Lemma 3.9 Rudin's functional analysis












1












$begingroup$


I need help in understanding the proof of the following




Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
functionals on a vector space $X$. Let
$$ N = left{x : Lambda_1x =
ldots = Lambda_n = 0 right} $$ The following three are equivalent



a) There are scalars $alpha_1,ldots,alpha_n$ such that
$$ Lambda =
alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$



b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
> max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$



c) $Lambda x = 0$ for every $x in N$.




Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.



That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)



So again



Why (a) implies (b)? Is my proof of (b) implies (c) correct?






functional-analysis topological-vector-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I need help in understanding the proof of the following




    Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
    functionals on a vector space $X$. Let
    $$ N = left{x : Lambda_1x =
    ldots = Lambda_n = 0 right} $$     The following three are equivalent



    a) There are scalars $alpha_1,ldots,alpha_n$ such that
    $$ Lambda =
    alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$



    b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
    > max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$



    c) $Lambda x = 0$ for every $x in N$.




    Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.



    That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)



    So again



    Why (a) implies (b)? Is my proof of (b) implies (c) correct?






    functional-analysis topological-vector-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I need help in understanding the proof of the following




      Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
      functionals on a vector space $X$. Let
      $$ N = left{x : Lambda_1x =
      ldots = Lambda_n = 0 right} $$       The following three are equivalent



      a) There are scalars $alpha_1,ldots,alpha_n$ such that
      $$ Lambda =
      alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$



      b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
      > max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$



      c) $Lambda x = 0$ for every $x in N$.




      Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.



      That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)



      So again



      Why (a) implies (b)? Is my proof of (b) implies (c) correct?






      functional-analysis topological-vector-spaces







      share|cite|improve this question









      $endgroup$




      I need help in understanding the proof of the following




      Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
      functionals on a vector space $X$. Let
      $$ N = left{x : Lambda_1x =
      ldots = Lambda_n = 0 right} $$       The following three are equivalent



      a) There are scalars $alpha_1,ldots,alpha_n$ such that
      $$ Lambda =
      alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$



      b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
      > max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$



      c) $Lambda x = 0$ for every $x in N$.




      Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.



      That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)



      So again



      Why (a) implies (b)? Is my proof of (b) implies (c) correct?






      functional-analysis topological-vector-spaces




      functional-analysis topological-vector-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 10:02









      user8469759user8469759

      1,4031618




      1,4031618






















          2 Answers
          2






          active

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          1












          $begingroup$

          Your proof of (b) implies (c) is correct.



          As for (a) implies (b):



          We have
          $$
          | Lambda x |
          = | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
          \
          leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
          \
          leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
          $$
          This means we can choose
          $gamma=|alpha_1|+dots + |alpha_n|$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

              votes









              1












              $begingroup$

              Your proof of (b) implies (c) is correct.



              As for (a) implies (b):



              We have
              $$
              | Lambda x |
              = | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
              \
              leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
              \
              leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
              $$
              This means we can choose
              $gamma=|alpha_1|+dots + |alpha_n|$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Your proof of (b) implies (c) is correct.



                As for (a) implies (b):



                We have
                $$
                | Lambda x |
                = | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
                \
                leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
                \
                leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
                $$
                This means we can choose
                $gamma=|alpha_1|+dots + |alpha_n|$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Your proof of (b) implies (c) is correct.



                  As for (a) implies (b):



                  We have
                  $$
                  | Lambda x |
                  = | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
                  \
                  leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
                  \
                  leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
                  $$
                  This means we can choose
                  $gamma=|alpha_1|+dots + |alpha_n|$.






                  share|cite|improve this answer









                  $endgroup$



                  Your proof of (b) implies (c) is correct.



                  As for (a) implies (b):



                  We have
                  $$
                  | Lambda x |
                  = | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
                  \
                  leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
                  \
                  leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
                  $$
                  This means we can choose
                  $gamma=|alpha_1|+dots + |alpha_n|$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 10:17









                  supinfsupinf

                  6,3111028




                  6,3111028























                      1












                      $begingroup$

                      a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.






                          share|cite|improve this answer









                          $endgroup$



                          a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 10 '18 at 10:13









                          Kavi Rama MurthyKavi Rama Murthy

                          57.5k42160




                          57.5k42160






























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