Linear transformation in space of infinite dimension

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Problem.



(a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.



(b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?




Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?




real-analysis functional-analysis continuity




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    up vote
    1
    down vote

    favorite













    Problem.



    (a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.



    (b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?




    Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?




    real-analysis functional-analysis continuity




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Problem.



      (a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.



      (b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?




      Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?




      real-analysis functional-analysis continuity




      share|cite|improve this question














      Problem.



      (a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.



      (b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?




      Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?





      real-analysis functional-analysis continuity





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      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago









      José Carlos Santos

      111k1695171




      111k1695171









      asked 12 hours ago









      Lucas Corrêa

      1,041219




      1,041219




















          1 Answer
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          A classical example consists in taking:



          • $X=Cbigl([0,1]bigr)$;

          • if $fin X$, $|f|=int_0^1|f|$;

          • $psi(f)=f(1)$.

          Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.






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          • Nice! Thank you!
            – Lucas Corrêa
            12 hours ago










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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          oldest

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          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          A classical example consists in taking:



          • $X=Cbigl([0,1]bigr)$;

          • if $fin X$, $|f|=int_0^1|f|$;

          • $psi(f)=f(1)$.

          Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.






          share|cite|improve this answer





















          • Nice! Thank you!
            – Lucas Corrêa
            12 hours ago














          up vote
          3
          down vote



          accepted










          A classical example consists in taking:



          • $X=Cbigl([0,1]bigr)$;

          • if $fin X$, $|f|=int_0^1|f|$;

          • $psi(f)=f(1)$.

          Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.






          share|cite|improve this answer





















          • Nice! Thank you!
            – Lucas Corrêa
            12 hours ago












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          A classical example consists in taking:



          • $X=Cbigl([0,1]bigr)$;

          • if $fin X$, $|f|=int_0^1|f|$;

          • $psi(f)=f(1)$.

          Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.






          share|cite|improve this answer













          A classical example consists in taking:



          • $X=Cbigl([0,1]bigr)$;

          • if $fin X$, $|f|=int_0^1|f|$;

          • $psi(f)=f(1)$.

          Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 12 hours ago









          José Carlos Santos

          111k1695171




          111k1695171











          • Nice! Thank you!
            – Lucas Corrêa
            12 hours ago
















          • Nice! Thank you!
            – Lucas Corrêa
            12 hours ago















          Nice! Thank you!
          – Lucas Corrêa
          12 hours ago




          Nice! Thank you!
          – Lucas Corrêa
          12 hours ago












           

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