1

Let $y=f(x)$ be a continuous function, and let $f(x)=f(x')$ with $x \neq x'$. Say that $x$ and $x'$ have the same sequel if $f(x+k)=f(x'+k)$ for all $k \ge 0$. Now, say that $f$ is determinate if $x$ and $x'$ have the same sequel whenever $f(x)=f(x')$. It's clear that a function is determinate if it is: a) injective, b) constant, or c) composed of an injective part followed by a constant part (e.g., $f(x)$ = $x$ when $x \le 0$ and $0$ when $ x \ge 0$).

Question: Is there a determinate function that that is not one of the aforementioned three types?

  • Caution: your title is misleading. –  Sep 28 '17 at 08:34
  • If I am right, it suffices for $f$ to be periodic. –  Sep 28 '17 at 08:34
  • sin and cos should be examples, just as @YvesDaoust pointed out (they are periodic). – Dirk Sep 28 '17 at 08:38
  • 1
    @Dirk A simple periodic function like sin or cos doesn't satisfy the condition. For example, let $x = \pi / 4$ and $x' = 3 \pi / 4$. Then $sin(\pi / 4)=sin(3 \pi / 4)$, but $sin(\pi / 4 + \pi / 4) = 1$ which is not equal to $sin(3 \pi / 4 + \pi / 4) = 0$. So $x$ and $x'$ have the same value, but not the same sequel. The definition of "determinate" is trickier than it looks at first glance. –  Sep 29 '17 at 03:46
  • @Yves As I pointed out to Dirk, simple periodic functions (like sin or cos) don't satisfy the condition. In fact, any function with a simple hump (a section which rises monotonically from $y=k$ to a maximum and then falls monotonically back to $y=k$) fails based on a argument like that for the sin function. –  Sep 29 '17 at 03:53
  • Please elaborate on the "hump sections". i can't figure out how monotonicity can play any role here. –  Sep 29 '17 at 06:31
  • @Yves A "hump" means a section of the curve that simply rises and falls like sin(x) between 0 and $\pi$. Any function that has such a hump can't be determinate because points at the same height on the rising and falling sides of the hump have different sequels (as in the example I mentioned). So if a non-constant determinate function exists, it can't have a hump anywhere, and must have characteristics like a fractal (e.g., the Weierstrass function, which has no simple humps). –  Sep 29 '17 at 21:40
  • @KevinKarn: you didn't specify "for all $x,x'$ such that $f(x)=f(x')$". Review your problem statement. –  Sep 30 '17 at 12:09

1 Answers1

0

I assume the final problem statement excludes injective functions, because these are automatically determinate.

But take a function $f:\Bbb R\to\Bbb R$ which is injective only for $x\le 0$ and constant for $x\ge 0$. This function is not constant everywhere but still determinate, because all possible $(x,x')$-pairs can only be found on the constant part and their sequels are only to the right of zero.

As it turns out, all counterexamples are of the form above. Any determinate function becomes constant as soon as it becomes non-injective. More precisely:

Theorem. If $f$ is determinate and $a\not=b$ with $f(a)=f(b)$. Then $f(a)=f(x)$ for all $x\ge a$.

Proof.

So assume $\alpha:=f(a)=f(b)$ for $a>b$. The proof goes in two steps. At first we show that $f$ is constant on $[a,b]$, and then that it is constant everywhere right of $b$.

Step 1. We show that $f^{-1}(\alpha)$ is dense in $[a,b]$. Assume it is not. Because $f^{-1}(\alpha)$ is closed, we can find an interval $[s,t]\subseteq[a,b]$ with $f(s)=f(t)=\alpha$ but $f(x)\not=\alpha$ for all $x\in(s,t)$. IVT shows that there are $s < s'< 1/2 < t' < t$ with $f(s')=f(t')$. Note that $s'+(t-t')\in(s,t)$. But then

$$f(s'+(t-t'))=f(t'+(t-t'))=f(t)=\alpha$$

in contradiction to the choice of $[s,t]$. Therefor $f^{-1}(\alpha)$ is dense in $[a,b]$ and because $f$ is continuous $f$ is constant on $[a,b]$.

Step 2. Assume $f$ is not constant everywhere right of $b$. Because $f^{-1}(\alpha)$ is closed, we can find a maximal interval $[a,c]$ on which $f$ is constant. Note that $a+\xi\in[a,c]$ for all $\xi\in[0,c-a]$. Hence

$$f(c+\xi)=f(a+\xi)=\alpha.$$

This contradicts the choice of $[a,c]$ and hence $f$ must be constant on $[a,\infty)$. $\;\square$

M. Winter
  • 29,928
  • Beautiful, but I now see I'm wondering about something deeper, i.e.: Is there a continuous, non-determinate function that is none of the following:
    1. Injective
    2. Constant
    3. Injective part followed by constant part
     –  Oct 03 '17 at 21:17
  • @KevinKarn Added a proof. Your conjecture is right. – M. Winter Oct 04 '17 at 08:06
  • I understand your proof except for the following two steps. Can you indicate the justification in a little more detail? I understand why $f^{-1}(\alpha)$ must be closed, but not why that implies existence of closed intervals with the given characteristics.

    1. "Because $f^{-1}(\alpha)$ is closed, we can find an interval $[s,t]\subseteq[a,b]$ with $f(s)=f(t)=\alpha$ but $f(x)\not=\alpha$ for all $x\in(s,t)$."

    2. "Because $f^{-1}(\alpha)$ is closed, we can find a maximal interval $[a,c]$ on which $f$ is constant."

     –  Nov 25 '17 at 07:07
  • @KevinKarn It was some time ago since I wrote this and I think I have to go over it again! Some parts are truly unclear or unnecessarily complicated :D. But to your questions: note that since $A:=f^{-1}(\alpha)$ is closed, its complement is open, hence consists of unions of maximally long open intervals. In both cases, we just chose such an open interval and close it by inserting the endpoints from $A$. We do not really need the density assumption. – M. Winter Nov 25 '17 at 11:17