Supremum of norms of images of unit ball equals infimum of upper bounds of norms of images in the whole space












1












$begingroup$


For a linear operator $T in B(X,Y)$, there exists





  • $k in mathbb{R}_{>0}$ such that for all $x$ in the unit ball of $X: lVert T(x)rVert leq k $;


  • $k in mathbb{R}_{>0}$ such that for all $x in X: lVert T(x)rVert leq k lVert xrVert$.


This is supposed to imply that
$$text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } = text{inf}{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)},
$$
but this is not proved. I'd like to add rigour to the statement.



Also, the above should imply that for all $y in X$
$$lVert T(y)rVert leq text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } lVert yrVert.
$$






functional-analysis analysis operator-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    For a linear operator $T in B(X,Y)$, there exists





    • $k in mathbb{R}_{>0}$ such that for all $x$ in the unit ball of $X: lVert T(x)rVert leq k $;


    • $k in mathbb{R}_{>0}$ such that for all $x in X: lVert T(x)rVert leq k lVert xrVert$.


    This is supposed to imply that
    $$text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } = text{inf}{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)},
    $$
    but this is not proved. I'd like to add rigour to the statement.



    Also, the above should imply that for all $y in X$
    $$lVert T(y)rVert leq text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } lVert yrVert.
    $$






    functional-analysis analysis operator-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      For a linear operator $T in B(X,Y)$, there exists





      • $k in mathbb{R}_{>0}$ such that for all $x$ in the unit ball of $X: lVert T(x)rVert leq k $;


      • $k in mathbb{R}_{>0}$ such that for all $x in X: lVert T(x)rVert leq k lVert xrVert$.


      This is supposed to imply that
      $$text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } = text{inf}{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)},
      $$
      but this is not proved. I'd like to add rigour to the statement.



      Also, the above should imply that for all $y in X$
      $$lVert T(y)rVert leq text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } lVert yrVert.
      $$






      functional-analysis analysis operator-theory







      share|cite|improve this question











      $endgroup$




      For a linear operator $T in B(X,Y)$, there exists





      • $k in mathbb{R}_{>0}$ such that for all $x$ in the unit ball of $X: lVert T(x)rVert leq k $;


      • $k in mathbb{R}_{>0}$ such that for all $x in X: lVert T(x)rVert leq k lVert xrVert$.


      This is supposed to imply that
      $$text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } = text{inf}{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)},
      $$
      but this is not proved. I'd like to add rigour to the statement.



      Also, the above should imply that for all $y in X$
      $$lVert T(y)rVert leq text{sup}{lVert T(x) rVert mid lVert xrVert leq 1 } lVert yrVert.
      $$






      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 Jan 16 at 21:36







      Jos van Nieuwman

















      asked Jan 16 at 21:28









      Jos van NieuwmanJos van Nieuwman

      469




      469






















          1 Answer
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          $begingroup$

          Take any $kin {k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}$. If $|x|leq1$, then
          $$
          |Tx|leq k|x|=k.
          $$
          As we can do this for all such $k$, we get
          $$
          |Tx|leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
          $$
          And we can do this for any $x$ with $|x|leq1$, so
          $$
          sup{lVert T(x) rVert mid lVert xrVert leq 1 } leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
          $$
          Also, for any nonzero $xin X$,
          $$
          |Tx|=|Tleft(tfrac x{|x|}right)|,|x|leq sup{lVert T(x) rVert mid lVert xrVert leq 1 },|x|.
          $$
          Thus
          $$
          inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}leq sup{lVert T(x) rVert mid lVert xrVert leq 1 }.
          $$






          share|cite|improve this answer









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            $begingroup$

            Take any $kin {k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}$. If $|x|leq1$, then
            $$
            |Tx|leq k|x|=k.
            $$
            As we can do this for all such $k$, we get
            $$
            |Tx|leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
            $$
            And we can do this for any $x$ with $|x|leq1$, so
            $$
            sup{lVert T(x) rVert mid lVert xrVert leq 1 } leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
            $$
            Also, for any nonzero $xin X$,
            $$
            |Tx|=|Tleft(tfrac x{|x|}right)|,|x|leq sup{lVert T(x) rVert mid lVert xrVert leq 1 },|x|.
            $$
            Thus
            $$
            inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}leq sup{lVert T(x) rVert mid lVert xrVert leq 1 }.
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Take any $kin {k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}$. If $|x|leq1$, then
              $$
              |Tx|leq k|x|=k.
              $$
              As we can do this for all such $k$, we get
              $$
              |Tx|leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
              $$
              And we can do this for any $x$ with $|x|leq1$, so
              $$
              sup{lVert T(x) rVert mid lVert xrVert leq 1 } leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
              $$
              Also, for any nonzero $xin X$,
              $$
              |Tx|=|Tleft(tfrac x{|x|}right)|,|x|leq sup{lVert T(x) rVert mid lVert xrVert leq 1 },|x|.
              $$
              Thus
              $$
              inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}leq sup{lVert T(x) rVert mid lVert xrVert leq 1 }.
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Take any $kin {k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}$. If $|x|leq1$, then
                $$
                |Tx|leq k|x|=k.
                $$
                As we can do this for all such $k$, we get
                $$
                |Tx|leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
                $$
                And we can do this for any $x$ with $|x|leq1$, so
                $$
                sup{lVert T(x) rVert mid lVert xrVert leq 1 } leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
                $$
                Also, for any nonzero $xin X$,
                $$
                |Tx|=|Tleft(tfrac x{|x|}right)|,|x|leq sup{lVert T(x) rVert mid lVert xrVert leq 1 },|x|.
                $$
                Thus
                $$
                inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}leq sup{lVert T(x) rVert mid lVert xrVert leq 1 }.
                $$






                share|cite|improve this answer









                $endgroup$



                Take any $kin {k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}$. If $|x|leq1$, then
                $$
                |Tx|leq k|x|=k.
                $$
                As we can do this for all such $k$, we get
                $$
                |Tx|leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
                $$
                And we can do this for any $x$ with $|x|leq1$, so
                $$
                sup{lVert T(x) rVert mid lVert xrVert leq 1 } leq inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}.
                $$
                Also, for any nonzero $xin X$,
                $$
                |Tx|=|Tleft(tfrac x{|x|}right)|,|x|leq sup{lVert T(x) rVert mid lVert xrVert leq 1 },|x|.
                $$
                Thus
                $$
                inf{k in mathbb{R}_{>0} mid (forall x in X)(lVert T(x)rVert leq k lVert xrVert)}leq sup{lVert T(x) rVert mid lVert xrVert leq 1 }.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 21:44









                Martin ArgeramiMartin Argerami

                127k1182183




                127k1182183






























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