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Given that $f: \mathbb R \to \mathbb R$, and that for some $a \in \mathbb R$, $f(x+a)={1\over 2}+\sqrt{f(x)-f^3(x)}$; prove $f(x)$ is a periodic fucntion.

I know that to prove a function is periodic one needs to prove $f(x+bK)=f(x)$ for all $b \in \mathbb Z$, and the period is $K$

But how can I solve this problem?

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    $\begingroup$ As far as I'm concerned it's "for all $x\in \mathbb R$" and $b$ shouldn't be there. $\endgroup$ – Git Gud Nov 4 '15 at 13:45
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    $\begingroup$ @user3313320 Even if it helps, it's not necessarily true. $\endgroup$ – Git Gud Nov 4 '15 at 13:47
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    $\begingroup$ Wait a min...$f^3(x)$ means $f(f(f(x)))$ or $(f(x))^3$? $\endgroup$ – Brian Cheung Nov 4 '15 at 13:49
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    $\begingroup$ In fact it seems clear $f(x)\in [\frac12,1]$ for all reals. $\endgroup$ – Macavity Nov 4 '15 at 14:15
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    $\begingroup$ Seems similar to math.stackexchange.com/questions/1147980/… $\endgroup$ – Mohsen Shahriari Nov 5 '15 at 9:19
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Let $g(x) = \frac 12 + \sqrt{x - x^3}$.

Forall $x \in \Bbb R$, $f(x) = g(f(x-a)) \ge 1/2$. Also, since $f(x+a) = g(f(x))$ exists, we must have $f(x)-f(x)^3 \ge 0$, hence $f(x) \le 1$

Therefore, $f(x) \in [0.5 ; 1]$ forall $x \in \Bbb R$.

$g$ is increasing from $x=1/2$ (where $f(1/2)>1$) to $x=1/\sqrt 3$ then decreases to $x=1$ (where $f(1)=1/2$).

We can compute and see that $0.83 > 1/\sqrt 3$ and $g(0.83) > 1$, so for any $x \in [0.5;0.83]$, $g(x) > 1$.

Therefore, $f(x) \in [0.83 ; 1]$ forall $x \in \Bbb R$ (or else $f(x+2a)$ wouldn't exist)

$g'(x)$ is decreasing on that interval, and $g'(0.83) < -1$, so forall $x \in \Bbb [0.83;1]$, $g'(x) \le g'(0.83) < -1$.

In particular we have $|f(x+2a) - f(x+a)| \ge |f(x+a)-f(x)||g'(0.83)|$.

Let $M = \sup_{x \in \Bbb R} |f(x+a)-f(x)|$. Since $f(x) \in [0.5;1]$, we obviously have $0 \le M \le 0.5$. But we also have $M \ge M |g'(0.83)|$, which is impossible unless $M=0$.

Therefore $f(x+a) = f(x)$ forall $x \in \Bbb R$ . And in fact $f$ is constant and its value is the unique fixpoint of $g$ in $[0.83;1]$

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