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If $1\leq p<q$, show that the unit ball $l_{n}^p(\mathbb{R})$ is contained in the unit ball $l_{n}^q(\mathbb{R})$.

Well the definition of $l_{n}^p(\mathbb{R})$ is that for $d_{p}(x,y)=\max_{1\leq j\leq n}|x_{j}-y_{j}|=||x-y||_{p}$, then $l_{n}^p(\mathbb{R})$ is the space $(\mathbb{R^n},d_{p})$.

If $q>p$ intuitively the ball around $l_{n}^p(\mathbb{R})$ would be smaller than the ball around $l_{n}^q(\mathbb{R})$, though I am not exactly sure where to get started in proving that. Do I need to use Holder's Inequality somehow?

Srivatsan
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Emir
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1 Answers1

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If $x\in l_p$, then

$$\tag{1}\sum |x_i|^q=\sum |x_i|^p|x_i|^{q-p}\le \Vert x\Vert_\infty^{q-p}\sum|x_i|^p$$

Suppose now that $\Vert x \Vert_p\le 1$. Then
$$\sum|x_i|^p\le 1,\quad\text{ and } \Vert x \Vert_\infty\le 1.$$ Thus, since $q-p>0$, we have from inequality $(1)$ that $$ \sum |x_i|^q\le 1; $$ which implies that $\Vert x\Vert_q\le1$.

Note that this argument holds for $0<p<q<\infty$.

Also, one can prove a stronger result, by other means (see the post linked to by Srivatsan in the comments), that $\Vert x\Vert_q\le \Vert x\Vert_p$.

David Mitra
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  • I might be a bit slow right now, but shouldn't we instead prove that $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$? – Srivatsan Jan 29 '12 at 23:31
  • @Srivatsan I may be off. I thought we needed to show that if $\Vert x\Vert_p\le 1$, then $\Vert x\Vert_q\le 1$ (the unit ball of $l_p$ is contained in the unit ball of $l_q$). – David Mitra Jan 29 '12 at 23:34
  • Ah, I see it now. I think both the arguments work. (If we are able to show that $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$, then it again follows that $l^p \subseteq l^q$.) Sorry about the noise. By the way, you might be interested in AD.'s different proof in the linked post. – Srivatsan Jan 29 '12 at 23:38
  • Looking back at this I'm a little confused -- why is it the case that $||x||_q\leq||x||_p$ for $q>p$ and not $||x||_q\geq||x||_p$? – Emir Jan 30 '12 at 22:20
  • @Emir $B(\ell_p)\subset B(\ell_q)$ means if $\Vert x\Vert_p\le1$, then $\Vert x\Vert_q\le1$. This latter condition is implied by $\Vert x\Vert_q\le\Vert x\Vert_p$. – David Mitra Jan 30 '12 at 22:50
  • @Emir if you're thinking about a venn diagram with the "$B(\ell_p)$" circle inside the $B(\ell_q)$ circle", take a point between the two circles. It's $\ell_q$ norm is $\le 1$ but its $\ell_p$ norm is $\ge 1$ (since its outside $B(\ell_p)$). – David Mitra Jan 30 '12 at 22:58