I asked my students a question in connecting with Sarkovskii's theorem in midterm exam:
"Can a continuous function on $R$ have a priodic point of period 48 and not one of period 24?"
This is similar to one of exercises of Devaney (A first course in chaotic dynamical systems - page 151). We know that $24=2^33$ procedes $48=2^43$ in Sarkovskii's ordering. Hence if a function has periodic points of period $24$ then it has periodic points of $48$. But the converse is not nececcerily true. \newline The converse of Sarkovskii's theorem gives the existennce of a continuous function that if has priodic points of priod 48 and no cycles of perod $n$ which procedes 48 specifically 24 so we can say YES as the answer to the above question.
My Question: How we can answer to the above question without refering to converse of Sarkovskii' theorem?
My answer: Suppose that we say No. This means that it can not be possible that a continous function have periodic points of period 48 but no one's of period 24. But this means that any continuous function which has periodic points of period 48 then has periodic points of period 24. Well, if we apply this argument to 24 and 12, and then to 12 and 3 we will have the priod 3 which is the first number in Sarkovskii's ordering, i.e. we are saying that any periodic function of period 48 has periodic points of all periods and we know that this is not true. "Can I say YES as the answer of the above bold question by this argument or I am missing something? "
This part of the converse is elementary. Suppose $p$ has period $48$, i.e. $f^{48}(p) = p$. Letting $g = f^{24}$, that says $g^2(p) = p$. Suppose $p$ does not have period $24$, so $g(p) \ne p$. Now one of $g(p) - p $ and $g(g(p)) - g(p)$ is positive and the other negative, so by the Intermediate Value Theorem there is $t$ between $p$ and $g(p)$ such that $g(t) = t$, i.e. $t$ is a periodic point of $f$ of order $24$.