Unit circles of different norms- Subsets

**chrischen88** · October 13th 2009, 09:51 PM

Show that $B_1(0;1) \subset B_2(0;1) \subset B_{\infty}(0,1)$ in $\mathbb{R}^n$ for all $n$ . I should clarify some notation here $B(0,1)$ means the open ball with center 0, and radius 1.

Definition of norms:
$d_1(x,y) := ||x - y||_1 = \displaystyle\sum_{j=1}^n|x_j - y_j|;$
$d_2(x,y) := ||x - y||_2 = \left(\displaystyle\sum_{j=1}^n|x_j - y_j|^2\right)^{1/2}$ ;
$d_{\infty}(x,y) := \displaystyle\max_{1\leq j \leq n}|x_j - y_j|$

My attempt:
I know that to show they are subsets of each other, I have to show the implication $\forall b \in B_1(0,1) \Rightarrow b \in B_2(0,1)$ as well and next $\forall b \in B_2(0,1) \Rightarrow b \in B_{\infty}(0,1)$ as well.

So I have, let $b \in B_1(0,1)$ , then $|b_1| + |b_2| + \cdots + |b_n| < 1$ , I just don't see how this would imply $\left(|b_1|^2 + |b_2|^2 + \cdots + |b_n|^2\right)^{1/2} < 1$ or $|b_1|^2 + |b_2|^2 + \cdots + |b_n|^2 < 1$

And I'm also unsure as to how to show $\forall b \in B_2(0,1) \Rightarrow b \in B_{\infty}(0,1)$ .

Hints?

**CaptainBlack** · October 13th 2009, 11:15 PM

Originally Posted by chrischen88

Show that $B_1(0;1) \subset B_2(0;1) \subset B_{\infty}(0,1)$ in $\mathbb{R}^n$ for all $n$ . I should clarify some notation here $B(0,1)$ means the open ball with center 0, and radius 1.

Definition of norms:
$d_1(x,y) := ||x - y||_1 = \displaystyle\sum_{j=1}^n|x_j - y_j|;$
$d_2(x,y) := ||x - y||_2 = \left(\displaystyle\sum_{j=1}^n|x_j - y_j|^2\right)^{1/2}$ ;
$d_{\infty}(x,y) := \displaystyle\max_{1\leq j \leq n}|x_j - y_j|$

My attempt:
I know that to show they are subsets of each other, I have to show the implication $\forall b \in B_1(0,1) \Rightarrow b \in B_2(0,1)$ as well and next $\forall b \in B_2(0,1) \Rightarrow b \in B_{\infty}(0,1)$ as well.

So I have, let $b \in B_1(0,1)$ , then $|b_1| + |b_2| + \cdots + |b_n| < 1$ , I just don't see how this would imply $\left(|b_1|^2 + |b_2|^2 + \cdots + |b_n|^2\right)^{1/2} < 1$ or $|b_1|^2 + |b_2|^2 + \cdots + |b_n|^2 < 1$

And I'm also unsure as to how to show $\forall b \in B_2(0,1) \Rightarrow b \in B_{\infty}(0,1)$ .

Hints?

As $|b_i|<1, \ i=1,\ ..\ ,n$ we have $|b_i|^2 \le |b_i|$

Hence:

$\sum_i |b_i|^2\le \sum_i |b_i| <1$

The same idea applies to the next part, if the sum of a number of positive terms is less than 1 each term individually is less than 1.

CB

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