2
$\begingroup$

In the book Putnam & Beyond, it is proven (exercise 383) that there is no non-constant function $f:(1,+\infty)\to\mathbb{R}$ such that $f(x)=f\left(\frac{1+x^2}{2}\right),\forall x>1$ and $\lim_\limits{x\to\infty}{f(x)}$ exists.

Clearly, if we remove the constraint that $\lim_\limits{x\to\infty}{f(x)}$ exists, there exist infinitely many discontinuous functions that satisfy the rest of the hypothesis. So, the question here would be for continuous functions (and obviously $\lim_\limits{x\to\infty}{f(x)}$ shouldn't exist as is trivially proven from the above exercise). Therefore, let's state the modified version of the problem:

Is there any non-constant, continuous function $f:(1,+\infty)\to\mathbb{R}$ such that $$f(x)=f\left(\frac{1+x^2}{2}\right),\forall x>1$$ and if yes, is it expressible with elementary functions or not?

Note: I believe it all comes down to the density (or not) within the real numbers of the sequence: $x_0$:arbitrary and $$x_{n+1}=\frac{1+x_n^2}{2},\forall n\geq0$$

$\endgroup$
  • $\begingroup$ If $f$ has limit on $1$ then $f$ is constant $\endgroup$ – Youem Aug 2 '17 at 15:54
  • $\begingroup$ Let $g \colon [2, 5/2] \to \mathbb{R}$ a continuous function with $g(2) = g(5/2)$. Can you use $g$ to construct $f$? $\endgroup$ – Daniel Fischer Aug 2 '17 at 15:58
  • $\begingroup$ $g(x)=\frac{1+x^2}{2}$ is a monotone expansion, so its iterates are not dense. But it has a fixed point at $1$ and is smooth. Accordingly, for any $x>1$ and any $\epsilon>0$ there exists $y>1$ and $n$ such that $y-1<\epsilon$ and $g^n(y)=x$ (where the power denotes iteration). Thus an $f$ satisfying your functional equation is determined by its values on an arbitrarily small interval $(1,1+\delta)$. So in particular you get Youem's statement. $\endgroup$ – Ian Aug 2 '17 at 16:02
  • $\begingroup$ @Ian I was thinking if we consider the sequence $(x_n)$ defined by $x_0= x$ and $x_{n+1} = \sqrt{2x_{n} - 1}$. It is well defined since $x>1$, it is also decreasing (one can compute $x_{n+1}^2 - x_n^2$ to prove that) and its limit is $1$. Finally $f(x_n) = f(x_{n+1})$ $\endgroup$ – Youem Aug 2 '17 at 16:17
  • $\begingroup$ @Youem Yes, that's what I was getting at: this iteration is how you get $y$ as in my comment back from $x$. $\endgroup$ – Ian Aug 2 '17 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.