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I posted this question last time but that thread got closed unfortunately because I couldn't write the problem aptly and up to the point of proper understanding.

$4.8.3.$ Let $C>0$ be an arbitrary constant. Find all continuous functions $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x)=f(x^2+C),$ for all $x\in\mathbb{R}.$

This question was tried hard by a full group of students but with fail. I request everyone here to try it out. My ideas: After seeing this problem I was reminded of the famous problem of finding all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}, f(x^2) = f(x)$, which was easy. I tried the problem in the same and similar manner but eventually failed.

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  • $\begingroup$ Instead of posting a new quesiton, the correct thing should have been to edit your older closed post to improve it. If you have enough closed or deleted posts, you may be prevented from posting new questions. $\endgroup$ – Calvin Khor Aug 23 '20 at 5:21
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We first consider the case $c \leq 1/4$; we shall show in this case $f$ must be constant. The relation $$ f(x) = f(x^2 + c) = f((-x)^2 + c) = f(-x) $$ proves that $f$ is an even function. Let $r_1 \leq r_2$ be the roots of $x^2 + c - x$, both of which are real. If $x > r_{2}$, define $x_{0} = x$ and $x_{n+1} = \sqrt{x_{n} - c}$ for each positive integer $x$. By induction on $n$, $r_{2} < x_{n+1} < x_{n}$ for all $n$, so the sequence $\{x_{n}\}$ tends to a limit $L$ which is a root of $x^{2} + c = x$ not less than $r_{2}$. Of course this means $L = r_{2}$. Since $f(x) = f(x_{n})$ for all $n$ and $x_{n} \to r_{2}$, we conclude $f(x) = f(r_{2})$, so $f$ is constant on $x \geq r_{2}$.

If $r_{1} < x < r_{2}$ and $x_{n}$ is defined as before, then by induction, $x_{n} < x_{n+1} < r_{2}$. Note that the sequence can be defined because $r_{1} > c$; the latter follows by noting that the polynomial $x^{2} - x + c$ is positive at $x = c$ and has its minimum at $1/2 > c$, so both roots are greater than $c$. In any case, we deduce that $f(x)$ is also constant on $r_{1} \leq x \leq r_{2}$.

Finally, suppose $x < r_{1}$. Now define $x_{0} = x, x_{n+1} = x_{n}^{2} + c$. Given that $x_{n} < r_{1}$, we have $x_{n+1} > x_{n}$. Thus if we had $x_{n} < r_{1}$ for all $n$, by the same argument as in the first case we deduce $x_{n} \to r_{1}$ and so $f(x) = f(r_{1})$. Actually, this doesn't happen; eventually we have $x_{n} > r_{1}$, in which case $f(x) = f(x_{n}) = f(r_{1})$ by what we have already shown. We conclude that $f$ is a constant function. (Thanks to Marshall Buck for catching an inaccuracy in a previous version of this solution.)

Now suppose $c > 1/4$. Then the sequence $x_n$ defined by $x_0 = 0$ and $x_{n+1} = x_n^2 + c$ is strictly increasing and has no limit point. Thus if we define $f$ on $[x_0, x_1]$ as any continuous function with equal values on the endpoints, and extend the definition from $[x_n, x_{n+1}]$ to $[x_{n+1}, x_{n+2}]$ by the relation $f(x) = f(x^2 + c)$, and extend the definition further to $x < 0$ by the relation $f(x) = f(-x)$, the resulting function has the desired property. Moreover, any function with that property clearly has this form.

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    $\begingroup$ Thanks a lot. The first part you did was exactly the way I was approaching $\endgroup$ – Mycroft Holmes Aug 22 '20 at 17:17
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    $\begingroup$ There are many such functional equation problems with the equation given as $f(x)=f(g(x))$ for all $x$. Then the natural approach is to look at the behaviour of the sequence $g(x),g(g(x)),\ldots,g^{\circ n}(x)\ldots$. For more details read the book g.co/kgs/povr5Z $\endgroup$ – Shubhrajit Bhattacharya Aug 22 '20 at 17:22
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    $\begingroup$ Thanks for this reference $\endgroup$ – Mycroft Holmes Aug 23 '20 at 4:48
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Since $f(-x)=f((-x)^2+C)=f(x^2+C)=f(x)$, one may as well restrict to $\mathbb R^+=[0,\infty)$.

Let $T:x\mapsto x^2+C$. The topological dynamics of $T$ depends on the value of $C$, which determines the number of fixed points of $T$ in $\mathbb R^+$: $T$ has zero, one, or two fixed points, depending on whether $C>1/4$, $C=1/4$, or $C<1/4$.

Here is a sketch of an answer for the case $C>1/4$, when $T$ has no fixed points on $\mathbb R^+$. Let $t_0=0$ and define $t_n$ recursively by $t_{n+1}=T(t_n)$. Note that $\lim_{n\to\infty}t_n=\infty$. The sequence of half-open intervals $I_1=[t_0,t_1),\ldots,I_n=[t_{n-1},t_n),\ldots$ form a partition of $\mathbb R^+$. Let $g:[t_0,t_1]\to\mathbb R$ be any continuous function such that $g(t_0)=g(t_1)$. Now let $f(x)=g(x)$ on $I_1$, let $f(x)=g(T(x))$ on $I_2$, and so on, so $f(x)=f(T(x))$ on each $I_n$.

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