Continuous linear transformation on normed linear spaces with $T(x)=y$

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Let $X,Y$ be normed linear spaces and let $x(neq0)in X,y(neq0)in Y$. prove that there exists $Tin mathscrB(X,Y)$ such that $Tx=y$.
I thought of a constant map but that will be not linear and continuous. Please give me idea how to construct this type of function?
and thanks in advance




functional-analysis




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  • It's enough to solve it for the special case $Y=Bbb K$ (the base field) and $y=1$.
    – Berci
    Jul 20 at 20:16










  • i didn't got your hint
    – ravi yadav
    Jul 20 at 20:47














up vote
2
down vote

favorite
1












Let $X,Y$ be normed linear spaces and let $x(neq0)in X,y(neq0)in Y$. prove that there exists $Tin mathscrB(X,Y)$ such that $Tx=y$.
I thought of a constant map but that will be not linear and continuous. Please give me idea how to construct this type of function?
and thanks in advance




functional-analysis




share|cite|improve this question





















  • It's enough to solve it for the special case $Y=Bbb K$ (the base field) and $y=1$.
    – Berci
    Jul 20 at 20:16










  • i didn't got your hint
    – ravi yadav
    Jul 20 at 20:47












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





Let $X,Y$ be normed linear spaces and let $x(neq0)in X,y(neq0)in Y$. prove that there exists $Tin mathscrB(X,Y)$ such that $Tx=y$.
I thought of a constant map but that will be not linear and continuous. Please give me idea how to construct this type of function?
and thanks in advance




functional-analysis




share|cite|improve this question













Let $X,Y$ be normed linear spaces and let $x(neq0)in X,y(neq0)in Y$. prove that there exists $Tin mathscrB(X,Y)$ such that $Tx=y$.
I thought of a constant map but that will be not linear and continuous. Please give me idea how to construct this type of function?
and thanks in advance





functional-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 20:26









mechanodroid

22.2k52041




22.2k52041









asked Jul 20 at 19:48









ravi yadav

295




295











  • It's enough to solve it for the special case $Y=Bbb K$ (the base field) and $y=1$.
    – Berci
    Jul 20 at 20:16










  • i didn't got your hint
    – ravi yadav
    Jul 20 at 20:47
















  • It's enough to solve it for the special case $Y=Bbb K$ (the base field) and $y=1$.
    – Berci
    Jul 20 at 20:16










  • i didn't got your hint
    – ravi yadav
    Jul 20 at 20:47















It's enough to solve it for the special case $Y=Bbb K$ (the base field) and $y=1$.
– Berci
Jul 20 at 20:16




It's enough to solve it for the special case $Y=Bbb K$ (the base field) and $y=1$.
– Berci
Jul 20 at 20:16












i didn't got your hint
– ravi yadav
Jul 20 at 20:47




i didn't got your hint
– ravi yadav
Jul 20 at 20:47










1 Answer
1






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3
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accepted










Hint:



Define a bounded linear functional $f : operatornamespanx to mathbbK$ with $f(alpha x) = alpha, forall alpha in mathbbK$.



Extend $f$ to a bounded linear functional $F : X to mathbbK$ by Hahn-Banach.



Check that the desired map $T : X to Y$ is given by $Tz = F(z)y$.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Hint:



    Define a bounded linear functional $f : operatornamespanx to mathbbK$ with $f(alpha x) = alpha, forall alpha in mathbbK$.



    Extend $f$ to a bounded linear functional $F : X to mathbbK$ by Hahn-Banach.



    Check that the desired map $T : X to Y$ is given by $Tz = F(z)y$.






    share|cite|improve this answer

























      up vote
      3
      down vote



      accepted










      Hint:



      Define a bounded linear functional $f : operatornamespanx to mathbbK$ with $f(alpha x) = alpha, forall alpha in mathbbK$.



      Extend $f$ to a bounded linear functional $F : X to mathbbK$ by Hahn-Banach.



      Check that the desired map $T : X to Y$ is given by $Tz = F(z)y$.






      share|cite|improve this answer























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Hint:



        Define a bounded linear functional $f : operatornamespanx to mathbbK$ with $f(alpha x) = alpha, forall alpha in mathbbK$.



        Extend $f$ to a bounded linear functional $F : X to mathbbK$ by Hahn-Banach.



        Check that the desired map $T : X to Y$ is given by $Tz = F(z)y$.






        share|cite|improve this answer













        Hint:



        Define a bounded linear functional $f : operatornamespanx to mathbbK$ with $f(alpha x) = alpha, forall alpha in mathbbK$.



        Extend $f$ to a bounded linear functional $F : X to mathbbK$ by Hahn-Banach.



        Check that the desired map $T : X to Y$ is given by $Tz = F(z)y$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 20 at 20:25









        mechanodroid

        22.2k52041




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