Let $\rho:[0,+\infty)^{\mathbb{N}}\to[0,+\infty] $ satisfy the following properties:

  • $\rho(\lambda u)=\lambda\rho( u)$, for every $u\in [0,+\infty)^{\mathbb{N}}$ and $\lambda\ge 0$;

  • $\rho(u+v)\le \rho( u)+\rho( v)$, for every $u,v\in [0,+\infty)^{\mathbb{N}}$.

Define $\|\cdot\|:\mathbb{R}^{\mathbb{N}} \to[0,+\infty] $ by $\left\|\{u_n\}_{n\in\mathbb{N}}\right\|=\rho(\{|u_n|\}_{n\in\mathbb{N}})$.

Assume that we know that $E=\{u\in \mathbb{R}^{\mathbb{N}}, \|u\|<+\infty\}$ is a linear space and $\|\cdot\|$ is a norm on $E$. Does it follow that $\rho$ is monotone?

By monotonicity I mean that $\rho(\{u_n\}_{n\in\mathbb{N}})\le \rho(\{v_n\}_{n\in\mathbb{N}})$, once $0\le u_n\le v_n$, for all $n$, and $\rho(\{u_n\}_{n\in\mathbb{N}}), \rho(\{v_n\}_{n\in\mathbb{N}})<+\infty$.

Remark 1. It is not hard to show monotonicity, if we consider complex sequences instead of real.

Remark 2. It is very possible that $\rho(\{v_n\}_{n\in\mathbb{N}})<+\infty$, $\rho(\{u_n\}_{n\in\mathbb{N}})=+\infty$, and $0\le u_n\le v_n$, for all $n$.

Remark 3 I can show

$E$ is a Banach space $\Rightarrow$ $\rho$ is monotone $\Rightarrow$ $E$ is a normed space $\Rightarrow$ $\rho$ is "finitely" monotone.

By the latter i mean, that $\rho(\{u_n\}_{n\in\mathbb{N}})\le \rho(\{v_n\}_{n\in\mathbb{N}})$, once $0\le u_n\le v_n$, for all $n$, and the set$\{\frac{u_n}{v_n}, n\in \mathbb{N}\}$ is finite.


For $f\in\mathbb R^{\mathbb N}$ define $\|f\|$ to be the infimum of the reals $B$ such that the graph of $f$ is contained in a finite union of lines of the form $Y=mX+c$ with $|m|,|c|\leq B.$ The infimum is $\infty$ if no such $B$ exists. The indicator function of $[1,\infty)$ has norm $1$ because any line except $Y=1$ has a finite intersection with its graph. But it is bounded by $x\mapsto \tfrac 12(x+1)$ which has norm $\tfrac12.$

  • $\begingroup$ Thank you! Perhaps it is worth adding that in order to complete the solution one also has to show that $\|\cdot\|$ is indeed a norm, and is symmetric with respect to taking $\pm$ of the entries of the sequences. However, this is easy to see. $\endgroup$ – erz Feb 3 at 9:34
  • $\begingroup$ It seems to me that this would also work with complex scalars. Is that right? $\endgroup$ – Matthew Daws Feb 3 at 10:18
  • $\begingroup$ @MatthewDaws: no, you need to be able to multiply by unit complex numbers without changing the norm - this is hidden in the $\pm$ detail erz mentioned above. For example $n\mapsto \exp(in)$ would need to have the same norm as $x\mapsto 1.$ $\endgroup$ – Dap Feb 3 at 14:11
  • $\begingroup$ Sorry, I kept editing my comment. I think I understand finally: the construction given in the answer would work over $\mathbb C$. But to address the original question, we need that there is a $\rho$ occurring. This is equivalent to $\|(x_n)\| = \| \ ( |x_n| ) \ \|$ for any $(x_n)$ in the space (and that if $( |x_n| )$ is in the space, then so is $(x_n)$). This is why we need to be able to multiply by arbitrary unit modulus sequences! $\endgroup$ – Matthew Daws Feb 4 at 10:43

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