[proofplan] The proof treats the three parameter regimes separately. For $\mu < 1$, we show that $F_\mu$ is a contraction on $[0,1]$ with unique fixed point $0$, so all orbits converge to $0$. For $\mu = 1$, direct computation shows every point in $[0, \tfrac{1}{2}]$ is fixed and every point in $(\tfrac{1}{2}, 1]$ maps into $[0, \tfrac{1}{2})$ after one iteration. For $\mu > 1$, we invoke the previously established results: [Tent Map is Chaotic for $\mu > 2$](/theorems/2807) and [Tent Map is Chaotic for $1 < \mu < 2$](/theorems/2808), which together cover the entire range $\mu > 1$. [/proofplan]

[step:Define the tent map precisely as a piecewise-linear map]The tent map is the continuous function \begin{align*} F_\mu \colon [0,1] &\to [0, \mu/2] \\ x &\mapsto \begin{cases} \mu x & \text{if } 0 \leq x \leq \tfrac{1}{2}, \\ \mu(1 - x) & \text{if } \tfrac{1}{2} \leq x \leq 1. \end{cases} \end{align*} For $\mu \leq 2$, $F_\mu$ maps $[0,1]$ into $[0,1]$ (since $F_\mu(1/2) = \mu/2 \leq 1$), so $F_\mu \colon [0,1] \to [0,1]$. For $\mu > 2$, $F_\mu$ maps $[0,1]$ into $[0, \mu/2]$ with $\mu/2 > 1$, so orbits may escape $[0,1]$; the restriction to the invariant set is relevant for the chaotic dynamics.[/step]

[guided]The tent map $F_\mu$ is the simplest piecewise-linear unimodal map. It is defined by two linear pieces: \begin{align*} F_\mu(x) = \begin{cases} \mu x & \text{if } 0 \leq x \leq 1/2, \\ \mu(1-x) & \text{if } 1/2 \leq x \leq 1, \end{cases} \end{align*} with slopes $+\mu$ on the left half and $-\mu$ on the right half. The parameter $\mu > 0$ controls the height of the tent: the peak value is $F_\mu(1/2) = \mu/2$. At $\mu = 1$, the peak $F_\mu(1/2) = 1/2$ sits exactly at the midpoint of $[0,1]$. For $\mu < 1$, the peak sags below $1/2$, meaning $F_\mu([0,1]) \subset [0, \mu/2] \subsetneq [0, 1/2)$ -- the map is a contraction. For $\mu > 1$, the peak rises above $1/2$: the slopes $|\mu| > 1$ make $F_\mu$ locally expanding, which is the source of chaotic behaviour. For $\mu \leq 2$, the image $[0, \mu/2] \subset [0,1]$, so $F_\mu \colon [0,1] \to [0,1]$ maps the unit interval into itself. For $\mu > 2$, the peak exceeds $1$ and orbits may escape $[0,1]$; the chaotic dynamics then lives on a Cantor-like invariant subset. This trichotomy in the peak height -- below, at, or above $1/2$ -- drives the qualitative trichotomy in dynamics that the theorem characterises.[/guided]

[step:Case $\mu < 1$: show all orbits converge to $0$]For $\mu < 1$ and any $x \in [0,1]$, the piecewise formula gives \begin{align*} F_\mu(x) = \begin{cases} \mu x & \text{if } x \leq \tfrac{1}{2}, \\ \mu(1-x) & \text{if } x \geq \tfrac{1}{2}. \end{cases} \end{align*} On $[0, 1/2]$, $F_\mu(x) = \mu x \leq \mu/2 < 1/2$. On $[1/2, 1]$, $F_\mu(x) = \mu(1-x) \leq \mu/2 < 1/2$. Hence \begin{align*} 0 \leq F_\mu(x) \leq \frac{\mu}{2} < \frac{1}{2} \quad \text{for all } x \in [0,1]. \end{align*} Since $F_\mu(x) < 1/2$ for every $x \in [0,1]$, the image always falls in the first branch $[0, 1/2]$, so the second iterate satisfies $F_\mu^2(x) = \mu \cdot F_\mu(x)$. By induction, for all $n \geq 1$: \begin{align*} F_\mu^n(x) = \mu^{n-1} \cdot F_\mu(x) \leq \frac{\mu^n}{2}. \end{align*} Since $0 < \mu < 1$, we have $\mu^n \to 0$ as $n \to \infty$, so $F_\mu^n(x) \to 0$ for every $x \in [0,1]$.[/step]

[guided]For $\mu < 1$, the tent map is a contraction: the slopes $\pm \mu$ satisfy $|\mu| < 1$, so $F_\mu$ shrinks distances. Let us trace through the argument in detail. On both branches, $F_\mu(x) \leq \mu/2 < 1/2$: \begin{align*} \text{Left branch: } F_\mu(x) = \mu x \leq \mu/2 < 1/2. \qquad \text{Right branch: } F_\mu(x) = \mu(1-x) \leq \mu/2 < 1/2. \end{align*} After one application, every point in $[0,1]$ lands in $[0, \mu/2] \subset [0, 1/2)$. Once in $[0, 1/2)$, the point is in the left branch, so the map acts as $x \mapsto \mu x$ (pure linear contraction). By induction, $F_\mu^n(x) = \mu^{n-1} \cdot F_\mu(x) \leq \mu^n / 2$ for all $n \geq 1$. Since $0 < \mu < 1$, $\mu^n \to 0$ geometrically, so $F_\mu^n(x) \to 0$ for every $x \in [0,1]$. Every orbit converges to $0$. We verify that $x = 0$ is the unique fixed point. On $[0, 1/2]$: $F_\mu(x) = x$ gives $\mu x = x$, hence $x(\mu - 1) = 0$, so $x = 0$ (since $\mu \neq 1$). On $[1/2, 1]$: $F_\mu(x) = x$ gives $\mu(1-x) = x$, hence $x = \mu/(1+\mu)$. Since $\mu < 1$, we have $\mu/(1+\mu) < 1/2$, so this candidate falls outside $[1/2, 1]$ and is not a valid fixed point on this branch. Hence $x = 0$ is the unique fixed point of $F_\mu$ for $\mu < 1$, and every orbit converges to it. There are no periodic orbits of any period $> 1$, and the dynamics is as simple as possible.[/guided]

[step:Case $\mu = 1$: verify that every point in $[0, 1/2]$ is fixed] For $\mu = 1$, the tent map becomes \begin{align*} F_1(x) = \begin{cases} x & \text{if } 0 \leq x \leq \tfrac{1}{2}, \\ 1 - x & \text{if } \tfrac{1}{2} \leq x \leq 1. \end{cases} \end{align*} Every $x \in [0, 1/2]$ satisfies $F_1(x) = x$, so $[0, 1/2]$ consists entirely of fixed points. For $x \in (1/2, 1]$, $F_1(x) = 1 - x \in [0, 1/2)$. Hence $F_1^2(x) = F_1(1 - x) = 1 - x$ (since $1 - x < 1/2$, so $F_1(1-x) = 1 - x$). Thus every point in $(1/2, 1]$ reaches the fixed point $1 - x \in [0, 1/2)$ after exactly one iteration and remains there.[/step]

[guided]At $\mu = 1$, the tent map becomes piecewise: \begin{align*} F_1(x) = \begin{cases} x & \text{if } 0 \leq x \leq 1/2, \\ 1-x & \text{if } 1/2 \leq x \leq 1. \end{cases} \end{align*} The left branch is the identity: $F_1(x) = x$ for all $x \in [0, 1/2]$. Every such $x$ is a fixed point. The right branch is the reflection $x \mapsto 1-x$, which maps $(1/2, 1]$ bijectively onto $[0, 1/2)$. For $x \in (1/2, 1]$: $F_1(x) = 1-x \in [0, 1/2)$, so the point immediately enters the left branch. On the next iteration, $F_1^2(x) = F_1(1-x) = 1-x$ (since $1-x < 1/2$, the identity branch applies). The point is now fixed at $1-x$ for all subsequent iterations. Note that $x = 1/2$ is also a fixed point, since $F_1(1/2) = 1/2$ from either branch. The full fixed-point set is therefore the closed interval $[0, 1/2]$. The dynamics is as tame as possible: every orbit reaches a fixed point in at most one step. There is no chaos, no periodicity beyond period $1$, and no sensitive dependence on initial conditions. The map $F_1$ is precisely at the boundary between the contracting regime ($\mu < 1$) and the expanding regime ($\mu > 1$).[/guided]

[step:Case $\mu > 1$: invoke the established chaotic results for both sub-regimes]For $\mu > 2$: the [Tent Map is Chaotic for $\mu > 2$](/theorems/2807) establishes that $F_\mu$ has a horseshoe (with $K_1 = (0, 1/\mu)$ and $K_2 = (1 - 1/\mu, 1)$ both mapping onto $J = (0,1)$), so $F_\mu$ is chaotic in the sense of Glendinning. For $\mu = 2$: the tent map with $\mu = 2$ is conjugate to the sawtooth map $x \mapsto 2x \pmod{1}$ via the conjugacy $h(x) = \frac{1}{\pi}\sin^{-1}(\sqrt{x})$ (or equivalently, $F_2$ on $[0,1]$ is topologically conjugate to the doubling map on $[0,1)$). Since the sawtooth map is Glendinning-chaotic (it has a horseshoe with $n = 1$), the conjugate $F_2$ is also chaotic. For $1 < \mu < 2$: the [Tent Map is Chaotic for $1 < \mu < 2$](/theorems/2808) establishes that $F_\mu^{2^n}$ has a horseshoe for some $n \geq 1$ (specifically, for $n$ chosen so that $\mu^{2^n} > 2$, which is possible since $\mu > 1$ implies $\mu^{2^n} \to \infty$). Hence $F_\mu$ is chaotic in the sense of Glendinning. In all cases with $\mu > 1$, $F_\mu$ is chaotic.[/step]

[guided]The case $\mu > 1$ is where the dynamics become rich. The mechanism is uniform across the entire range: the slopes $\pm \mu$ satisfy $|\mu| > 1$, so $F_\mu$ is locally expanding. Local expansion is the hallmark of chaotic behaviour -- nearby points are pushed apart at each iteration, creating sensitive dependence on initial conditions. The difference between the sub-cases $\mu > 2$, $\mu = 2$, and $1 < \mu < 2$ is merely how quickly the horseshoe structure becomes visible. **For $\mu > 2$**: the horseshoe is immediate. The peak $F_\mu(1/2) = \mu/2 > 1$ exceeds the interval $[0,1]$. The [Tent Map is Chaotic for $\mu > 2$](/theorems/2807) identifies two disjoint sub-intervals $K_1 = (0, 1/\mu)$ and $K_2 = (1 - 1/\mu, 1)$ such that $F_\mu(K_1) = F_\mu(K_2) = (0, 1)$. Each interval maps onto all of $(0,1)$ in a single iteration -- this is a horseshoe with $n = 1$, directly establishing Glendinning chaos. **For $\mu = 2$**: the peak $F_2(1/2) = 1$ is exactly at the boundary. The horseshoe construction uses $K_1 = (0, 1/2)$ and $K_2 = (1/2, 1)$, with $F_2(K_1) = F_2(K_2) = (0, 1)$. Alternatively, $F_2$ is topologically conjugate to the doubling map $x \mapsto 2x \pmod{1}$ via the conjugacy $h(x) = (1/\pi)\sin^{-1}(\sqrt{x})$. Since topological conjugacy preserves chaotic behaviour, $F_2$ is chaotic. **For $1 < \mu < 2$**: the tent stays within $[0,1]$ ($F_\mu(1/2) = \mu/2 < 1$), so no single-step horseshoe exists. However, the [Tent Map is Chaotic for $1 < \mu < 2$](/theorems/2808) shows that the iterate $F_\mu^{2^n}$ has a horseshoe for some $n \geq 1$. The key observation is that $F_\mu^2$ restricted to an appropriate sub-interval behaves like a tent map with effective parameter $\mu^2$. If $\mu^2 > 2$ (i.e., $\mu > \sqrt{2}$), then $F_\mu^2$ has a direct horseshoe. If not, iterate again: $F_\mu^4$ has effective parameter $\mu^4$. Since $\mu > 1$, the sequence $\mu^{2^n}$ grows without bound and eventually exceeds $2$ at the smallest $n$ with $\mu^{2^n} > 2$. A horseshoe for $F_\mu^{2^n}$ implies chaos for $F_\mu$ itself (Glendinning chaos is inherited by roots of iterates). In all cases with $\mu > 1$, $F_\mu$ is chaotic in the sense of Glendinning.[/guided]

[step:Combine the three cases to complete the characterisation] Collecting the results: - **$\mu < 1$:** Every orbit converges to $x = 0$ (Step 2). - **$\mu = 1$:** Every point in $[0, 1/2]$ is a fixed point, and every point in $(1/2, 1]$ maps to a fixed point in one iteration (Step 3). - **$\mu > 1$:** $F_\mu$ is chaotic in the sense of Glendinning, by [Tent Map is Chaotic for $\mu > 2$](/theorems/2807) and [Tent Map is Chaotic for $1 < \mu < 2$](/theorems/2808) (Step 4). This exhausts all values of the parameter $\mu > 0$ and completes the characterisation of the tent map's dynamics. [/step]