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If $1\leq p, 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?

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    See [this post](http://math.stackexchange.com/q/4094); specifically AD.'s answer.2012-01-29

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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.

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

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    @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 $B(\ell_p)$).2012-01-30