How many elements can the set S(f)={f(x+y)−f(x)−f(y) | x,y∈R}S(f)=\{f(x+y)-f(x)-f(y)\ |\ x,y\in R\} have?

Question:

For a surjective function fRR,where R denotes the set of real numbers, define the set
S(f)={f(x+y)f(x)f(y) | x,yR} .
Assume that S(f) is finite and |S(f)|1. Find the possible values of |S(f)|.

I think |S(f)|=2 is possible because of the example
f(x)=12(x{x}) .

It is clear f is surjective,because for any tR, if we let x=2t+2{2t}, then we have
f(x)=12(2t+2{2t}2{2t})=t ,
and we then have
S(f)=f(x+y)f(x)f(y)={x}+{y}{x+y}={0,1} ,
so we have|S(f)|=2 .

Using this method, Can we find examples such that |S(f)|=3,4,5,? Maybe there are other methods to solve this problem.

Answer

One example for more than 2.

f(x)={1if x=122if x=144xif x12,14

f(x)=4xsgn|x12|sgn|x14|+1(1sgn|x12|)+2(1sgn|x12|)
S(f)={3,1,0,1,2}

And the method to generate more of these functions with finite |S(f)| values.

Given surjective real function g:RR, such that |S(g)|=1, and a finite permutation on real numbers h:RR with n>0 values permuted, and the composition surjective real function f(x)=g(h(x)), the value |S(f)| is finite, at least 2 and with the upper bound 1+12n(n+5). (Working at the bottom)

Looking at your function and the other answer, I think the kinds of functions F such that S(F) is finite can be defined by composition of linear functions, finite permutations and surjective sawtooth functions.

F=f1fn

where for each i=1,,n, surjection is assumed and one of the following is true:

  • m,cR:m0xR:fi(x)=mx+c (linear)
  • bR>0:xR:fi(x)=x+(x%b) (surjective sawtooth)
  • |{xRfi(x)x}|<0yR:!xR:fi(x)=y (finite permutation)

Working out for the finite permutation case:
FS={f:RRyR:xR:f(x)=y}FB={f:RRyR:!xR:f(x)=y}S:FSRS(f)={f(x+y)f(x)f(y)x,yR}gFS|S(g)|=1cR:x,yR:g(x+y)g(x)g(y)=cm,cR:xR:g(x)=mx+c|S(g)|=1g(x)=mx+cS(g)={c}T:FBRT(f)={xRf(x)x}hFBfFSf(x)=g(h(x))Sf:=S(f)Sg:=S(g)Th:=T(h)0<|Th|<01<|Sf|<0F(x,y):=f(x+y)f(x)f(y)G(x,y):=g(x+y)g(x)g(y)Sf={c}{F(x,y)x,yRF(x,y)c}F(x,y)cf(x)g(x)f(y)g(y)f(x+y)g(x+y)xThyTh(x+y)Th|Sf|>1

\begin{align}
S_f &= \{-c\} \cup P_1 \cup P_2 \cup P_3 \\
P_1 &= \left\{F(t - x, x) \mid t \in T_h \land x \in \mathbb{R} \land x \notin T_h \land t - x \notin T_h \right\} \\
&= \left\{f(t) - f(t-x) - f(x) \mid \Phi(x,t) \right\} \\
&= \left\{mh(t) + c - (m(t-x) + c + mx + c) \mid \Phi(x, t) \right\} \\
&= \left\{m(h(t) - t) - c \mid t \in T_h \right\} \\
(\forall t \in h : h(t) \neq t) &\vdash \exists x \in P_1: x \neq -c \\
&\vdash \left| \{-c\} \cup P_1 \right| > 1 \\
0 < &\left|P_1\right| \leq \left|T_h\right| \\
P_2 &= \left\{ F(x, y) \mid x,y \in T_h \land (x > y \lor x = y) \right\} \\
&\vdash \left|P_2\right| \leq \frac12 \left|T_h\right|\left(\left|T_h\right| + 1\right) \\
P_3 &= \left\{ F(t, x) \mid t \in T_h \land x \in \mathbb{R} \setminus T_h \land t + x \notin T_h \right\} \\
&= \left\{ f(t + x) - f(t) - f(x) \mid \phi(t, x) \right\} \\
&= \left\{ m(t+x) + c - (mh(t) + c + mx + c) \mid \phi(t, x) \right\} \\
&= \left\{ m(t - h(t)) - c \mid t \in T_h \right\} \\
&\vdash \left|P_3\right| \leq \left|T_h\right| \\
\left|S_f\right| &\leq 1 + 2\left|T_h\right| + \frac12 \left|T_h\right|\left(\left|T_h\right| + 1\right) \\
&\leq 1 + \frac12 \left|T_h\right|\left( \left|T_h\right| + 5 \right) \\
1 < &\left|S_f\right| \leq 1 + \frac12 \left|T_h\right|\left( \left|T_h\right| + 5 \right)
\end{align}

Attribution
Source : Link , Question Author : math110 , Answer Author : AkariAkaori

Leave a Comment