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I am interested in functions of the form $\psi: F^n \to \mathbb{R}^+$, where $F$ is a finite field, that have norm-like properties, e.g., $\psi(x+y) \le \psi(x) + \psi(y)$. Does anybody know if there is any literature on this area?

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marked as duplicate by Alex M., user99914, José Carlos Santos, Saad, Community Mar 9 '18 at 2:06

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    $\begingroup$ Exactly which properties do you want? If you want something that behaves like the usual $\lVert \alpha\mathbf{v}\rVert =|\alpha|\,\lVert\mathbf{v}\rVert$ for scalars $\alpha$, then the "absolute value" function is going to have to map everything to $1$. Then it's just a matter of mapping the basis vectors and making sure any vector that has nonzero $i_1,\ldots,i_k$ coordinates has image less than $\psi(\mathbf{e}_{i_1})+\cdots+\psi(\mathbf{e}_{i_k})$, and that would satisfy the conditions. Seems like there is just too much freedom, so there would be very little to be said. $\endgroup$ – Arturo Magidin Aug 27 '11 at 0:28
  • $\begingroup$ @Arturo: I seem to have gotten that positive homogeneity yields $\psi(x)=0$ for all $x\in F$. Are you mixing up identities, or did I mess up? $\endgroup$ – robjohn Aug 27 '11 at 0:39
  • $\begingroup$ @robjohn: I may not be considering all of them. But positive homogeneity is derived from several of the properties of the norm so I don't know which ones we would have "on hand". $\endgroup$ – Arturo Magidin Aug 27 '11 at 3:10
  • $\begingroup$ @Arturo: you said that $\lVert \alpha\mathbf{v}\rVert =|\alpha|\,\lVert\mathbf{v}\rVert$ implies that "absolute value" maps everything to $1$. I got that $\psi(rx)=|r|\psi(x)$ implies that $\psi(x)=0$. The difference between $0$ and $1$ was my concern. $\endgroup$ – robjohn Aug 27 '11 at 3:45
  • $\begingroup$ @robjohn: Dratted comments. Originally, I had that since $\alpha^{p^n}=\alpha$ for some $n$, then it follows that you will have $|\alpha|^{p^n}=|\alpha|$ for whatever function $|\cdot|$ will represent. This forces $|\alpha|=0$ or $|\alpha|=1$, and if you want non-triviality, one needs the latter (but it got lost in the shuffle). Note that you cannot really talk about arbitrary $r\in\mathbb{Q}$ in what you do, because $rx$ only makes sense for "rationals" with denominator prime to $p$... $\endgroup$ – Arturo Magidin Aug 27 '11 at 3:47
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If you are truly only interested in the triangle inequality, then there is the Hamming weight $w(x_1,x_2,\ldots,x_n)=m$ where $m$ is simply the number of non-zero components. This gives you a metric. Mind you, the space $F^n$ is finite, so any metric on it is going to give you the discrete topology. Adding any kind of norm-like requirements (on top of the triangle inequality) is problematic for several reason, as others have pointed out.

The Hamming weight obviously depends on the choice of basis, which may restrict its usefulness (depending on what you wanted to do with this 'norm').

I'm sad to say there isn't an awful lot of analysis happening in this space.

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The triangle inequality might allow some interesting functions, but positive homogeneity doesn't. A finite field has a prime characteristic, $p$. Positive homogeneity says $\psi(rx)=|r|\psi(x)$ for $r\in\mathbb{Q}$. To satisfy this property, $\psi(0)=\psi(px)=p\psi(x)$, which first says $\psi(0)=0$ and then $\psi(x)=0$ for all $x\in F$.

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  • $\begingroup$ As Arturo Magidin notes, we can't use all $r\in\mathbb{Q}$, just the ones whose denominator is not divisible by $p$. $\endgroup$ – robjohn Aug 27 '11 at 4:12
  • $\begingroup$ You shouldn't use $\mathbb Q$, but the copy of $\mathbb Q$ that lives inside $F$, which is $\mathbb Z /(p)$ with $p = \operatorname {char } F$ (the prime field of $F$). $\endgroup$ – Alex M. Mar 8 '18 at 20:20
  • $\begingroup$ @AlexM.: did you see my last comment in the thread after the question? In it, I mention that the subfield is actually $\mathbb{Z}_p$. $\endgroup$ – robjohn Mar 9 '18 at 4:41

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