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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.

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    $\begingroup$ For any invertible matrix $M$ and any norm $\|\cdot\|$ on $\mathbb{R}^n$, the map $x \mapsto \|Mx\|$ is a norm, too. This is usually a good way to build counterexamples to this kind of statements. $\endgroup$ – Federico Poloni Jun 22 '18 at 18:22
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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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    $\begingroup$ Thank you very much. Do you have a reference for the theorem you are quoting? $\endgroup$ – Alphonse Jun 22 '18 at 16:01
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    $\begingroup$ @Alphonse The article Absolute and monotonic norms by Bauer, Stoer and Witzgall. $\endgroup$ – mechanodroid Jun 22 '18 at 16:03
  • $\begingroup$ Thanks. Do you know an analogous result for $\Bbb R^m$ instead of $\Bbb C^n$ (typically if $m$ is odd...) ? $\endgroup$ – Alphonse Jun 22 '18 at 16:10
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    $\begingroup$ @Alphonse I believe the equivalence still holds. $\endgroup$ – mechanodroid Jun 22 '18 at 16:17
  • $\begingroup$ @mechanodroid Do you know anything about, whether a not absolut monotonic norm can be transformed by a linear mapping into a absolut monotonic norm, more exact: Given a norm $N$, does there always exist a lin mapping $A:\mathbb{C}^n\rightarrow\mathbb{C}^n$ such that $N\circ A$ is absolut monotonic? $\endgroup$ – tommsch Jun 23 '18 at 8:01

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