Are Inequalities of Norms of Vectors Affected by the Choice of Norm

Question

Suppose we have two norms,

$p_{1}:V\to{}\mathbb{R}, \qquad{} \text{ and } \qquad{} p_{2}:V\to{}\mathbb{R}$,

defined on a finite dimensional vector space $V$.

Since all norms on a finite dimensional vector space are equivalent, $p_{1}$ and $p_{2}$ are equivalent.

However, when we let $x,y\in{}V$, is it also correct that if $p_{1}(x)\geq{}p_{1}(y)$ then $p_{2}(x)\geq{}p_{2}(y)$ ?

Answer 1

No, in general you need some futher scaling. If $c, C > 0$ are constants so that $cp_1 \le p_2 \le C p_1$ is satisfied, then $p_1(x) \ge p_1(y)$ implies $$p_2(x) \ge cp_1(x) \ge cp_1(y) \ge \frac{c}{C} p_2(y).$$

For example, consider $\mathbb{R}^2$ with $p_1( \cdot) = \| \cdot \|_\infty$, $p_2(\cdot) = \|\cdot\|_1$, $x = (1, 0)$ and $y = (1, 1)$. Then $p_1(x) = 1 \ge 1 = p_1(y)$, but $p_2(x) = 1 < 2 = p_2(y)$.

Answer 2

No. Take $V= \mathbb R^2$, $p_1(x)= \max\{|x_1|,|x_2|\}$ and $p_2(x)=\sqrt{x_1^2+x_2^2}$

Let $x=(1,1)$ and $y=(1,0)$. Then we have

$p_1(x)=1$ and $p_1(y)=1$. But

$p_2(x)=1$ and $p_2(y)=\sqrt{2}$