Evaluating the value of the norm of a function












0












$begingroup$


I am trying to show:
$$|f_alpha | = | alpha |_{ell^infty} $$
Given that:
$$f_alpha:mathbb{R}^2 rightarrow mathbb{R}, quad f_alpha = alpha_1 x_1 + alpha_2 x_2$$



($mathbb{R^2}$ is equiped with the $|.|_infty$)



I am struggling to get passed the definition of the operator norm:



$$|f|= sup limits_{substack{x in mathbb{R}^2\ |x|_{ell^infty} = 1}} |alpha_1 x_1 + alpha_2 x_2 | = sup limits_{substack{x in mathbb{R}^2\ x neq 0}} frac{ |alpha_1 x_1 + alpha_2 x_2 |}{|x |_{ell^infty}}$$
Could someone point me in the right direction?






real-analysis functional-analysis norm







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  • 1




    $begingroup$
    Is $alpha =(alpha_1,alpha_2)$?
    $endgroup$
    – herb steinberg
    Dec 2 '18 at 1:44












  • $begingroup$
    @herbsteinberg yes!
    $endgroup$
    – CucumberCactus
    Dec 2 '18 at 2:39










  • $begingroup$
    I don't think this is true. Let $alpha=(1,2)$. Then $|alpha|_infty=2$. However, note that for any $x in Bbb{R}^2$ with $|x|_{infty}=1$, we have $|f_alpha(x)| = |x_1+2x_2| leq 3$ and $|f_alpha(1,1)|=3$. Thus, $|f_alpha|=3 > 2 = |alpha|_infty$
    $endgroup$
    – Alonso Delfín
    Dec 2 '18 at 2:52


















0












$begingroup$


I am trying to show:
$$|f_alpha | = | alpha |_{ell^infty} $$
Given that:
$$f_alpha:mathbb{R}^2 rightarrow mathbb{R}, quad f_alpha = alpha_1 x_1 + alpha_2 x_2$$



($mathbb{R^2}$ is equiped with the $|.|_infty$)



I am struggling to get passed the definition of the operator norm:



$$|f|= sup limits_{substack{x in mathbb{R}^2\ |x|_{ell^infty} = 1}} |alpha_1 x_1 + alpha_2 x_2 | = sup limits_{substack{x in mathbb{R}^2\ x neq 0}} frac{ |alpha_1 x_1 + alpha_2 x_2 |}{|x |_{ell^infty}}$$
Could someone point me in the right direction?






real-analysis functional-analysis norm







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Is $alpha =(alpha_1,alpha_2)$?
    $endgroup$
    – herb steinberg
    Dec 2 '18 at 1:44












  • $begingroup$
    @herbsteinberg yes!
    $endgroup$
    – CucumberCactus
    Dec 2 '18 at 2:39










  • $begingroup$
    I don't think this is true. Let $alpha=(1,2)$. Then $|alpha|_infty=2$. However, note that for any $x in Bbb{R}^2$ with $|x|_{infty}=1$, we have $|f_alpha(x)| = |x_1+2x_2| leq 3$ and $|f_alpha(1,1)|=3$. Thus, $|f_alpha|=3 > 2 = |alpha|_infty$
    $endgroup$
    – Alonso Delfín
    Dec 2 '18 at 2:52
















0












0








0





$begingroup$


I am trying to show:
$$|f_alpha | = | alpha |_{ell^infty} $$
Given that:
$$f_alpha:mathbb{R}^2 rightarrow mathbb{R}, quad f_alpha = alpha_1 x_1 + alpha_2 x_2$$



($mathbb{R^2}$ is equiped with the $|.|_infty$)



I am struggling to get passed the definition of the operator norm:



$$|f|= sup limits_{substack{x in mathbb{R}^2\ |x|_{ell^infty} = 1}} |alpha_1 x_1 + alpha_2 x_2 | = sup limits_{substack{x in mathbb{R}^2\ x neq 0}} frac{ |alpha_1 x_1 + alpha_2 x_2 |}{|x |_{ell^infty}}$$
Could someone point me in the right direction?






real-analysis functional-analysis norm







share|cite|improve this question









$endgroup$




I am trying to show:
$$|f_alpha | = | alpha |_{ell^infty} $$
Given that:
$$f_alpha:mathbb{R}^2 rightarrow mathbb{R}, quad f_alpha = alpha_1 x_1 + alpha_2 x_2$$



($mathbb{R^2}$ is equiped with the $|.|_infty$)



I am struggling to get passed the definition of the operator norm:



$$|f|= sup limits_{substack{x in mathbb{R}^2\ |x|_{ell^infty} = 1}} |alpha_1 x_1 + alpha_2 x_2 | = sup limits_{substack{x in mathbb{R}^2\ x neq 0}} frac{ |alpha_1 x_1 + alpha_2 x_2 |}{|x |_{ell^infty}}$$
Could someone point me in the right direction?






real-analysis functional-analysis norm




real-analysis functional-analysis norm






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 0:59









CucumberCactusCucumberCactus

305




305








  • 1




    $begingroup$
    Is $alpha =(alpha_1,alpha_2)$?
    $endgroup$
    – herb steinberg
    Dec 2 '18 at 1:44












  • $begingroup$
    @herbsteinberg yes!
    $endgroup$
    – CucumberCactus
    Dec 2 '18 at 2:39










  • $begingroup$
    I don't think this is true. Let $alpha=(1,2)$. Then $|alpha|_infty=2$. However, note that for any $x in Bbb{R}^2$ with $|x|_{infty}=1$, we have $|f_alpha(x)| = |x_1+2x_2| leq 3$ and $|f_alpha(1,1)|=3$. Thus, $|f_alpha|=3 > 2 = |alpha|_infty$
    $endgroup$
    – Alonso Delfín
    Dec 2 '18 at 2:52
















  • 1




    $begingroup$
    Is $alpha =(alpha_1,alpha_2)$?
    $endgroup$
    – herb steinberg
    Dec 2 '18 at 1:44












  • $begingroup$
    @herbsteinberg yes!
    $endgroup$
    – CucumberCactus
    Dec 2 '18 at 2:39










  • $begingroup$
    I don't think this is true. Let $alpha=(1,2)$. Then $|alpha|_infty=2$. However, note that for any $x in Bbb{R}^2$ with $|x|_{infty}=1$, we have $|f_alpha(x)| = |x_1+2x_2| leq 3$ and $|f_alpha(1,1)|=3$. Thus, $|f_alpha|=3 > 2 = |alpha|_infty$
    $endgroup$
    – Alonso Delfín
    Dec 2 '18 at 2:52










1




1




$begingroup$
Is $alpha =(alpha_1,alpha_2)$?
$endgroup$
– herb steinberg
Dec 2 '18 at 1:44






$begingroup$
Is $alpha =(alpha_1,alpha_2)$?
$endgroup$
– herb steinberg
Dec 2 '18 at 1:44














$begingroup$
@herbsteinberg yes!
$endgroup$
– CucumberCactus
Dec 2 '18 at 2:39




$begingroup$
@herbsteinberg yes!
$endgroup$
– CucumberCactus
Dec 2 '18 at 2:39












$begingroup$
I don't think this is true. Let $alpha=(1,2)$. Then $|alpha|_infty=2$. However, note that for any $x in Bbb{R}^2$ with $|x|_{infty}=1$, we have $|f_alpha(x)| = |x_1+2x_2| leq 3$ and $|f_alpha(1,1)|=3$. Thus, $|f_alpha|=3 > 2 = |alpha|_infty$
$endgroup$
– Alonso Delfín
Dec 2 '18 at 2:52






$begingroup$
I don't think this is true. Let $alpha=(1,2)$. Then $|alpha|_infty=2$. However, note that for any $x in Bbb{R}^2$ with $|x|_{infty}=1$, we have $|f_alpha(x)| = |x_1+2x_2| leq 3$ and $|f_alpha(1,1)|=3$. Thus, $|f_alpha|=3 > 2 = |alpha|_infty$
$endgroup$
– Alonso Delfín
Dec 2 '18 at 2:52












1 Answer
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$begingroup$

As @Alonso Delfín mentioned, the desired conclusion is wrong. Actually this question is related to the so-called dual norm. By Hölder inequality, for $a,xinmathbb{R}^n$, there holds
$$
sum |a_ix_i|leq|a|_{ell^p}|x|_{ell^q},
$$
where $p^{-1}+q^{-1}=1$ for $1leq p,qleqinfty$ and for each $p,q$ and $a$ there exists $x$ establishing the equality. Therefore we have
$$
|langle a,cdotrangle|_{ell^p}=|a|_{ell^q},
$$
and that's why it is called the dual norm. As your case, it actually should be
$$
|langle a,cdotrangle|_{ell^infty}=|a|_{ell^1}.
$$






share|cite|improve this answer











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    2












    $begingroup$

    As @Alonso Delfín mentioned, the desired conclusion is wrong. Actually this question is related to the so-called dual norm. By Hölder inequality, for $a,xinmathbb{R}^n$, there holds
    $$
    sum |a_ix_i|leq|a|_{ell^p}|x|_{ell^q},
    $$
    where $p^{-1}+q^{-1}=1$ for $1leq p,qleqinfty$ and for each $p,q$ and $a$ there exists $x$ establishing the equality. Therefore we have
    $$
    |langle a,cdotrangle|_{ell^p}=|a|_{ell^q},
    $$
    and that's why it is called the dual norm. As your case, it actually should be
    $$
    |langle a,cdotrangle|_{ell^infty}=|a|_{ell^1}.
    $$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      As @Alonso Delfín mentioned, the desired conclusion is wrong. Actually this question is related to the so-called dual norm. By Hölder inequality, for $a,xinmathbb{R}^n$, there holds
      $$
      sum |a_ix_i|leq|a|_{ell^p}|x|_{ell^q},
      $$
      where $p^{-1}+q^{-1}=1$ for $1leq p,qleqinfty$ and for each $p,q$ and $a$ there exists $x$ establishing the equality. Therefore we have
      $$
      |langle a,cdotrangle|_{ell^p}=|a|_{ell^q},
      $$
      and that's why it is called the dual norm. As your case, it actually should be
      $$
      |langle a,cdotrangle|_{ell^infty}=|a|_{ell^1}.
      $$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        As @Alonso Delfín mentioned, the desired conclusion is wrong. Actually this question is related to the so-called dual norm. By Hölder inequality, for $a,xinmathbb{R}^n$, there holds
        $$
        sum |a_ix_i|leq|a|_{ell^p}|x|_{ell^q},
        $$
        where $p^{-1}+q^{-1}=1$ for $1leq p,qleqinfty$ and for each $p,q$ and $a$ there exists $x$ establishing the equality. Therefore we have
        $$
        |langle a,cdotrangle|_{ell^p}=|a|_{ell^q},
        $$
        and that's why it is called the dual norm. As your case, it actually should be
        $$
        |langle a,cdotrangle|_{ell^infty}=|a|_{ell^1}.
        $$






        share|cite|improve this answer











        $endgroup$



        As @Alonso Delfín mentioned, the desired conclusion is wrong. Actually this question is related to the so-called dual norm. By Hölder inequality, for $a,xinmathbb{R}^n$, there holds
        $$
        sum |a_ix_i|leq|a|_{ell^p}|x|_{ell^q},
        $$
        where $p^{-1}+q^{-1}=1$ for $1leq p,qleqinfty$ and for each $p,q$ and $a$ there exists $x$ establishing the equality. Therefore we have
        $$
        |langle a,cdotrangle|_{ell^p}=|a|_{ell^q},
        $$
        and that's why it is called the dual norm. As your case, it actually should be
        $$
        |langle a,cdotrangle|_{ell^infty}=|a|_{ell^1}.
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 3:05

























        answered Dec 2 '18 at 3:20









        JRenJRen

        1844




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