I'm looking for triples $(K,z, |-|)$ where $K$ is a local field, $z \in K$, and $|-|$ is an absolute value on $K$, such that, for all $f(x),g(x),h(x) \in \mathbb N[x]$, the following inequality is satisfied:


$$|f(z) + g(z)| \leq |f(z) + h(z)| + |g(z) + zh(z)|$$


I know that the following triples work:


* $(\mathbb R, z, |-|)$ whenever $z \geq 0$ and $|-|$ is arbitrary.


(and by extension $(\mathbb C, z, |-|)$ works when $z$ is on the positive real line)


* $(K, -1, |-|)$ when $K$ and $|-|$ are arbitrary.


Are there other triples which work? What are all such triples?



When can I satisfy the following twisted subadditivity inequality?
Surprisingly to me, these are the only possibilities! We will consider polynomials $f(z), g(z), h(z)$ as above. Set $a(z) = f(z)/h(z)$, and $b(z) = g(z)/h(z)$. So the desired inequality is


$$|a(z) + b(z)| \leq |a(z) + 1| + |b(z) + z| \qquad (\ast)$$


where $a(z),b(z)$ are rational functions of $z$ with natural number coefficients.


**Case 1:** Suppose that $\mathbb R \ni z < 0$. We claim that $z = -1$. Since $z$ is negative, the possible values of $a(z), b(z)$ are dense in $\mathbb R$, so in order for $(\ast)$ to be satisfied, we must have $|a + b| \leq |a+1| + |b+z|$ for all $a,b \in \mathbb R$. Taking $a = -1$, $b = -z$, we see that $|-1 - z| \leq 0$, so that $z = -1$.


**Case 2:** Suppose that $z \in \mathbb C \setminus \mathbb R$. Then the possible values of $a(z),b(z)$ are dense in $\mathbb C$, so the same argument as in Case 1 shows that $z = -1$, a contradiction.


**Case 3:** Suppose that the field is nonarchimedean. Then since $-1$ is in the closure of $\mathbb N$, in some limit we may take $a(z) = -1$ and $b(z) = -z$ and we conclude as in Case 1.


