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Do there exist "nice" (maybe analytic?) functions $f_0,f_1:\mathbb R \to \mathbb R$ such that

$\forall n\in\mathbb N,\forall \sigma\in\{0,1\}^n,\exists x\in\mathbb R, \forall \tau\in\{0,1\}^n, (\tau\ne \sigma \to (f_\sigma(x)>f_\tau(x))$ ?

Here, for $\sigma,\tau\in\{0,1\}^*$, concatenation is composition: $f_{\sigma \tau}=f_\sigma\circ f_\tau$.

For example, if $f_0(x)=ax+b$ and $f_1(x)=cx+d$ then $$f_{01}(x)=a(cx+d)+b=acx + (ad+b)$$ $$f_{10}(x)=c(ax+b)+d=acx + (bc+d)$$ $$f_{00}(x)=a(ax+b)+b=a^2 x + (ab+b)$$ $$f_{11}(x)=c(cx+d)+d=c^2 x + (cd+d)$$ For $\sigma=01$ we would need $ad+b>bc+d$ and an $x$ with $$acx + ad > a^2 x + ab,$$ $$acx + (ad+b) > c^2 x + (cd+d).$$ For example, $d=0$, $a=c<0$, $b>0$ suffices. On the other hand, surely no $a,b,c,d$ work for all $\sigma$.

But what if $f_0,f_1$ can be rather arbitrary functions?

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  • $\begingroup$ If I grasped this correctly, then for any given $n,\sigma$, one can satisfy this at a predetermined $x$ by manipulating $f_j$ at only finitely many points. Let's say $n=2$, $\sigma =00$, $x=0$. Then we could set $f_0(0)=1$, $f_0(1)=2$, $f_1(0)=3$, $f_0(3)=4$, $f_1(3)=5$. The point is to never reuse values. Since for any discrete set (here, the integers), there is an entire function that takes any prescribed values on that set, we can make $f_j$ entire. $\endgroup$ Commented May 22, 2024 at 19:46
  • $\begingroup$ @ChristianRemling that sounds good! The next question would be, can you use polynomials? I expect not, of course. $\endgroup$ Commented May 22, 2024 at 19:55

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