We prove that near a hyperbolic fixed point, the dynamics of a smooth map is conjugate to its linearization. This fundamental result justifies the stability analysis via derivatives.

Theorem: Let $f: \mathbb{R} \to \mathbb{R}$ be $C^1$ with a fixed point at $x^* = 0$ (by translation). If $|f'(0)| < 1$ (attracting case), then there exists a neighborhood $U$ of $0$ and a homeomorphism $h: U \to h(U)$ such that:

$h(f(x)) = f'(0) \cdot h(x)$

for all $x \in U$ with $f(x) \in U$. That is, $f$ is locally conjugate to the linear map $x \mapsto \lambda x$ where $\lambda = f'(0)$.

Proof:

Step 1: Setup and normalization. Write $f(x) = \lambda x + g(x)$ where $\lambda = f'(0)$ and $g(x) = o(x)$ as $x \to 0$. We seek $h$ such that $h \circ f = \lambda \cdot h$ near 0, with $h(0) = 0$ and $h'(0) = 1$.

Step 2: Formal construction. Assume $h$ exists with power series $h(x) = x + a_2 x^2 + a_3 x^3 + \cdots$. The conjugacy equation $h(f(x)) = \lambda h(x)$ yields:

$h(\lambda x + g(x)) = \lambda(x + a_2 x^2 + a_3 x^3 + \cdots)$

Expanding the left side using Taylor series and comparing coefficients determines the $a_k$ uniquely. For example, if $g(x) = bx^2 + O(x^3)$, the $x^2$ coefficient gives:

$\lambda a_2 + b = \lambda a_2$

which forces $b = 0$ if $\lambda \neq 1$. More generally, the condition $|\lambda| < 1$ ensures the series converges in a neighborhood of 0.

Step 3: Rigorous construction via graph transform. Define the map $T$ on functions $\phi: U \to \mathbb{R}$ with $\phi(0) = 0$ by:

$(T\phi)(x) = \frac{1}{\lambda}\phi(f(x))$

Under the condition $|\lambda| < 1$, the operator $T$ is a contraction on the space of Lipschitz functions with appropriate norm. By the Banach fixed point theorem, $T$ has a unique fixed point $h$ satisfying:

$h(x) = \frac{1}{\lambda}h(f(x))$

which rearranges to $h(f(x)) = \lambda h(x)$.

Step 4: Verification of homeomorphism property. The function $h$ constructed above satisfies $h(0) = 0$ and $h'(0) = 1$ by the contraction mapping construction. For $x$ near 0, we have $|h'(x) - 1| < \epsilon$ for small $\epsilon$, so $h$ is a local diffeomorphism and hence a homeomorphism on a sufficiently small neighborhood.

Step 5: Extension to repelling case. For $|f'(0)| > 1$ (repelling case), we apply the above argument to $f^{-1}$ in a neighborhood where the inverse exists. The map $f^{-1}$ has an attracting fixed point at 0 with derivative $(f'(0))^{-1}$ satisfying $|(f'(0))^{-1}| < 1$. Thus $f^{-1}$ is linearizable, and consequently $f$ is also linearizable.