# Does the norm on a sequence space have to be monotone?

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 $$\tfrac12.$$
• 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. – erz Feb 3 at 9:34
• @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 $x\mapsto 1.$ – Dap Feb 3 at 14:11
• 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! – Matthew Daws Feb 4 at 10:43