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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?


$||x||_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + (|x_{1}|^2 + |x_{2}|^2) = 2|x_{1}|^2 + 2|x_{2}|^2 = 2 ||x||_{2}^{2}$, 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.

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1 Answer 1

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  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.
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Could you please comment on why C is the best possible constant? – Arun Jun 10 at 14:18
@Arun, because for $x=(1,1,1,\ldots,1)$ this bound is attained – Norbert Jun 10 at 14:27
@Norbert: I'm trying to derive a similar bound for the case $p>q$. Any ideas about how to proceed ? – Ashok Vardhan Jun 22 at 12:19
@AshokVardhan what does "similar" mean? Do you want to prove something like $\Vert x\Vert_p \leq C\Vert x\Vert_q$ for $p>q$ ? – Norbert Jun 23 at 14:24
@Norbert: Yes, by similar, I meant a bound that you mentioned in the above comment. Thanks. – Ashok Vardhan Jun 23 at 14:43

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