Continuous Maps with Every Point Periodic

Detailed Solution

First, we show that $f$ is injective. Suppose that $f(a)=f(b)$. Since every point is periodic, there exist $m,n\in\mathbb N$ such that $f^m(a)=a$ and $f^n(b)=b$. Choose a common multiple $N$ of $m$ and $n$. Then $f^N(a)=a$ and $f^N(b)=b$. Applying $f^{N-1}$ to $f(a)=f(b)$ gives $f^N(a)=f^N(b)$, and hence $a=b$. Thus $f$ is injective. A continuous injective function on an interval is strictly monotone. Therefore $f$ is either strictly increasing or strictly decreasing. If $f$ is strictly increasing, suppose that $f(x)>x$ for some $x$. Then \[ x<f(x)<f^2(x)<f^3(x)<\cdots, \] so no positive iterate of $x$ can equal $x$, contradicting the hypothesis. Similarly, $f(x)<x$ is impossible. Hence $f(x)=x$ for every $x$, so $f$ is the identity. Now suppose that $f$ is strictly decreasing. Define \[ g=f^2. \] Then $g$ is strictly increasing. Moreover, every point is periodic under $g$: if $f^n(x)=x$, then \[ g^n(x)=(f^2)^n(x)=f^{2n}(x)=x. \] Applying the increasing-case argument to $g$, we obtain \[ g(x)=x\qquad\text{for all }x\in\mathbb R. \] Therefore \[ f^2=\operatorname{id}_{\mathbb R}. \] Thus $f$ is a strictly decreasing involution. Consequently, the complete classification is \[ \boxed{ \text{$f$ is either the identity map or a strictly decreasing involution.} } \] Hence the correct option is \[ \boxed{\mathrm{C}}. \] Option (A) is false because $f(x)=-x$ is a continuous strictly decreasing involution and satisfies $f^2(x)=x$ for every $x$. Option (B) is false because although $f$ is indeed a bijection, continuity and injectivity force it to be strictly monotone. Option (D) is false because $f(x)=|x|$ does not satisfy the hypothesis. For example, if $x<0$, then \[ f(x)=|x|>0,\qquad f^k(x)=|x| \quad (k\ge1), \] so the orbit never returns to the original negative point $x$.