Prove that $T$ has an orbit of period 3

Suppose that $T$ is continuous map from an interval $I$ to itself. Moreover, suppose that there exists $x_1 < x_2 < x_3 < x_4 $ such that

$$T(x_1) = x_2, T(x_2) = x_3, T(x_3) = x_4\ \ \text{ and }\ \ T(x_4) \le x_1.$$

Prove that $T$ has an orbit of period 3.

I don't really know how to start the problem. I am wondering how do I apply Sharkovskii's Theorem to the problem? Helps are much appreciated!

As we are given $x_1 < x_2 < x_3 < x_4 $ such that $T(x_1) = x_2, T(x_2) = x_3, T(x_3) = x_4 $ and $T(x_4) \le x_1$, we get

$ x_1 < x_2 = T(x_1) < x_3 = T(x_2) < x_4 = T(x_3)$. Also note that $T(x_4) \le x_1$, then $T(x_4) \le x_1 < T(x_3)$.

Solution 1:

$$f^3(x_1) = x_4> x_2 > x_1$$. $$f^3(x_2) = f(x_4) < x_1$$

Let $g(x) = f^3(x) -x$. Then $g(x)$ is continuous (since a continuous map on a continuous map is continuous, and subtraction preserves continuity), and: $$ g(x_1) = x_4 - x_1 > 0 \\ g(x_2) = f(x_4) - x_1 < 0$$

Then by the intermediate value theorem on $g(x)$ which is continuous and positive at $x_1$ and negative at $x_2$, there is some $x_0 : x_1 < x_20 < x_2 : g(x_0) = 0$.

Now consider $$f^3(x_0) = g(x_0) + x_0 = x_0$$ so $x_0$ is a fixed point of $f^3(x)$ which means that it is part of an orbit of period 3 or period 1.

But if it is of period 1, then the same argument using $x_0$ in place of $x_1$ results in another fixed point of $f^3(x)$ between $x_0$ and $x_2$. Now there must be some largest point $p$ between $x_1$ and $x_2$ such that $f(p) = p$ and in fact $p$ cannot be $x_2$ because then the map would not be continuous at $x_2$ as $f(x_2) > x_2$. Apply that same argument again for $p$, and you obtain a point between $p$ and $x_2$ which must be a fixed point of $f^3(x)$ but cannot be a fixed point of $f(x)$; this is the desired point on an orbit of period 3.