The $p$-norm on $\mathbb R^n$ is given by $\|x\|_{p}=\big(\sum_{k=1}^n |x_{k}|^p\big)^{1/p}$. For $0 < p < q$ it can be shown that $\|x\|_p\geq\|x\|_q$ (1, 2). It appears that in $\mathbb{R}^n$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$(3), $\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $\mathbb R^n$. For instance, for $n=2$ and $n=3$ one can see that for $0 < p < q$, the spheres with radius $\sqrt n$ with $\|\cdot\|_p$ inscribe spheres with radius $1$ with $\|\cdot\|_q$.

It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $n=2$ it is easily proven (see below), but not for $n>2$. So my questions are:

  1. How can relation (3) be proven for arbitrary $n\,$?
  2. Can this be generalized into something of the form $\|x\|_{p} \leq C \|x\|_{q}$ for arbitrary $0<p < q\,$?
  3. Do any of the relations also hold for infinite-dimensional spaces, i.e. in $l^p$ spaces?


$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$, hence $=2\|x\|_{2}^{2}$
$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$. This works because $|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$, but only because $(|x_{1}| - |x_{2}|)^2 \geq 0$, while for more than two terms $\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$ gives an inequality that never gives the right signs for the cross terms.

  • $\begingroup$ Your first link (1) has exactly the equation you seek. Actually a version of Norberts excellent answer, generalized to measure spaces. $\endgroup$ Mar 31, 2016 at 20:57
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    $\begingroup$ Probably a dead thread, but norms in finite dimensional spaces are equivalent, so one can always find c ad C such that $c\|x\|_p\leq \|x\|_q\leq C\|x\|_p$. $\endgroup$
    – BigM
    Dec 1, 2019 at 20:23

1 Answer 1

  1. Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$ $$ \Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2 $$
  2. Such a bound does exist. Recall Hölder's inequality $$ \sum\limits_{i=1}^n |a_i||b_i|\leq \left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}} $$ Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$ $$ \sum\limits_{i=1}^n |x_i|^p= \sum\limits_{i=1}^n |x_i|^p\cdot 1\leq \left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}} \left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}} $$ Then $$ \Vert x\Vert_p= \left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq \left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\= n^{1/p-1/q}\Vert x\Vert_q $$ In fact $C=n^{1/p-1/q}$ is the best possible constant.
  3. For infinite dimensional case such inequality doesn't hold. For explanation see this answer.
  • $\begingroup$ Could you please comment on why C is the best possible constant? $\endgroup$
    – Arun
    Jun 10, 2015 at 14:18
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    $\begingroup$ @Arun, because for $x=(1,1,1,\ldots,1)$ this bound is attained $\endgroup$
    – Norbert
    Jun 10, 2015 at 14:27
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    $\begingroup$ @Norbert: I'm trying to derive a similar bound for the case $p>q$. Any ideas about how to proceed ? $\endgroup$ Jun 22, 2015 at 12:19
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    $\begingroup$ @AshokVardhan what does "similar" mean? Do you want to prove something like $\Vert x\Vert_p \leq C\Vert x\Vert_q$ for $p>q$ ? $\endgroup$
    – Norbert
    Jun 23, 2015 at 14:24
  • $\begingroup$ @Norbert: Yes, by similar, I meant a bound that you mentioned in the above comment. Thanks. $\endgroup$ Jun 23, 2015 at 14:43

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