Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The p-norm is given by $||x||_{p} = (\sum_{n=1}^{\infty }|x_{n}|^{p})^{1/p}$. For $0 < p < q$, it can be shown that $||x||_{p} \geq ||x||_{q}$ (1, 2). It sppears that in $R^{n}$ 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 $R^{n}$. For instance, for n=2 and n=3 one can see that for $0 < p < q$, the spheres with radius $\sqrt n$ with $||.||_{p}$ inscribe spheres with radius 1 with $||.||_{q}$.

It is not hard to prove 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 question is:

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

Notes:

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

share|improve this question

1 Answer 1

up vote 9 down vote accepted
  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.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.