The operator norm of the composition of linear bounded operators between Banach spaces.











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The set $B(X, Y )$ is a normed linear space with the operator norm.



If $T ∈ B(X, Y)$ and $S ∈ B(Y, Z)$ for $X, Y , Z$ normed linear spaces, then the composition
$ST ∈ B(X, Z)$ and $|ST| ≤ |S| |T|.$



I don't know how to prove this.



Here is my trial.



Since the composition if exist of 2 linear operators is a linear operator then $ST$ is a linear operator. To prove it is bounded we prove that $|ST x| < infty$.



if $|T| = sup_{|x|=1}|Tx| < infty$.



and $|S| = sup_{|y|=1}|Sy| < infty$.



I don't know how to continue.






functional-analysis proof-verification operator-theory linear-transformations







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    up vote
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    down vote

    favorite












    The set $B(X, Y )$ is a normed linear space with the operator norm.



    If $T ∈ B(X, Y)$ and $S ∈ B(Y, Z)$ for $X, Y , Z$ normed linear spaces, then the composition
    $ST ∈ B(X, Z)$ and $|ST| ≤ |S| |T|.$



    I don't know how to prove this.



    Here is my trial.



    Since the composition if exist of 2 linear operators is a linear operator then $ST$ is a linear operator. To prove it is bounded we prove that $|ST x| < infty$.



    if $|T| = sup_{|x|=1}|Tx| < infty$.



    and $|S| = sup_{|y|=1}|Sy| < infty$.



    I don't know how to continue.






    functional-analysis proof-verification operator-theory linear-transformations







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      The set $B(X, Y )$ is a normed linear space with the operator norm.



      If $T ∈ B(X, Y)$ and $S ∈ B(Y, Z)$ for $X, Y , Z$ normed linear spaces, then the composition
      $ST ∈ B(X, Z)$ and $|ST| ≤ |S| |T|.$



      I don't know how to prove this.



      Here is my trial.



      Since the composition if exist of 2 linear operators is a linear operator then $ST$ is a linear operator. To prove it is bounded we prove that $|ST x| < infty$.



      if $|T| = sup_{|x|=1}|Tx| < infty$.



      and $|S| = sup_{|y|=1}|Sy| < infty$.



      I don't know how to continue.






      functional-analysis proof-verification operator-theory linear-transformations







      share|cite|improve this question















      The set $B(X, Y )$ is a normed linear space with the operator norm.



      If $T ∈ B(X, Y)$ and $S ∈ B(Y, Z)$ for $X, Y , Z$ normed linear spaces, then the composition
      $ST ∈ B(X, Z)$ and $|ST| ≤ |S| |T|.$



      I don't know how to prove this.



      Here is my trial.



      Since the composition if exist of 2 linear operators is a linear operator then $ST$ is a linear operator. To prove it is bounded we prove that $|ST x| < infty$.



      if $|T| = sup_{|x|=1}|Tx| < infty$.



      and $|S| = sup_{|y|=1}|Sy| < infty$.



      I don't know how to continue.






      functional-analysis proof-verification operator-theory linear-transformations




      functional-analysis proof-verification operator-theory linear-transformations






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      share|cite|improve this question













      share|cite|improve this question




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      edited 2 days ago









      Yaddle

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      asked 2 days ago









      HybridAlien

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          Let $0 neq x in X$. Then you have that $Vert frac{x}{Vert x Vert} Vert = 1$. Thus we obtain
          $$frac{1}{Vert x Vert} Vert Tx Vert = Vert Tfrac{x}{Vert x Vert} Vert leq Vert T Vert.$$
          This implies shows that $Vert Tx Vert leq Vert T Vert Vert x Vert$ for all $x in X$. (Note: For $x = 0$ this is trivial.) Now from this it follows that
          $$ Vert STx Vert leq Vert S Vert Vert T x Vert leq Vert S Vert Vert T Vert Vert x Vert qquad text{for all } x in X.$$
          Finally, this implies
          $$ Vert ST Vert = sup_{Vert x Vert = 1} Vert STx Vert leq sup_{Vert x Vert = 1} Vert S Vert Vert T Vert Vert x Vert = Vert S Vert Vert T Vert$$
          and you are since the equality above implies $Vert ST Vert leq Vert S Vert Vert T Vert < infty$.






          share|cite|improve this answer





















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            Let $0 neq x in X$. Then you have that $Vert frac{x}{Vert x Vert} Vert = 1$. Thus we obtain
            $$frac{1}{Vert x Vert} Vert Tx Vert = Vert Tfrac{x}{Vert x Vert} Vert leq Vert T Vert.$$
            This implies shows that $Vert Tx Vert leq Vert T Vert Vert x Vert$ for all $x in X$. (Note: For $x = 0$ this is trivial.) Now from this it follows that
            $$ Vert STx Vert leq Vert S Vert Vert T x Vert leq Vert S Vert Vert T Vert Vert x Vert qquad text{for all } x in X.$$
            Finally, this implies
            $$ Vert ST Vert = sup_{Vert x Vert = 1} Vert STx Vert leq sup_{Vert x Vert = 1} Vert S Vert Vert T Vert Vert x Vert = Vert S Vert Vert T Vert$$
            and you are since the equality above implies $Vert ST Vert leq Vert S Vert Vert T Vert < infty$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Let $0 neq x in X$. Then you have that $Vert frac{x}{Vert x Vert} Vert = 1$. Thus we obtain
              $$frac{1}{Vert x Vert} Vert Tx Vert = Vert Tfrac{x}{Vert x Vert} Vert leq Vert T Vert.$$
              This implies shows that $Vert Tx Vert leq Vert T Vert Vert x Vert$ for all $x in X$. (Note: For $x = 0$ this is trivial.) Now from this it follows that
              $$ Vert STx Vert leq Vert S Vert Vert T x Vert leq Vert S Vert Vert T Vert Vert x Vert qquad text{for all } x in X.$$
              Finally, this implies
              $$ Vert ST Vert = sup_{Vert x Vert = 1} Vert STx Vert leq sup_{Vert x Vert = 1} Vert S Vert Vert T Vert Vert x Vert = Vert S Vert Vert T Vert$$
              and you are since the equality above implies $Vert ST Vert leq Vert S Vert Vert T Vert < infty$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Let $0 neq x in X$. Then you have that $Vert frac{x}{Vert x Vert} Vert = 1$. Thus we obtain
                $$frac{1}{Vert x Vert} Vert Tx Vert = Vert Tfrac{x}{Vert x Vert} Vert leq Vert T Vert.$$
                This implies shows that $Vert Tx Vert leq Vert T Vert Vert x Vert$ for all $x in X$. (Note: For $x = 0$ this is trivial.) Now from this it follows that
                $$ Vert STx Vert leq Vert S Vert Vert T x Vert leq Vert S Vert Vert T Vert Vert x Vert qquad text{for all } x in X.$$
                Finally, this implies
                $$ Vert ST Vert = sup_{Vert x Vert = 1} Vert STx Vert leq sup_{Vert x Vert = 1} Vert S Vert Vert T Vert Vert x Vert = Vert S Vert Vert T Vert$$
                and you are since the equality above implies $Vert ST Vert leq Vert S Vert Vert T Vert < infty$.






                share|cite|improve this answer












                Let $0 neq x in X$. Then you have that $Vert frac{x}{Vert x Vert} Vert = 1$. Thus we obtain
                $$frac{1}{Vert x Vert} Vert Tx Vert = Vert Tfrac{x}{Vert x Vert} Vert leq Vert T Vert.$$
                This implies shows that $Vert Tx Vert leq Vert T Vert Vert x Vert$ for all $x in X$. (Note: For $x = 0$ this is trivial.) Now from this it follows that
                $$ Vert STx Vert leq Vert S Vert Vert T x Vert leq Vert S Vert Vert T Vert Vert x Vert qquad text{for all } x in X.$$
                Finally, this implies
                $$ Vert ST Vert = sup_{Vert x Vert = 1} Vert STx Vert leq sup_{Vert x Vert = 1} Vert S Vert Vert T Vert Vert x Vert = Vert S Vert Vert T Vert$$
                and you are since the equality above implies $Vert ST Vert leq Vert S Vert Vert T Vert < infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Yaddle

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