Real Analysis – Is Every Norm Increasing?

normed-spacesreal-analysis

Let $N$ be any norm on $\Bbb R^n$. Is it true that if $0 \leq a_i \leq b_i$ for $1 \leq i \leq n$, then $N(a_1, \ldots, a_n) \leq N(b_1, \ldots, b_n)$ ?
Clearly this is true for the norms $\| \cdot \|_p$ where $1 \leq p \leq \infty$, being defined as a composition of increasing functions. Any two norms are equivalent, but I don't see how my property is preserved under equivalence.

Best Answer

The answer is no.

Consider the norm $\|(x,y)\| = |x| + |x-y|$ on $\mathbb{R}^2$.

We have $(1,0) \le (1,1)$ in the sense you defined but

$$\|(1,0)\| = 2, \quad \|(1,1)\| = 1$$

There is a nontrivial theorem characterizing this property:

Let $\|\cdot\|$ be a norm on $\mathbb{C}^n$. The following is equivalent:

  • $|x_i| \le |y_i|, \forall i=1, \ldots, n$ implies that $\|(x_1, \ldots, x_n)\| \le \|(y_1, \ldots, y_n)\|$.
  • $\|(x_1, \ldots, x_n)\| = \|(\left|x_1\right|, \ldots, \left|x_n\right|)\|, \forall (x_1, \ldots, x_n) \in \mathbb{C}^n$.
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