Equivalence of definitions of operator norm over general normed vector spaces A normed module over a general normed ring $(R, |\cdot|)$ is a module with a norm $(V,\|\cdot\|)$ satisfying $\|rx\|=|r|\|x\|$; the norm on the ring is an absolute value in the usual sense, i.e. $|rs|=|r||s|$ with triangle equality and positive-definiteness for both norms. Let $A:V\to W$ be a linear operator on normed modules, and define the following operator norms, which take values in $[0,\infty]$:
$$\|A\|_1=\inf\{c\ge 0: \forall x\in V,\|Ax\|\le c\|x\|\}$$
$$\|A\|_2=\sup\{\|Ax\|:x\in V,\|x\|\le 1\}$$
$$\|A\|_3=\sup\{\|Ax\|:x\in V,\|x\|=1\}$$
$$\|A\|_4=\sup\left\{\frac{\|Ax\|}{\|x\|}:x\in V,x\ne0\right\}$$
Under what conditions on $R$ are these norms equal? If $R=\Bbb R$ or $\Bbb C$ with the usual absolute value (the usual setting), then they are all equal, but I think that some of the proofs can go through under weaker assumptions. Specifically, I am considering one of the following:
  (0. No additional assumptions)


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*There exist elements of $R$ of (a) arbitrarily small / (b) arbitrarily large norm.

*The set $\{|r|:r\in R\}$ is dense in $[0,\infty)$.

*$\{|r|:r\in R\}=[0,\infty)$.
(One can also throw in assumption F: $R$ is a field, i.e. these are vector spaces.) My guess is that $\|A\|_1=\|A\|_2$ under (1a) or maybe (2), $\|A\|_1=\|A\|_4$ under (0), and $\|A\|_1=\|A\|_3$ under (3). It should be clear that $\|A\|_3\le\|A\|_2\le\|A\|_1$ is true generally.

A proof that $\|A\|_1=\|A\|_4$: First, we show $\|A\|_1$ achieves its minimum, that is $\|Ax\|\le\|A\|_1\|x\|$. If $x=0$ then this is trivial, so assume that $\|x\|>0$, and suppose that $c\ge0$ satisfies $\forall y\in V,\|Ay\|\le c\|y\|$. Then $\|Ax\|\le c\|x\|\to\frac{\|Ax\|}{\|x\|}\le c$, so $\frac{\|Ax\|}{\|x\|}$ is less than the infimum, which is $\|A\|_1$, so $\|Ax\|\le\|A\|_1\|x\|$.
Thus, for all $x\ne0$, we have $\frac{\|Ax\|}{\|x\|}\le\|A\|_1$, so $\|A\|_4\le\|A\|_1$. Conversely, since $\|Ax\|\le\|A\|_4\|x\|$ is trivially true for $x=0$ and $\frac{\|Ax\|}{\|x\|}\le\|A\|_4\to\|Ax\|\le\|A\|_4\|x\|$ if $x\ne0$, we have $\|A\|_4$ in the set for which $\|A\|_1$ is defined as an infimum, so $\|A\|_1\le\|A\|_4$. Thus $\|A\|_1=\|A\|_4$ with no additional assumptions.
 A: A counterexample for (1b)
Take the finitely supported subset of $\Bbb Z^\Bbb N$ as a $\Bbb Z$-module, with the norm given by $\|x\|=\max\{|x_n|t_n:n\in\Bbb N\}$, where $t_{2n}=\pi^{-n}$ (the relevant properties being that $t_{2n}\to 0$ is a sequence of irrationals, except for $t_0=1$) and $t_{2n+1}=2$. Letting $e_n$ be the sequence with a $1$ in the $n$th position, the only elements of norm $1$ are $\pm e_0$, and all the elements of norm $\le1$ are zero in odd positions. Thus the operator defined by $(Ax)_n=a_nx_n$, where $s_0=1$, $a_{2n}=2$ for $n>0$, and $a_{2n+1}=n$, has $\|A\|_3=1$, $\|A\|_2=2$, and $\|A\|_1=\|A\|_4=\infty$; similarly the operator $(Bx)_n=b_nx_n$, where $b_n=n$, has $\|B\|_3=0$ and $\|A\|_2=\|A\|_1=\|A\|_4=\infty$. Thus under hypothesis (1b) there is still the possibility of strict inequality $\|A\|_3<\|A\|_2<\|A\|_1=\|A\|_4$, and the norms can even disagree about whether an operator is bounded or zero. Indeed $\|A\|_2$ and $\|A\|_3$ are not norms in the usual sense because of this.
A proof for (3)
The separation of norms collapses under hypothesis (3), which we will show by proving $\|A\|_3\le\|A\|_4$, which by the definition of $\|A\|_4$ reduces to $\frac{\|Ax\|}{\|x\|}\le\|A\|_3$ for $x\ne0$. Pick some $r\in R$ with $|r|=\frac1{\|x\|}$, which exists by hypothesis (3). Then $\|rx\|=1$, so $\|Arx\|\le\|A\|_3$. But by linearity of $A$ and the distribution property for norm, $\|Arx\|=|r|\|Ax\|=\frac{\|Ax\|}{\|x\|}\le\|A\|_3$.
A proof and counterexample for (2)
Under (2) we have $\|A\|_3<\|A\|_2=\|A\|_1$. To show that $\|A\|_3$ is still badly behaved, we consider the free module on $\Bbb N$ again, this time over $\Bbb Q$. The norm is now $\|x\|=\sum_{n\in\Bbb N}|x_n|t_n$, where $t_{2n}=\pi$ and $t_{2n+1}=1$. Then for all $x$, $\|x\|=\sum_{n\in\Bbb N}|x_{2n+1}|+\sum_{n\in\Bbb N}|x_{2n}|\pi$ where the two sums are in $\Bbb Q$, and so since $\pi$ is irrational if $\|x\|=1$ we have $\sum_{n\in\Bbb N}|x_{2n}|=0$ so $x_{2n}=0$ for all $n$. Then we can just set $(Ax)_n=a_nx_n$ with $a_{2n}=0$ and $a_{2n+1}=n$ to get $\|A\|_3=0$, $\|A\|_1=\|A\|_2=\infty$.
Nevertheless, $\|A\|_2$ is now valid under (2); we will show $\|A\|_2\le\|A\|_1$. Assume by contradiction that $\|Ax\|>\|A\|_2\|x\|$. Then $x\ne0$, so we can rewrite this as $\frac{\|A\|_2}{\|Ax\|}<\frac1{\|x\|}$. Using assumption (2), choose some $r\in R$ such that $\frac{\|A\|_2}{\|Ax\|}<|r|<\frac1{\|x\|}$. Then $\|rx\|<1$ so $\|Arx\|=|r|\|Ax\|\le\|A\|_2$, so $|r|\le\frac{\|A\|_2}{\|Ax\|}$, a contradiction.
This leaves open the question of whether $\|A\|_1,\|A\|_2,\|A\|_3$ are equal or not under hypothesis (1a) or (1a+1b).
