Proving that $Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert$ defines a norm on $B(X,Y). $












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Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y,$ normed linear maps. Then, $Vert cdot Vert,$ defined by $Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert,$ for arbitrary $Tin B(X,Y), $ is a norm on $B(X,Y). $



we have for arbitrary $Tin B(X,Y), $
begin{align} Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert= suplimits_{Vert x Vert = 1}Vert T x Vert=suplimits_{xneq 0} frac{Vert T x Vert}{Vert x Vert}.end{align}



MY TRIAL



1.
begin{align} Vert T Vert =0&iff suplimits_{Vert x Vertleq 1}Vert T x Vert =0iff Vert T x Vert =0,;;forall ,xin X, ,Tin B(X,Y)\& iff T x=0,;;forall ,xin X, ,Tin B(X,Y) \& iff T =0,;;forall , ,Tin B(X,Y)end{align}
2.
begin{align} Vert kT Vert =& suplimits_{Vert x Vertleq 1}Vert k T x Vert \=& |k|suplimits_{Vert x Vertleq 1}Vert T x Vert \=& |k|Vert T Vert,;;forall ,,kin K, ,Tin B(X,Y)end{align}



3.
begin{align} Vert T+S Vert =& suplimits_{Vert x Vertleq 1}Vert T x + S x Vert \leq & suplimits_{Vert x Vertleq 1}left(Vert T x Vert + Vert S x Vert right) \=&suplimits_{Vert x Vertleq 1}Vert T x Vert +suplimits_{Vert x Vertleq 1} Vert S x Vert \=&Vert T Vert+Vert S Vert,;;forall ,T,Sin B(X,Y)end{align}
Kindly help check if this is correct. If not, corrections and alternative proofs will be highly welcome.






functional-analysis norm normed-spaces







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    $begingroup$


    Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y,$ normed linear maps. Then, $Vert cdot Vert,$ defined by $Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert,$ for arbitrary $Tin B(X,Y), $ is a norm on $B(X,Y). $



    we have for arbitrary $Tin B(X,Y), $
    begin{align} Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert= suplimits_{Vert x Vert = 1}Vert T x Vert=suplimits_{xneq 0} frac{Vert T x Vert}{Vert x Vert}.end{align}



    MY TRIAL



    1.
    begin{align} Vert T Vert =0&iff suplimits_{Vert x Vertleq 1}Vert T x Vert =0iff Vert T x Vert =0,;;forall ,xin X, ,Tin B(X,Y)\& iff T x=0,;;forall ,xin X, ,Tin B(X,Y) \& iff T =0,;;forall , ,Tin B(X,Y)end{align}
    2.
    begin{align} Vert kT Vert =& suplimits_{Vert x Vertleq 1}Vert k T x Vert \=& |k|suplimits_{Vert x Vertleq 1}Vert T x Vert \=& |k|Vert T Vert,;;forall ,,kin K, ,Tin B(X,Y)end{align}



    3.
    begin{align} Vert T+S Vert =& suplimits_{Vert x Vertleq 1}Vert T x + S x Vert \leq & suplimits_{Vert x Vertleq 1}left(Vert T x Vert + Vert S x Vert right) \=&suplimits_{Vert x Vertleq 1}Vert T x Vert +suplimits_{Vert x Vertleq 1} Vert S x Vert \=&Vert T Vert+Vert S Vert,;;forall ,T,Sin B(X,Y)end{align}
    Kindly help check if this is correct. If not, corrections and alternative proofs will be highly welcome.






    functional-analysis norm normed-spaces







    share|cite|improve this question









    $endgroup$















      0












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      0


      1



      $begingroup$


      Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y,$ normed linear maps. Then, $Vert cdot Vert,$ defined by $Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert,$ for arbitrary $Tin B(X,Y), $ is a norm on $B(X,Y). $



      we have for arbitrary $Tin B(X,Y), $
      begin{align} Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert= suplimits_{Vert x Vert = 1}Vert T x Vert=suplimits_{xneq 0} frac{Vert T x Vert}{Vert x Vert}.end{align}



      MY TRIAL



      1.
      begin{align} Vert T Vert =0&iff suplimits_{Vert x Vertleq 1}Vert T x Vert =0iff Vert T x Vert =0,;;forall ,xin X, ,Tin B(X,Y)\& iff T x=0,;;forall ,xin X, ,Tin B(X,Y) \& iff T =0,;;forall , ,Tin B(X,Y)end{align}
      2.
      begin{align} Vert kT Vert =& suplimits_{Vert x Vertleq 1}Vert k T x Vert \=& |k|suplimits_{Vert x Vertleq 1}Vert T x Vert \=& |k|Vert T Vert,;;forall ,,kin K, ,Tin B(X,Y)end{align}



      3.
      begin{align} Vert T+S Vert =& suplimits_{Vert x Vertleq 1}Vert T x + S x Vert \leq & suplimits_{Vert x Vertleq 1}left(Vert T x Vert + Vert S x Vert right) \=&suplimits_{Vert x Vertleq 1}Vert T x Vert +suplimits_{Vert x Vertleq 1} Vert S x Vert \=&Vert T Vert+Vert S Vert,;;forall ,T,Sin B(X,Y)end{align}
      Kindly help check if this is correct. If not, corrections and alternative proofs will be highly welcome.






      functional-analysis norm normed-spaces







      share|cite|improve this question









      $endgroup$




      Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y,$ normed linear maps. Then, $Vert cdot Vert,$ defined by $Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert,$ for arbitrary $Tin B(X,Y), $ is a norm on $B(X,Y). $



      we have for arbitrary $Tin B(X,Y), $
      begin{align} Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert= suplimits_{Vert x Vert = 1}Vert T x Vert=suplimits_{xneq 0} frac{Vert T x Vert}{Vert x Vert}.end{align}



      MY TRIAL



      1.
      begin{align} Vert T Vert =0&iff suplimits_{Vert x Vertleq 1}Vert T x Vert =0iff Vert T x Vert =0,;;forall ,xin X, ,Tin B(X,Y)\& iff T x=0,;;forall ,xin X, ,Tin B(X,Y) \& iff T =0,;;forall , ,Tin B(X,Y)end{align}
      2.
      begin{align} Vert kT Vert =& suplimits_{Vert x Vertleq 1}Vert k T x Vert \=& |k|suplimits_{Vert x Vertleq 1}Vert T x Vert \=& |k|Vert T Vert,;;forall ,,kin K, ,Tin B(X,Y)end{align}



      3.
      begin{align} Vert T+S Vert =& suplimits_{Vert x Vertleq 1}Vert T x + S x Vert \leq & suplimits_{Vert x Vertleq 1}left(Vert T x Vert + Vert S x Vert right) \=&suplimits_{Vert x Vertleq 1}Vert T x Vert +suplimits_{Vert x Vertleq 1} Vert S x Vert \=&Vert T Vert+Vert S Vert,;;forall ,T,Sin B(X,Y)end{align}
      Kindly help check if this is correct. If not, corrections and alternative proofs will be highly welcome.






      functional-analysis norm normed-spaces




      functional-analysis norm normed-spaces






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      asked Dec 18 '18 at 9:20









      Omojola MichealOmojola Micheal

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          Your proof is fine except that you have to fill in some details on why $sup{|Tx|:|x|leq 1}$ implies $Tx=0$ for all $x$. You get $Tx=0$ for $|x|leq 1$ and then you have to argue that if $x neq 0$ then $frac x {|x|}$ has norm $1$ so $Tfrac x {|x|}=0$; finally, linearity of $T$ gives $frac {Tx} {|x|}=0$, so $Tx=0$.






          share|cite|improve this answer









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          • $begingroup$
            Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
            $endgroup$
            – Omojola Micheal
            Dec 18 '18 at 12:06










          • $begingroup$
            Isn't it obvious that $T=0$ implies $|T|=0$?
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:13











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          $begingroup$

          Your proof is fine except that you have to fill in some details on why $sup{|Tx|:|x|leq 1}$ implies $Tx=0$ for all $x$. You get $Tx=0$ for $|x|leq 1$ and then you have to argue that if $x neq 0$ then $frac x {|x|}$ has norm $1$ so $Tfrac x {|x|}=0$; finally, linearity of $T$ gives $frac {Tx} {|x|}=0$, so $Tx=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
            $endgroup$
            – Omojola Micheal
            Dec 18 '18 at 12:06










          • $begingroup$
            Isn't it obvious that $T=0$ implies $|T|=0$?
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:13
















          2












          $begingroup$

          Your proof is fine except that you have to fill in some details on why $sup{|Tx|:|x|leq 1}$ implies $Tx=0$ for all $x$. You get $Tx=0$ for $|x|leq 1$ and then you have to argue that if $x neq 0$ then $frac x {|x|}$ has norm $1$ so $Tfrac x {|x|}=0$; finally, linearity of $T$ gives $frac {Tx} {|x|}=0$, so $Tx=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
            $endgroup$
            – Omojola Micheal
            Dec 18 '18 at 12:06










          • $begingroup$
            Isn't it obvious that $T=0$ implies $|T|=0$?
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:13














          2












          2








          2





          $begingroup$

          Your proof is fine except that you have to fill in some details on why $sup{|Tx|:|x|leq 1}$ implies $Tx=0$ for all $x$. You get $Tx=0$ for $|x|leq 1$ and then you have to argue that if $x neq 0$ then $frac x {|x|}$ has norm $1$ so $Tfrac x {|x|}=0$; finally, linearity of $T$ gives $frac {Tx} {|x|}=0$, so $Tx=0$.






          share|cite|improve this answer









          $endgroup$



          Your proof is fine except that you have to fill in some details on why $sup{|Tx|:|x|leq 1}$ implies $Tx=0$ for all $x$. You get $Tx=0$ for $|x|leq 1$ and then you have to argue that if $x neq 0$ then $frac x {|x|}$ has norm $1$ so $Tfrac x {|x|}=0$; finally, linearity of $T$ gives $frac {Tx} {|x|}=0$, so $Tx=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 9:24









          Kavi Rama MurthyKavi Rama Murthy

          60k42161




          60k42161












          • $begingroup$
            Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
            $endgroup$
            – Omojola Micheal
            Dec 18 '18 at 12:06










          • $begingroup$
            Isn't it obvious that $T=0$ implies $|T|=0$?
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:13


















          • $begingroup$
            Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
            $endgroup$
            – Omojola Micheal
            Dec 18 '18 at 12:06










          • $begingroup$
            Isn't it obvious that $T=0$ implies $|T|=0$?
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 23:13
















          $begingroup$
          Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
          $endgroup$
          – Omojola Micheal
          Dec 18 '18 at 12:06




          $begingroup$
          Please, how do I get an "if and only if" statement from this? From what you've stated, I think I can only get implication from this. Can you, please, help?
          $endgroup$
          – Omojola Micheal
          Dec 18 '18 at 12:06












          $begingroup$
          Isn't it obvious that $T=0$ implies $|T|=0$?
          $endgroup$
          – Kavi Rama Murthy
          Dec 18 '18 at 23:13




          $begingroup$
          Isn't it obvious that $T=0$ implies $|T|=0$?
          $endgroup$
          – Kavi Rama Murthy
          Dec 18 '18 at 23:13


















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