Open Mapping Theorem fails when space is not complete












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I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely



For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?






functional-analysis complete-spaces open-map







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    $begingroup$


    I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely



    For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?






    functional-analysis complete-spaces open-map







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely



      For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?






      functional-analysis complete-spaces open-map







      share|cite|improve this question









      $endgroup$




      I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely



      For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?






      functional-analysis complete-spaces open-map




      functional-analysis complete-spaces open-map






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      share|cite|improve this question











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      asked Jan 8 at 17:34









      JamesJames

      1,002320




      1,002320






















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          I might have come up with the answer myself, I wonder if it is correct.



          Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate



          $$
          |T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
          $$
          It follows that $T$ is bounded.



          We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
          $$
          |T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
          $$



          So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.



          Is this proof correct?






          share|cite|improve this answer









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            1 Answer
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            $begingroup$

            I might have come up with the answer myself, I wonder if it is correct.



            Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate



            $$
            |T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
            $$
            It follows that $T$ is bounded.



            We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
            $$
            |T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
            $$



            So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.



            Is this proof correct?






            share|cite|improve this answer









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              0












              $begingroup$

              I might have come up with the answer myself, I wonder if it is correct.



              Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate



              $$
              |T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
              $$
              It follows that $T$ is bounded.



              We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
              $$
              |T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
              $$



              So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.



              Is this proof correct?






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I might have come up with the answer myself, I wonder if it is correct.



                Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate



                $$
                |T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
                $$
                It follows that $T$ is bounded.



                We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
                $$
                |T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
                $$



                So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.



                Is this proof correct?






                share|cite|improve this answer









                $endgroup$



                I might have come up with the answer myself, I wonder if it is correct.



                Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate



                $$
                |T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
                $$
                It follows that $T$ is bounded.



                We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
                $$
                |T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
                $$



                So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.



                Is this proof correct?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 18:10









                JamesJames

                1,002320




                1,002320






























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