Does a continuous mapping $f\colon \mathbb R \to \mathbb R$ which satisfies $f(f(x))=x$ for each $x \in \mathbb R$ necessarily have a fixed point?
Suppose $f(a) = b$ for some $a \neq b$, then $f(b) = f(f(a)) = a$. Define $g(x) = f(x) -x$, and we have $$g(a) = f(a) - a = b-a$$ and $$g(b) = f(b) - b = a-b$$ from $g$ is continuous, by intermediate value theorem, there exists $c$ between $a,b$ such that $$g(c) = f(c) - c = 0$$ because $b-a$ and $a-b$ have opposite signs.
- We can assume that $f(0)\neq0$. Without loss of generality we can then also assume that $f(0)>0$. This is because if $f(f(x))=x$ for all reals $x$, then also $F(F(x))=x$ for all $x$, where I define $F(x)=-f(-x)$. Furthermore, $f(-x)=-x$ iff $F(x)=x$, so if one has a fixed point so does the other. All this amounts to is that we can study $F$ instead of $f$ to get $f(0)>0$.
- Look at the restriction of $f$ to the interval $[0,f(0)]$. Notice that $f$ maps the endpoints of this interval to each other. Plot the graphs of both $f$ and the identity function $id(x)=x$. Why must they intersect in this interval? Bolzano's theorem (or intermediate value theorem) on $g(x):=f(x)-x$.