Norm of a bounded linear functional.












1












$begingroup$


Let $X=(mathbb R^2, |.|_3)$ be a real normed space, where $|(x_1,x_2)|_3=[|x_1|^3+|x_2|^3]^{1/3}$. How to find the norm of bounded linear functional $ax+by$?



I tried this way:
$|ax+by|leq |a||x|+|b||y|leq max{|a|,|b|}(|x|+|y|)leqmax{|a|,|b|}([|x|^3+|y|^3]^{1/3}+|y|)leq max{|a|,|b|}([|x|^3+|y|^3]^{1/3}+[|x|^3+|y|^3]^{1/3})= 2max{|a|,|b|}([|x|^3+|y|^3]^{1/3})=2max{|a|,|b|} |(x,y)|_3 $.
This implies $ax+by$ is bounded linear functional. Now how should I find the norm of this functional?






functional-analysis banach-spaces norm normed-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    try to make all your inequalities equalities
    $endgroup$
    – mathworker21
    Dec 10 '18 at 7:24










  • $begingroup$
    @mathworker21 I could not do. Please give some more hints.
    $endgroup$
    – Infinity
    Dec 10 '18 at 7:28


















1












$begingroup$


Let $X=(mathbb R^2, |.|_3)$ be a real normed space, where $|(x_1,x_2)|_3=[|x_1|^3+|x_2|^3]^{1/3}$. How to find the norm of bounded linear functional $ax+by$?



I tried this way:
$|ax+by|leq |a||x|+|b||y|leq max{|a|,|b|}(|x|+|y|)leqmax{|a|,|b|}([|x|^3+|y|^3]^{1/3}+|y|)leq max{|a|,|b|}([|x|^3+|y|^3]^{1/3}+[|x|^3+|y|^3]^{1/3})= 2max{|a|,|b|}([|x|^3+|y|^3]^{1/3})=2max{|a|,|b|} |(x,y)|_3 $.
This implies $ax+by$ is bounded linear functional. Now how should I find the norm of this functional?






functional-analysis banach-spaces norm normed-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    try to make all your inequalities equalities
    $endgroup$
    – mathworker21
    Dec 10 '18 at 7:24










  • $begingroup$
    @mathworker21 I could not do. Please give some more hints.
    $endgroup$
    – Infinity
    Dec 10 '18 at 7:28
















1












1








1





$begingroup$


Let $X=(mathbb R^2, |.|_3)$ be a real normed space, where $|(x_1,x_2)|_3=[|x_1|^3+|x_2|^3]^{1/3}$. How to find the norm of bounded linear functional $ax+by$?



I tried this way:
$|ax+by|leq |a||x|+|b||y|leq max{|a|,|b|}(|x|+|y|)leqmax{|a|,|b|}([|x|^3+|y|^3]^{1/3}+|y|)leq max{|a|,|b|}([|x|^3+|y|^3]^{1/3}+[|x|^3+|y|^3]^{1/3})= 2max{|a|,|b|}([|x|^3+|y|^3]^{1/3})=2max{|a|,|b|} |(x,y)|_3 $.
This implies $ax+by$ is bounded linear functional. Now how should I find the norm of this functional?






functional-analysis banach-spaces norm normed-spaces







share|cite|improve this question









$endgroup$




Let $X=(mathbb R^2, |.|_3)$ be a real normed space, where $|(x_1,x_2)|_3=[|x_1|^3+|x_2|^3]^{1/3}$. How to find the norm of bounded linear functional $ax+by$?



I tried this way:
$|ax+by|leq |a||x|+|b||y|leq max{|a|,|b|}(|x|+|y|)leqmax{|a|,|b|}([|x|^3+|y|^3]^{1/3}+|y|)leq max{|a|,|b|}([|x|^3+|y|^3]^{1/3}+[|x|^3+|y|^3]^{1/3})= 2max{|a|,|b|}([|x|^3+|y|^3]^{1/3})=2max{|a|,|b|} |(x,y)|_3 $.
This implies $ax+by$ is bounded linear functional. Now how should I find the norm of this functional?






functional-analysis banach-spaces norm normed-spaces




functional-analysis banach-spaces norm normed-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 7:22









InfinityInfinity

607414




607414












  • $begingroup$
    try to make all your inequalities equalities
    $endgroup$
    – mathworker21
    Dec 10 '18 at 7:24










  • $begingroup$
    @mathworker21 I could not do. Please give some more hints.
    $endgroup$
    – Infinity
    Dec 10 '18 at 7:28




















  • $begingroup$
    try to make all your inequalities equalities
    $endgroup$
    – mathworker21
    Dec 10 '18 at 7:24










  • $begingroup$
    @mathworker21 I could not do. Please give some more hints.
    $endgroup$
    – Infinity
    Dec 10 '18 at 7:28


















$begingroup$
try to make all your inequalities equalities
$endgroup$
– mathworker21
Dec 10 '18 at 7:24




$begingroup$
try to make all your inequalities equalities
$endgroup$
– mathworker21
Dec 10 '18 at 7:24












$begingroup$
@mathworker21 I could not do. Please give some more hints.
$endgroup$
– Infinity
Dec 10 '18 at 7:28






$begingroup$
@mathworker21 I could not do. Please give some more hints.
$endgroup$
– Infinity
Dec 10 '18 at 7:28












1 Answer
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Hints: use the inequality $|ax+by| leq (|x|^{3}+|y|^{3} )^{1/3} (|a|^{3/2}+|b|^{3/2} )^{2/3}$ (This is Holder's Inequality) to see that the norm of the functional is less than or equal to $(|a|^{3/2}+|b|^{3/2} )^{2/3}$. To show that the the norm is exactly equal to this number take $x=pm t|a|^{1/2},y=pm t|b|^{1/2}$ where $t =frac 1 {|(|a|^{1/2},|b|^{1/2})}|}$. (In the first term $pm$ stands for $1$ if $a >0$, $-1$ otherwise. Similarly choose $pm$ sign in the second term based on the sign of $b$).






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    $begingroup$

    Hints: use the inequality $|ax+by| leq (|x|^{3}+|y|^{3} )^{1/3} (|a|^{3/2}+|b|^{3/2} )^{2/3}$ (This is Holder's Inequality) to see that the norm of the functional is less than or equal to $(|a|^{3/2}+|b|^{3/2} )^{2/3}$. To show that the the norm is exactly equal to this number take $x=pm t|a|^{1/2},y=pm t|b|^{1/2}$ where $t =frac 1 {|(|a|^{1/2},|b|^{1/2})}|}$. (In the first term $pm$ stands for $1$ if $a >0$, $-1$ otherwise. Similarly choose $pm$ sign in the second term based on the sign of $b$).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hints: use the inequality $|ax+by| leq (|x|^{3}+|y|^{3} )^{1/3} (|a|^{3/2}+|b|^{3/2} )^{2/3}$ (This is Holder's Inequality) to see that the norm of the functional is less than or equal to $(|a|^{3/2}+|b|^{3/2} )^{2/3}$. To show that the the norm is exactly equal to this number take $x=pm t|a|^{1/2},y=pm t|b|^{1/2}$ where $t =frac 1 {|(|a|^{1/2},|b|^{1/2})}|}$. (In the first term $pm$ stands for $1$ if $a >0$, $-1$ otherwise. Similarly choose $pm$ sign in the second term based on the sign of $b$).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hints: use the inequality $|ax+by| leq (|x|^{3}+|y|^{3} )^{1/3} (|a|^{3/2}+|b|^{3/2} )^{2/3}$ (This is Holder's Inequality) to see that the norm of the functional is less than or equal to $(|a|^{3/2}+|b|^{3/2} )^{2/3}$. To show that the the norm is exactly equal to this number take $x=pm t|a|^{1/2},y=pm t|b|^{1/2}$ where $t =frac 1 {|(|a|^{1/2},|b|^{1/2})}|}$. (In the first term $pm$ stands for $1$ if $a >0$, $-1$ otherwise. Similarly choose $pm$ sign in the second term based on the sign of $b$).






        share|cite|improve this answer









        $endgroup$



        Hints: use the inequality $|ax+by| leq (|x|^{3}+|y|^{3} )^{1/3} (|a|^{3/2}+|b|^{3/2} )^{2/3}$ (This is Holder's Inequality) to see that the norm of the functional is less than or equal to $(|a|^{3/2}+|b|^{3/2} )^{2/3}$. To show that the the norm is exactly equal to this number take $x=pm t|a|^{1/2},y=pm t|b|^{1/2}$ where $t =frac 1 {|(|a|^{1/2},|b|^{1/2})}|}$. (In the first term $pm$ stands for $1$ if $a >0$, $-1$ otherwise. Similarly choose $pm$ sign in the second term based on the sign of $b$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 7:34









        Kavi Rama MurthyKavi Rama Murthy

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