Norm of a bounded linear functional.
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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try to make all your inequalities equalities
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– mathworker21
Dec 10 '18 at 7:24
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@mathworker21 I could not do. Please give some more hints.
$endgroup$
– Infinity
Dec 10 '18 at 7:28
add a comment |
$begingroup$
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
$endgroup$
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
asked Dec 10 '18 at 7:22
InfinityInfinity
607414
607414
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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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
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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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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $b$).
$endgroup$
add a comment |
$begingroup$
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 $b$).
$endgroup$
add a comment |
$begingroup$
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 $b$).
$endgroup$
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 $b$).
answered Dec 10 '18 at 7:34
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
58.2k42161
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$begingroup$
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