📂Dynamics

Proof of the Lee-Yang Theorem

Theorem

If a periodic-$3$ orbit exists for the continuous map $f: [a,b] \to [a,b]$, then $f$ is chaotic.

Description

The Li-Yorke Theorem also known as the Period-$3$ Theorem, is often mentioned as a statement that periodic-$3$ induces chaos. Although this theorem seems to be limited to $1$-dimensional maps, the mere existence of a periodic-$3$ orbit ensuring the existence of all periodic orbits is mathematically astonishing. Typically in mathematics, if a property holds for $n$, it often only applies to its divisors, multiples, or smaller numbers.

History

A generalization of the Li-Yorke theorem is the Sharkovsky’s Theorem, but actually, Li-Yorke’s paper was published in 1975 and Sharkovsky’s in 1964, so technically, the Li-Yorke theorem is a special case of Sharkovsky’s. Due to the Cold War, Sharkovsky’s work was recognized late, and by that time, the Li-Yorke theorem had already established itself as a central theorem in chaos theory.

Proof

Strategy: The reason we need a periodic-$3$ orbit is that if there are three periodic points $x_{0} < x_{1} < x_{2}$, the space can be divided into two parts, $L=[x_{0} , x_{1}]$ and $R=[x_{1} , x_{2} ]$. To prove the existence of a periodic-$m$ orbit, we will find a sequence of subintervals where a fixed point $p$ starts from the left $L$, stays at $R$, and finally returns to $L$.

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Part 1. If we define a closed interval $I,i '$ and assume $I’ \subset f(I)$, there exists a subinterval $S \subset I$ that satisfies $f(S) = i '$.

$i '$ being a closed interval, by the Extreme Value Theorem and Intermediate Value Theorem, there exists $s_{0} , s_{1} \in I$ that satisfies $f(s_{0}) = \min i '$ and $f(s_{1}) = \max i '$. If we define the smallest closed interval containing $s_{0}$ and $s_{1}$ as $S$, then $S \subset I$ satisfies $f(S) = i '$ due to the continuity of $f$.

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Part 2. If $I_{0} \subset f ( I_{0} )$, there exists a sequence of closed intervals $\left\{ I_{n} \right\}_{n \in \mathbb{N}}$ that satisfies $I_{n} \subset I_{0}$ and $I_{n-1} = f ( I_{n} )$, and $I_{0} = f^{n} ( I_{n} )$ holds.

If $n=1$, Part 1 implies the existence of $I_{1} \subset I_{0}$ that satisfies $I_{0} = f(I_{1})$.

If $n>1$, let’s assume that $I_{n} \subset I_{0}$ and $I_{n-1} \subset f ( I_{n} )$ hold. Then, because $I_{n} \subset I_{0} \subset f ( I_{0} )$, Part 1 implies the existence of $I_{n+1} \subset I_{0}$ that satisfies $I_{n} = f(I_{n+1})$, and by mathematical induction, the sequence $\left\{ I_{n} \right\}_{n \in \mathbb{N}}$ exists.

Unfolding the recursive containment yields $$I_{0} = f ( I_{1} ) = f \left( f(I_{2} ) \right) = f^{2} ( I_{2} ) = \cdots = f^{n} ( I_{n} )$$, thus obtaining $I_{0} = f^{n} ( I_{n} )$.

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Part 3. $f^{m}$ has a fixed point for all $m \in \mathbb{N}$.

Let’s say the three points $x_{0} < x_{1} < x_{2}$ are a periodic-$3$ orbit of $f$ that satisfy $x_{1} := f(x_{0})$ and $x_{2} := f^{2} (x_{0} )$. For convenience, define $L := [x_{0} ,x_{1} ]$ as left and $R := [x_{1} , x_{2}]$ as right.

• Part 3-1. $R \subset f ( R)$
  We get $$[x_{1} , x_{2} ] \subset [x_{0} , x_{2} ]$$, and $$[x_{0} , x_{2}] = [ f( x_{2} ) , f ( x_{1} ) ] \subset f [ x_{1} , x_{2} ]$$, which yields $$[x_{1} , x_{2} ] \subset [x_{0} , x_{2}] \subset f [ x_{1} , x_{2} ]$$. To summarize, $R \subset f ( R)$ holds.
• Part 3-2. $R \subset f ( L)$ It holds that $$[x_{1} , x_{2}] = [ f( x_{0} ) , f ( x_{1} ) ] \subset f [ x_{0} , x_{1} ]$$. To summarize, $R \subset f ( L )$ holds.
• Part 3-3. A fixed point $p$ of $f^{m}$ exists.

Fixed Point Form of the Intermediate Value Theorem: If $I \subset f (I)$, the continuous function $f$ has a fixed point at the interval $I$.

By the Fixed Point Form of the Intermediate Value Theorem, if we can show the existence of an interval $I$ that satisfies $I \subset f^{m} (I)$ for $f^{m}$, then a fixed point exists.

• Part 3-3-1. $m=1$

  • Since we proved $R \subset f (R )$ in Part 3-1., the intermediate value theorem implies that $f^{1}$ has a fixed point at $R$.

• Part 3-3-2. $m \ne 1$

  • If we define $I_{0} := R$, then Part 3-1. implies that $I_{0} \subset f ( I_{0} )$ holds. Following Part 2, for all $n =1 , \cdots, m-2$, there exists $I_{n} \subset I_{0} = R$ that satisfies $I_{n-1} = f ( I_{n} )$.
  • However, for $n = m-1$, we utilize $R \subset f ( L)$ from Part 3-2 to apply Part 1. Then, because $I_{m-2} \subset R \subset f(L)$, there exists $I_{m-1} \subset L$ that satisfies $f(I_{m-1}) = I_{m-2}$ and $$\begin{align*} I_{m-1} \subset& L \\ \subset& [x_{0} , x_{2} ] \\ =& [ f( x_{2} ) , f( x_{1} ) ] \\ \subset& f [ x_{1} , x_{2} ] \\ =& f ( R ) \\ =& f ( I_{0} ) \\ \subset& f \left( f^{m-2} ( I_{m-2} ) \right) \\ \subset& f \left( f^{m-2} ( f ( I_{m-1} ) ) \right) \\ \subset& f^{m} \left( I_{m-1} \right) \end{align*}$$ To summarize, $I_{m-1} \subset f^{m} (I_{m-1})$ holds, thus by the intermediate value theorem, $f^{m}$ has a fixed point at $I_{m-1} \subset L$.

Let’s call this fixed point of all $m \in \mathbb{N}$ at $f^{m}$ as $p \in I_{m-1} \subset L$.

• Part 3-4. The fixed point $p$ of $f^{m}$ is a periodic-$m$ point of $f$.
  • If $m=1$, the fixed point is essentially a periodic-$1$ point. So, it suffices to show that for $m>1$, there can’t be a fixed point $p$ for natural numbers $1 \le k < m$ according to the definition of $\left\{ I_{k} \right\}_{k=0}^{m-1}$, $$f^{k} ( I_{m-1} ) = f^{k-1} \left( f ( I_{m-1} ) \right) = f^{k-1} \left( I_{m-2} \right) \subset R$$. But since $p \in L$, $f^{k} (p) \in R$ can not be equal. In other words, for $k= 1 , \cdots , m-1$, $f^{k} (p) \ne p$ holds, therefore $p$ is a periodic-$m$ point of $f$.

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Summarizing Part 1.~Part 3., $f$ has a periodic-$m$ orbit for all $m \in \mathbb{N}$, therefore it is chaotic.

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Note that the auxiliary lemma proven in Part 1 provides the preimage $S$ for a given image $i '$, hence it is also referred to as the Preimage Lemma.