Metrics Spaces Inequalities

1. Oct 26, 2011

Ted123

1. The problem statement, all variables and given/known data

(a) Show that for $x,y\in \mathbb{R}^N$:

(i) $d_{\infty} (x,y) \leq d_1 (x,y) \leq N d_{\infty} (x,y)$ ;

(ii) $d_{\infty} (x,y) \leq d_2 (x,y) \leq \sqrt{N} d_{\infty} (x,y)$.

(b) Find constants $A$ and $B$ such that $A d_1 (x,y) \leq d_2 (x,y) \leq B d_1 (x,y)$.

2. Relevant equations

Sum metric: $\displaystyle d_1 (x,y) = \sum^N_{j=1} |x_j - y_j|$

Euclidean metric: $\displaystyle d_2 (x,y) = \left( \sum^N_{j=1} (x_j - y_j)^2 \right)^{1/2}$

Maximum metric: $\displaystyle d_{\infty} (x,y) = \text{max}^N_{j=1} |x_j - y_j |$

3. The attempt at a solution

For any $a_1,\cdots , a_N \geq 0$ and $p\geq 1$:

$\sum_j (a_j)^p \geq (a_j)^p$ for all $j \in \mathbb{N}$. Taking the maximum over $j$ of both sides yields the lower bound.

Notice that $(a_j)^p \leq (\text{max}_ja_j)^p$ for all $j\in\mathbb{N}$. Summing both sides over $j$ yields the upper bound.

So these lead to the inequalities $$\sum_j (a_j)^p \geq (\text{max}_j a_j )^p$$ and $$\displaystyle \sum_j (a_j)^p \leq N( \text{max}_j a_j )^p$$ which leads to (a)(i) and (a)(ii) for $p=1,2$.

For part (b) is the following correct:

By (a)(i), $d_1(x,y) \leq Nd_{\infty}(x,y) \iff N^{-1} d_1(x,y)\leq d_{\infty}(x,y)$ (noting that $N\in\mathbb{N}$ so non-zero and non-negative).

But by (a)(ii), $d_{\infty}(x,y) \leq d_2(x,y) \leq N^{1/2}d_{\infty}(x,y)$.

So we have $N^{-1}d_1(x,y) \leq d_2(x,y) \leq N^{1/2} d_1(x,y) \iff N^{-3/2}d_1(x,y) \leq d_2(x,y) \leq d_1(x,y)$.

Therefore $A=N^{-3/2}$ and $B=1$ are 2 constants satisfying part (b).

Last edited: Oct 26, 2011