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My question is mostly: is there a name for this kind of things. I am mostly interested by finding book or articles about what follows, but without even a word or a name, it is quite hard to search for information.

Let $F$ be the smallest class of functions $f(x_1,\dots,x_n)$ which contains the functions: -real constants -projection on one component (that is, $x_i$ for some $1\le i\le n$, -$\lfloor f\rfloor$, for some $f\in F$, -$f_1+f_2$, for $f_1,f_2\in F$ -$cf$ for some $c\in\mathbb R$ and $f\in F$

As stated by Dmitri Zaitsev, the functions in $F$ are piecewise affine functions. But this description is too general, there are piecewise affine function which lacks the periodicity provided by the floor function. Therefore, I would like a name for $F$, or at least, to know which kind of property is satisfied by functions of $F$.

(Of course, those functions are interpreted on $\mathbb R$, but if it helps it could be $\mathbb Q$, as I don't know any other field where $\lfloor\rfloor$ is defined, apart from $\mathbb Z$ where the question becomes trivial.)

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  • $\begingroup$ In what sense "functions are interpreted on $\mathbb R$" here? $\endgroup$ – Dmitri Zaitsev Apr 17 '17 at 2:14
  • $\begingroup$ The codomain of the functions is the set of real, and the variables are interpreted as reals. $\endgroup$ – Arthur Milchior Apr 17 '17 at 6:24
  • $\begingroup$ I think your question needs a possibly more applied context where such functions arise. I have never seen them treated theoretically, they are basically composed of two very different types of functions, each class is important but their composition looks random to me out of context. $\endgroup$ – Dmitri Zaitsev Apr 17 '17 at 9:16
  • $\begingroup$ @DmitriZaitsev I consider the logic FO[R,Z,+,<]. This logic admits the elimination of quantifiers when you add the modular relations and the floor functions. (Weispfenning 1999). Alas, this paper only characterize the set of reals defined in this logic, and I can't find any reference concerning sets of vectors. $\endgroup$ – Arthur Milchior Apr 19 '17 at 2:30
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Such functions are special cases of the so-called piecewise affine (or not very accurately "piecewise linear" as their pieces are affine, not linear, see here for an explanation of the difference between affine and linear).

Formally, a function is piecewise affine if its domain of definition can be decomposed into a disjoint union of polytopes, on each of which it is affine. Here the polytopes are sets defined by finitely many affine equations and inequalities.

To get a better understanding, it is instructive to look at the simplest case of a floor $g(x) = \lfloor f(x)\rfloor$ of a single affine function $f$. Then $g$ is constant on the affine stripes $n\le f(x) < n+1$, and the level set $g=const$ defines one of the stripes or empty set, depending on the constant.

Slightly more generally, $g(x) = f_0(x) + a\lfloor f(x)\rfloor$ is affine on each stripe, whose values on each two stripes differ by a constant. Now the level sets (where the constructed function is constant) are parallel affine hyperplanes inside the stripes (for functions in general position).

More generally, for an arbitrary function $f(x)$ as in the question, the level sets of each $\lfloor f_j(x)\rfloor$ is the family of parallel stripes $n\le f_j(x) < n+1$. Then $f$ is affine on each polytope obtained by intersecting the stripes. Again, values of $f(x)$ on each two such intersections must differ by a constant. Hence, functions constructed in this way are very special among piecewise affine ones.

Given the above interpretation, each level set $g(x)=const$ consists of a family of parallel hyperplanes on each polytope.

More abstractly, the floor function $\lfloor t \rfloor$ can be replaced by any piecewise constant function valued in any set, leading to the same conclusion.

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  • $\begingroup$ Thank you for this answer (I forgot I asked this 3 years ago). I do know the difference between affine and linear, I do research in mathematics. I have edited the question in order to make it more clear. $\endgroup$ – Arthur Milchior Apr 15 '17 at 3:05

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