Banach spaces $X,Y$ and a mapping $f:Xto Y$












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Let $X,Y$ be Banach spaces and a mapping $fin mathcal{L}(X,Y)$. Suppose that $f$ is also an injective map and an open map from $X$ to $Y$. Show that then $f in mathcal{B}(X,Y).$



Here I denoted the set of all linear mappings from $X$ to $Y$ by $mathcal{L}(X,Y)$, and the set of all bounded mappings from $X$ to $Y$ by $mathcal{B}(X,Y)$.
Does anyone have an idea about this statement? I appreciate any hints.






linear-transformations banach-spaces







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    0












    $begingroup$


    Let $X,Y$ be Banach spaces and a mapping $fin mathcal{L}(X,Y)$. Suppose that $f$ is also an injective map and an open map from $X$ to $Y$. Show that then $f in mathcal{B}(X,Y).$



    Here I denoted the set of all linear mappings from $X$ to $Y$ by $mathcal{L}(X,Y)$, and the set of all bounded mappings from $X$ to $Y$ by $mathcal{B}(X,Y)$.
    Does anyone have an idea about this statement? I appreciate any hints.






    linear-transformations banach-spaces







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X,Y$ be Banach spaces and a mapping $fin mathcal{L}(X,Y)$. Suppose that $f$ is also an injective map and an open map from $X$ to $Y$. Show that then $f in mathcal{B}(X,Y).$



      Here I denoted the set of all linear mappings from $X$ to $Y$ by $mathcal{L}(X,Y)$, and the set of all bounded mappings from $X$ to $Y$ by $mathcal{B}(X,Y)$.
      Does anyone have an idea about this statement? I appreciate any hints.






      linear-transformations banach-spaces







      share|cite|improve this question











      $endgroup$




      Let $X,Y$ be Banach spaces and a mapping $fin mathcal{L}(X,Y)$. Suppose that $f$ is also an injective map and an open map from $X$ to $Y$. Show that then $f in mathcal{B}(X,Y).$



      Here I denoted the set of all linear mappings from $X$ to $Y$ by $mathcal{L}(X,Y)$, and the set of all bounded mappings from $X$ to $Y$ by $mathcal{B}(X,Y)$.
      Does anyone have an idea about this statement? I appreciate any hints.






      linear-transformations banach-spaces




      linear-transformations banach-spaces






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 17 '18 at 9:31







      mandella

















      asked Dec 17 '18 at 7:40









      mandellamandella

      787521




      787521






















          1 Answer
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          $begingroup$

          $textbf{Hint:}$ Using given conditions, show that $f$ is also surjective. Then appeal to the inverse mapping theorem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you help me to show that $f$ is surjective? I do not see it right away
            $endgroup$
            – mandella
            Dec 17 '18 at 8:01






          • 1




            $begingroup$
            Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
            $endgroup$
            – Song
            Dec 17 '18 at 8:10










          • $begingroup$
            Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 8:11






          • 1




            $begingroup$
            Actually, since $f$ is open, $f^{-1}$ is continuous.
            $endgroup$
            – Song
            Dec 17 '18 at 8:53










          • $begingroup$
            So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 9:21













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          1 Answer
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          active

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          active

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          2












          $begingroup$

          $textbf{Hint:}$ Using given conditions, show that $f$ is also surjective. Then appeal to the inverse mapping theorem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you help me to show that $f$ is surjective? I do not see it right away
            $endgroup$
            – mandella
            Dec 17 '18 at 8:01






          • 1




            $begingroup$
            Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
            $endgroup$
            – Song
            Dec 17 '18 at 8:10










          • $begingroup$
            Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 8:11






          • 1




            $begingroup$
            Actually, since $f$ is open, $f^{-1}$ is continuous.
            $endgroup$
            – Song
            Dec 17 '18 at 8:53










          • $begingroup$
            So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 9:21


















          2












          $begingroup$

          $textbf{Hint:}$ Using given conditions, show that $f$ is also surjective. Then appeal to the inverse mapping theorem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you help me to show that $f$ is surjective? I do not see it right away
            $endgroup$
            – mandella
            Dec 17 '18 at 8:01






          • 1




            $begingroup$
            Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
            $endgroup$
            – Song
            Dec 17 '18 at 8:10










          • $begingroup$
            Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 8:11






          • 1




            $begingroup$
            Actually, since $f$ is open, $f^{-1}$ is continuous.
            $endgroup$
            – Song
            Dec 17 '18 at 8:53










          • $begingroup$
            So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 9:21
















          2












          2








          2





          $begingroup$

          $textbf{Hint:}$ Using given conditions, show that $f$ is also surjective. Then appeal to the inverse mapping theorem.






          share|cite|improve this answer











          $endgroup$



          $textbf{Hint:}$ Using given conditions, show that $f$ is also surjective. Then appeal to the inverse mapping theorem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 8:06

























          answered Dec 17 '18 at 7:44









          SongSong

          18.2k21449




          18.2k21449












          • $begingroup$
            Can you help me to show that $f$ is surjective? I do not see it right away
            $endgroup$
            – mandella
            Dec 17 '18 at 8:01






          • 1




            $begingroup$
            Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
            $endgroup$
            – Song
            Dec 17 '18 at 8:10










          • $begingroup$
            Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 8:11






          • 1




            $begingroup$
            Actually, since $f$ is open, $f^{-1}$ is continuous.
            $endgroup$
            – Song
            Dec 17 '18 at 8:53










          • $begingroup$
            So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 9:21




















          • $begingroup$
            Can you help me to show that $f$ is surjective? I do not see it right away
            $endgroup$
            – mandella
            Dec 17 '18 at 8:01






          • 1




            $begingroup$
            Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
            $endgroup$
            – Song
            Dec 17 '18 at 8:10










          • $begingroup$
            Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 8:11






          • 1




            $begingroup$
            Actually, since $f$ is open, $f^{-1}$ is continuous.
            $endgroup$
            – Song
            Dec 17 '18 at 8:53










          • $begingroup$
            So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
            $endgroup$
            – mandella
            Dec 17 '18 at 9:21


















          $begingroup$
          Can you help me to show that $f$ is surjective? I do not see it right away
          $endgroup$
          – mandella
          Dec 17 '18 at 8:01




          $begingroup$
          Can you help me to show that $f$ is surjective? I do not see it right away
          $endgroup$
          – mandella
          Dec 17 '18 at 8:01




          1




          1




          $begingroup$
          Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
          $endgroup$
          – Song
          Dec 17 '18 at 8:10




          $begingroup$
          Well, $f$ being surjective follows from openness. See how an open set looks like in Banach spaces.
          $endgroup$
          – Song
          Dec 17 '18 at 8:10












          $begingroup$
          Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
          $endgroup$
          – mandella
          Dec 17 '18 at 8:11




          $begingroup$
          Ah after spending some time I see this. So with the fact that $f$ is surjective then its inverse $f^{-1}$ is bounded?
          $endgroup$
          – mandella
          Dec 17 '18 at 8:11




          1




          1




          $begingroup$
          Actually, since $f$ is open, $f^{-1}$ is continuous.
          $endgroup$
          – Song
          Dec 17 '18 at 8:53




          $begingroup$
          Actually, since $f$ is open, $f^{-1}$ is continuous.
          $endgroup$
          – Song
          Dec 17 '18 at 8:53












          $begingroup$
          So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
          $endgroup$
          – mandella
          Dec 17 '18 at 9:21






          $begingroup$
          So then by the inverse mapping theorem, $f^{-1}$ since bijective and continuous linear map it will have a bounded inverse i.e. $f$ will be bounded?
          $endgroup$
          – mandella
          Dec 17 '18 at 9:21




















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