# Is there a non discrete norm on $\mathbb{Z}$ (the group of integers) which is unbounded?

Is there a non discrete norm on $\mathbb{Z}$ (the group of integers) which is unbounded?

A norm on $\mathbb{Z}$ is a function $f$ from $\mathbb{Z}$ (the group of integers) to $\mathbb{R}$ (the real numbers) which satisfies the following axioms:

1. $f(x) \ge 0$ for all $x \in \mathbb{Z}$ and $f(x)=0$ iff $x=0$,
2. $f(x)=f(-x)$ for all $x \in \mathbb{Z}$,
3. $f(x+y)\le f(x)+f(y)$ for all $x,y \in \mathbb{Z}$.

Notice that $f$ doesn't have to be multiplicative, otherwise by Ostrowski's theorem it follows that there isn't such norm.

• MathOverflow recognizes basic LaTeX symbols. You clearly know LaTeX, since you're using \le and \in. Please edit your post to use LaTeX. So for example, you can type x \in \mathbb{Z} inside dollar symbols to get $x\in\mathbb{Z}$. – Joe Silverman Nov 5 '16 at 21:47
• The absolute value function satisfies your three axioms, and is unbounded, so what do you mean by "non-discrete"? – Gerry Myerson Nov 5 '16 at 21:51
• @GerryMyerson I might be wrong, but I suppose they mean that the image of $\mathbb Z$ under $f$ is not a discrete subset of $\mathbb R$. – Wojowu Nov 5 '16 at 21:55
• It is non-discrete if $\inf_{x > 0} f(x)=0$. – Andreas Thom Nov 5 '16 at 21:56
• So we need to partition the integers into a set $A$ with big norm and a set $B$ with small norm, such that no two $B$-numbers add up to a $A$-number. Maybe try giving the even a small norm and the odds a big one. – Gerry Myerson Nov 5 '16 at 22:02

## 2 Answers

In fact the previously deleted sawtooth-like answer is almost right, but the periods of each successive sawtooth must grow much faster than exponentially.

Let $d_n(x)=d(x,2^{n^2}\mathbb{Z})$ be the distance of $x$ to $2^{n^2}\mathbb{Z}$, and let $f=\sum_{n=1}^\infty 2^{-n^2+n}d_n$. Then clearly $f$ is a norm on $\mathbb{Z}$. We have $f(2^{n^2-1})\ge 2^{-n^2+n}d_n(2^{n^2-1})=2^{n-1}$, and $f(2^{n^2})=\sum_{k=n+1}^\infty d_k(2^{n^2})=2^{n^2}\sum_{k=n+1}^\infty 2^{-k^2+k}<2^{-n+1}$.

• ~ This works! ~ – YCor Nov 6 '16 at 2:43
• This is a nice answer. – Andreas Thom Nov 6 '16 at 13:27

Here's a classical construction.

Fix a sequence $(s_n)_{n\in\mathbf{N}}$ of nonzero real numbers in $[-\pi,\pi]$. On $\mathbf{C}^\mathbf{N}$, define a linear automorphism by $T((x_n))=(e^{is_n}x_n)$. Note that the unique fixed point of $T$ is 0. Endow $\mathbf{C}^\mathbf{N}$ with the $\ell^2$-"norm", allowing infinite values. Then $T$ preserves this norm. In particular, it permutes the cosets of $\ell^2$, the subspace of elements in $\mathbf{C}^\mathbf{N}$ with finite norm, and the restriction of $T$ to any of this coset is an isometry.

Now assume that $(s_n)$ is in $\ell^2$. Since $e^{is_n}=1+is_n+o(s_n)$, we deduce that $T(1)-1\in\ell^2$. Thus $T$ maps the coset $V=1+\ell^2$ into itself, and thus induces a (surjective) self-isometry of $V$.

Define a length function on $\mathbf{Z}$ by $f(n)=\|T^n(1)-1\|$. (Length function is the axioms that you call norm, except that nonzero elements are allowed to have zero length). Since no power of $T$ fixes 1, we see that this length is indeed nonzero outside zero.

Since $T$ has no fixed point and $V$ is isometric to a Hilbert space, $T$ has unbounded orbits (by Jung's center lemma); in particular, the above length function is unbounded.

To ensure that the length is non-discrete, we need stronger hypotheses on $(s_n)$. Say, $s_n=2\pi/(\max(2,n)!$. Then it is easy to check that $T^{n!}(1)$ tends to 1, and hence the length $f(n!)$ tends to 0.

If one wishes a more explicit argument argument avoiding the center lemma, one can replace the constant sequence 1 by the sequence c=(n), and keep the same choice of $(s_n)$, so $\|T(c)-c\|<\infty$, and define $f'(n)=\|T^nc-c\|$. Then the $n$-th coordinate of $T^{n!/2}c$ is $-n$. So for $n\ge 2$, $f'(n!/2)=\|T^{n!/2}c-c\|\ge 2n$, and hence $f'$ is unbounded. The argument showing that the length of $n!$ tends to zero is essentially the same computation.