牛顿法的几乎必然收敛性证明技术问询

Alright, let's work through this problem step by step. We're dealing with a real polynomial $f$ that has at least one real root, and we need to show that all but a zero-measure set of starting points $x_0 \in \mathbb{R}$ will produce a Newton iteration sequence that converges to some real root of $f$.

1. Local Convergence Around Real Roots #

First, let's recall the local behavior of Newton's method near a real root $\alpha$:

• For a simple root ($f'(\alpha) \neq 0$), there's an open neighborhood $U_\alpha$ around $\alpha$ where any initial point $x_0 \in U_\alpha$ will generate a sequence that converges quadratically to $\alpha$. This comes directly from the error bound provided:
  $$|\alpha - x_{n+1}| = \frac{|f''(\zeta_n)|}{2|f'(x_n)|} \cdot |\alpha - x_n|^2$$
  Since $f$ is a polynomial, $f'$ and $f''$ are continuous everywhere. Close to $\alpha$, $|f'(x_n)|$ stays above some positive bound (because $f'(\alpha) \neq 0$) and $|f''(\zeta_n)|$ is bounded, so the error shrinks extremely quickly—guaranteeing convergence.
• For a multiple root (multiplicity $k \geq 2$), the convergence is linear instead of quadratic, but the key point still holds: there's a neighborhood around $\alpha$ where any starting point will eventually converge to $\alpha$.

2. Identifying "Bad" Initial Points #

The only starting points that fail to converge to a real root fall into a few categories, and every single one of these categories has measure zero (i.e., they're negligible in terms of real numbers):

• Singular points: Any $x_0$ where $f'(x_0) = 0$, or any initial point whose iteration eventually hits such a point. Since $f'$ is a polynomial, it has finitely many zeros. Plus, each singular point has finitely many pre-images under the Newton map (because solving $x_{n} = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}$ for $x_{n-1}$ gives a polynomial equation with finite solutions). The entire set of these points is finite, so measure zero.
• Cycle points: Initial points that enter a periodic loop (e.g., $x_2 = x_0$, $x_1 \neq x_0$) without ever reaching a root. For polynomials, these cycles correspond to solutions of polynomial equations (like $x_0 = N(N(x_0))$, where $N(x)$ is the Newton map). These equations have finite solutions, so the set of cycle points is finite—again, measure zero.
• Divergent points: Starting points where the sequence blows up to $\pm\infty$. For real polynomials, these regions are "thin" (they don't contain any intervals of positive length) and make up a zero-measure set. Intuitively, the basins of attraction around real roots dominate the real line, leaving only tiny, negligible regions where iteration diverges.

3. Wrapping It Up #

The set of "good" initial points is the entire real line minus the union of these zero-measure sets. Since the union of even countably many zero-measure sets is still zero-measure, the complement (our good points) has full measure in $\mathbb{R}$.

In short: almost every initial point will lead to a Newton sequence that converges to some real root of $f$.

内容的提问来源于stack exchange，提问作者syn3449