11

Prove that there exists no differentiable real function $g(x)$ such that $g(g(x))=-x^3+x+1$.

I have googled it but find nothing useful.

Now I know it's a Iterated function problem.

It's an exercise problem after the chapter DERIVATIVE, so I guess maybe it's not too difficult.

Could you give me a hint to solve this problem?

Could you give me a book list about the systematic introduction about Iterated function?

闫嘉琦
  • 1,376
  • 1
    I am not aware of a solution, but if you are in a basic calculus course, it may want for you to apply the chain rule. There are reasons why every analytic function may look like a polynomial, that is Taylor's Theorem and Taylors' Series, but I am not sure about how one could extend this to all real functions without using analysis. – theREALyumdub Aug 15 '18 at 13:53
  • 1
    Note: the problem does not specify that $g(x)$ is differentiable, nor even continuous. – lulu Aug 15 '18 at 13:53
  • @lulu maybe we can just assume it's third order differentiable or even higher order to get a solution at least? – 闫嘉琦 Aug 15 '18 at 13:59
  • 1
    I don't see any basis at all for assuming that the function is even continuous (unless the problem specified that). – lulu Aug 15 '18 at 14:01
  • @lulu Ooops, Sorry I forgot a condition..... $g(x)$ shall be differentiable. – 闫嘉琦 Aug 15 '18 at 14:03
  • That should help a lot. Please add that condition to your problem statement. – lulu Aug 15 '18 at 14:04
  • While we are at it: What are the allowed domains of $g$? – Christian Blatter Aug 15 '18 at 14:08
  • @ChristianBlatter $(-\infty, +\infty)$. – 闫嘉琦 Aug 15 '18 at 14:09
  • 1
    Is there not a possible argument based on limiting behavior at $x \to \pm \infty$? Note that $g$ has to "send" ${-\infty, +\infty}$ into ${-\infty, +\infty}$, and none of the four possible maps gives the right limiting behavior $g \circ g(-\infty) = +\infty$, $g \circ g(+\infty) = -\infty$. – Connor Harris Aug 15 '18 at 14:28
  • 1
    This would show more generally that there's no (continuous!) functional square root of any odd-degree polynomial with a negative leading coefficient. – Connor Harris Aug 15 '18 at 14:28
  • I think the limiting argument works, and it's not hard to formalize it with the standard definition of limits at infinity. – Connor Harris Aug 15 '18 at 14:45

2 Answers2

9

We have $$ g(g(g(x)) = -g(x)^3 + g(x) + 1 \iff \\ g(-x^3+x+1) = -g(x)^3 + g(x) + 1 $$ For $x=1$ this turns into $$ g(1) = -g(1)^3 + g(1) + 1 \iff \\ g(1)^3 = 1 $$ So $g(1) = 1$, if $g$ is a real valued function.

Differentiating both sides of $g(g(x)) = -x^3+x+1$ gives $$ g'(g(x))\, g'(x) = -3x^2 + 1 $$ This gives $$ g'(g(1))\,g'(1) = - 2 \iff \\ g'(1)^2 = -2 $$ which is not possible for real valued $g'(x)$ and thus for $g(x)$, as the derivative of a real valued function is a real valued function.

mvw
  • 34,562
6

Recall that a fixed point of a function $f$ is a solution to $f(x) = x$.

The only fixed point of $g \circ g$ is $1$, so the only fixed point of $g$, if any, is also $1$, as any fixed point of $g$ is also a fixed point of $g \circ g$. If $g$ has no fixed point but instead interchanges $1$ and $\xi = g(1) \neq 1$, then the graph of $g$ must connect $(\xi, 1)$ and $(1, \xi)$. These points lie on opposite sides of the line $y=x$ regardless of the value of $\xi$, so $g$ has a fixed point not at $1$ by the Intermediate Value Theorem, a contradiction.

Therefore, $g$ has one fixed point, at $1$. Taking derivatives of $g \circ g$ gives $g'(x) g'(g(x)) = 1 - 3x^2$. Setting $x = g(x) = 1$ gives $g'(1)^2 = -2$, impossible. The conclusion follows.

Connor Harris
  • 7,070