Show that the $l^p$ norms with $1leq p,qleq infty$ are equivalent












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I want to show that for any $1leq p,qleq infty$, that there exists $a,b$ such that $aleft|xright|_p leq left|xright|_q leq bleft|xright|_p$ for all $x$ in $mathbb{R}^n$.

Is there a way to do this without Hölder's Inequalities?






inequality norm







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  • $begingroup$
    I am able to do this in the case of the 2-norm and infinity norm with a =1 and b =sqrt(n), is the proof similar to this specific case?
    $endgroup$
    – Sam.S
    Jan 16 at 22:31










  • $begingroup$
    Yes: it is quite elementary when $p=infty$, and the general case follows.
    $endgroup$
    – Mindlack
    Jan 16 at 22:32






  • 1




    $begingroup$
    What is $X$ here?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:34










  • $begingroup$
    Its the set of R^n
    $endgroup$
    – Sam.S
    Jan 16 at 22:36










  • $begingroup$
    Well, we do have to be careful. When e.g., $q =infty$ and $p=1$ the constants $a$ and $b$ may depend on $n$. I would imagine that the OP wants $a$ and $b$ to be of the form $a=a(p,q)$ and $b=b(p,q)$ where both $a,a^{-1},b,b^{-1}$ are functions of $p$ and $q$ only--no matter how large $n$ may be
    $endgroup$
    – Mike
    Jan 16 at 22:47


















1












$begingroup$


I want to show that for any $1leq p,qleq infty$, that there exists $a,b$ such that $aleft|xright|_p leq left|xright|_q leq bleft|xright|_p$ for all $x$ in $mathbb{R}^n$.

Is there a way to do this without Hölder's Inequalities?






inequality norm







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am able to do this in the case of the 2-norm and infinity norm with a =1 and b =sqrt(n), is the proof similar to this specific case?
    $endgroup$
    – Sam.S
    Jan 16 at 22:31










  • $begingroup$
    Yes: it is quite elementary when $p=infty$, and the general case follows.
    $endgroup$
    – Mindlack
    Jan 16 at 22:32






  • 1




    $begingroup$
    What is $X$ here?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:34










  • $begingroup$
    Its the set of R^n
    $endgroup$
    – Sam.S
    Jan 16 at 22:36










  • $begingroup$
    Well, we do have to be careful. When e.g., $q =infty$ and $p=1$ the constants $a$ and $b$ may depend on $n$. I would imagine that the OP wants $a$ and $b$ to be of the form $a=a(p,q)$ and $b=b(p,q)$ where both $a,a^{-1},b,b^{-1}$ are functions of $p$ and $q$ only--no matter how large $n$ may be
    $endgroup$
    – Mike
    Jan 16 at 22:47
















1












1








1


0



$begingroup$


I want to show that for any $1leq p,qleq infty$, that there exists $a,b$ such that $aleft|xright|_p leq left|xright|_q leq bleft|xright|_p$ for all $x$ in $mathbb{R}^n$.

Is there a way to do this without Hölder's Inequalities?






inequality norm







share|cite|improve this question











$endgroup$




I want to show that for any $1leq p,qleq infty$, that there exists $a,b$ such that $aleft|xright|_p leq left|xright|_q leq bleft|xright|_p$ for all $x$ in $mathbb{R}^n$.

Is there a way to do this without Hölder's Inequalities?






inequality norm




inequality norm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 22:42







Sam.S

















asked Jan 16 at 22:30









Sam.SSam.S

679




679












  • $begingroup$
    I am able to do this in the case of the 2-norm and infinity norm with a =1 and b =sqrt(n), is the proof similar to this specific case?
    $endgroup$
    – Sam.S
    Jan 16 at 22:31










  • $begingroup$
    Yes: it is quite elementary when $p=infty$, and the general case follows.
    $endgroup$
    – Mindlack
    Jan 16 at 22:32






  • 1




    $begingroup$
    What is $X$ here?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:34










  • $begingroup$
    Its the set of R^n
    $endgroup$
    – Sam.S
    Jan 16 at 22:36










  • $begingroup$
    Well, we do have to be careful. When e.g., $q =infty$ and $p=1$ the constants $a$ and $b$ may depend on $n$. I would imagine that the OP wants $a$ and $b$ to be of the form $a=a(p,q)$ and $b=b(p,q)$ where both $a,a^{-1},b,b^{-1}$ are functions of $p$ and $q$ only--no matter how large $n$ may be
    $endgroup$
    – Mike
    Jan 16 at 22:47




















  • $begingroup$
    I am able to do this in the case of the 2-norm and infinity norm with a =1 and b =sqrt(n), is the proof similar to this specific case?
    $endgroup$
    – Sam.S
    Jan 16 at 22:31










  • $begingroup$
    Yes: it is quite elementary when $p=infty$, and the general case follows.
    $endgroup$
    – Mindlack
    Jan 16 at 22:32






  • 1




    $begingroup$
    What is $X$ here?
    $endgroup$
    – Lorenzo Quarisa
    Jan 16 at 22:34










  • $begingroup$
    Its the set of R^n
    $endgroup$
    – Sam.S
    Jan 16 at 22:36










  • $begingroup$
    Well, we do have to be careful. When e.g., $q =infty$ and $p=1$ the constants $a$ and $b$ may depend on $n$. I would imagine that the OP wants $a$ and $b$ to be of the form $a=a(p,q)$ and $b=b(p,q)$ where both $a,a^{-1},b,b^{-1}$ are functions of $p$ and $q$ only--no matter how large $n$ may be
    $endgroup$
    – Mike
    Jan 16 at 22:47


















$begingroup$
I am able to do this in the case of the 2-norm and infinity norm with a =1 and b =sqrt(n), is the proof similar to this specific case?
$endgroup$
– Sam.S
Jan 16 at 22:31




$begingroup$
I am able to do this in the case of the 2-norm and infinity norm with a =1 and b =sqrt(n), is the proof similar to this specific case?
$endgroup$
– Sam.S
Jan 16 at 22:31












$begingroup$
Yes: it is quite elementary when $p=infty$, and the general case follows.
$endgroup$
– Mindlack
Jan 16 at 22:32




$begingroup$
Yes: it is quite elementary when $p=infty$, and the general case follows.
$endgroup$
– Mindlack
Jan 16 at 22:32




1




1




$begingroup$
What is $X$ here?
$endgroup$
– Lorenzo Quarisa
Jan 16 at 22:34




$begingroup$
What is $X$ here?
$endgroup$
– Lorenzo Quarisa
Jan 16 at 22:34












$begingroup$
Its the set of R^n
$endgroup$
– Sam.S
Jan 16 at 22:36




$begingroup$
Its the set of R^n
$endgroup$
– Sam.S
Jan 16 at 22:36












$begingroup$
Well, we do have to be careful. When e.g., $q =infty$ and $p=1$ the constants $a$ and $b$ may depend on $n$. I would imagine that the OP wants $a$ and $b$ to be of the form $a=a(p,q)$ and $b=b(p,q)$ where both $a,a^{-1},b,b^{-1}$ are functions of $p$ and $q$ only--no matter how large $n$ may be
$endgroup$
– Mike
Jan 16 at 22:47






$begingroup$
Well, we do have to be careful. When e.g., $q =infty$ and $p=1$ the constants $a$ and $b$ may depend on $n$. I would imagine that the OP wants $a$ and $b$ to be of the form $a=a(p,q)$ and $b=b(p,q)$ where both $a,a^{-1},b,b^{-1}$ are functions of $p$ and $q$ only--no matter how large $n$ may be
$endgroup$
– Mike
Jan 16 at 22:47












1 Answer
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$|x|_{infty} leq 1$ implies $|x|_{p} leq n^{1/p}$ for any $p in [1,infty)$. Use this and a scaling argument to show that $|x|_{p} leq n^{1/p} |x|_{infty}$. [You have to consider $y=frac x {|x|_{infty}}$ when $x neq 0$]. On the other hand, $ |x|_{infty} leq |x|_{p} $ because $|x_i| leq |x|_{p} $ for each $i$. Hence all of the $p$-norms with $p<infty$ are equivalent to $infty$ norm. This implies that any two of the $p$-norms are equivalent. [ $|x|' leq c|x|''$ and $|x|'' leq d|x|'''$ implies $|x|' leq cd|x|'''$ etc].






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    $begingroup$

    $|x|_{infty} leq 1$ implies $|x|_{p} leq n^{1/p}$ for any $p in [1,infty)$. Use this and a scaling argument to show that $|x|_{p} leq n^{1/p} |x|_{infty}$. [You have to consider $y=frac x {|x|_{infty}}$ when $x neq 0$]. On the other hand, $ |x|_{infty} leq |x|_{p} $ because $|x_i| leq |x|_{p} $ for each $i$. Hence all of the $p$-norms with $p<infty$ are equivalent to $infty$ norm. This implies that any two of the $p$-norms are equivalent. [ $|x|' leq c|x|''$ and $|x|'' leq d|x|'''$ implies $|x|' leq cd|x|'''$ etc].






    share|cite|improve this answer











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      1












      $begingroup$

      $|x|_{infty} leq 1$ implies $|x|_{p} leq n^{1/p}$ for any $p in [1,infty)$. Use this and a scaling argument to show that $|x|_{p} leq n^{1/p} |x|_{infty}$. [You have to consider $y=frac x {|x|_{infty}}$ when $x neq 0$]. On the other hand, $ |x|_{infty} leq |x|_{p} $ because $|x_i| leq |x|_{p} $ for each $i$. Hence all of the $p$-norms with $p<infty$ are equivalent to $infty$ norm. This implies that any two of the $p$-norms are equivalent. [ $|x|' leq c|x|''$ and $|x|'' leq d|x|'''$ implies $|x|' leq cd|x|'''$ etc].






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $|x|_{infty} leq 1$ implies $|x|_{p} leq n^{1/p}$ for any $p in [1,infty)$. Use this and a scaling argument to show that $|x|_{p} leq n^{1/p} |x|_{infty}$. [You have to consider $y=frac x {|x|_{infty}}$ when $x neq 0$]. On the other hand, $ |x|_{infty} leq |x|_{p} $ because $|x_i| leq |x|_{p} $ for each $i$. Hence all of the $p$-norms with $p<infty$ are equivalent to $infty$ norm. This implies that any two of the $p$-norms are equivalent. [ $|x|' leq c|x|''$ and $|x|'' leq d|x|'''$ implies $|x|' leq cd|x|'''$ etc].






        share|cite|improve this answer











        $endgroup$



        $|x|_{infty} leq 1$ implies $|x|_{p} leq n^{1/p}$ for any $p in [1,infty)$. Use this and a scaling argument to show that $|x|_{p} leq n^{1/p} |x|_{infty}$. [You have to consider $y=frac x {|x|_{infty}}$ when $x neq 0$]. On the other hand, $ |x|_{infty} leq |x|_{p} $ because $|x_i| leq |x|_{p} $ for each $i$. Hence all of the $p$-norms with $p<infty$ are equivalent to $infty$ norm. This implies that any two of the $p$-norms are equivalent. [ $|x|' leq c|x|''$ and $|x|'' leq d|x|'''$ implies $|x|' leq cd|x|'''$ etc].







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 at 23:37

























        answered Jan 16 at 23:24









        Kavi Rama MurthyKavi Rama Murthy

        59k42161




        59k42161






























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