaudreau
            Dec 19 '18 at 1:33










          • $begingroup$
            You're welcome. Glad to help!
            $endgroup$
            – Aweygan
            Dec 19 '18 at 1:35














          2












          2








          2





          $begingroup$

          The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.



          Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.






          share|cite|improve this answer









          $endgroup$



          The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.



          Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 4:25









          AweyganAweygan

          14.5k21442




          14.5k21442












          • $begingroup$
            This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:25










          • $begingroup$
            So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:35






          • 1




            $begingroup$
            I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
            $endgroup$
            – Aweygan
            Dec 18 '18 at 23:52










          • $begingroup$
            Yes, I've come to the same understanding. Thanks for your excellent example.
            $endgroup$
            – MichaelGaudreau
            Dec 19 '18 at 1:33










          • $begingroup$
            You're welcome. Glad to help!
            $endgroup$
            – Aweygan
            Dec 19 '18 at 1:35


















          • $begingroup$
            This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:25










          • $begingroup$
            So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
            $endgroup$
            – MichaelGaudreau
            Dec 18 '18 at 20:35






          • 1




            $begingroup$
            I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
            $endgroup$
            – Aweygan
            Dec 18 '18 at 23:52










          • $begingroup$
            Yes, I've come to the same understanding. Thanks for your excellent example.
            $endgroup$
            – MichaelGaudreau
            Dec 19 '18 at 1:33










          • $begingroup$
            You're welcome. Glad to help!
            $endgroup$
            – Aweygan
            Dec 19 '18 at 1:35
















          $begingroup$
          This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
          $endgroup$
          – MichaelGaudreau
          Dec 18 '18 at 20:25




          $begingroup$
          This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
          $endgroup$
          – MichaelGaudreau
          Dec 18 '18 at 20:25












          $begingroup$
          So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
          $endgroup$
          – MichaelGaudreau
          Dec 18 '18 at 20:35




          $begingroup$
          So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
          $endgroup$
          – MichaelGaudreau
          Dec 18 '18 at 20:35




          1




          1




          $begingroup$
          I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
          $endgroup$
          – Aweygan
          Dec 18 '18 at 23:52




          $begingroup$
          I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
          $endgroup$
          – Aweygan
          Dec 18 '18 at 23:52












          $begingroup$
          Yes, I've come to the same understanding. Thanks for your excellent example.
          $endgroup$
          – MichaelGaudreau
          Dec 19 '18 at 1:33




          $begingroup$
          Yes, I've come to the same understanding. Thanks for your excellent example.
          $endgroup$
          – MichaelGaudreau
          Dec 19 '18 at 1:33












          $begingroup$
          You're welcome. Glad to help!
          $endgroup$
          – Aweygan
          Dec 19 '18 at 1:35




          $begingroup$
          You're welcome. Glad to help!
          $endgroup$
          – Aweygan
          Dec 19 '18 at 1:35


















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          Verónica Boquete

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          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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