Rudin中Stone-Weierstrass推论7.27替换函数后的证明变化及疑问

Hey there, let's unpack your two questions clearly— I remember wrestling with these exact points when working through Rudin's Principles of Mathematical Analysis:

First, let's recap the context: Corollary 7.27 is the classic Weierstrass Approximation Theorem, which states every continuous function on a closed interval can be uniformly approximated by polynomials. Rudin's proof leans on the Stone-Weierstrass Theorem (Theorem 7.26), but he uses the specific case of approximating $|x|$ as intuitive scaffolding.

If you swap $|x|$ for an arbitrary continuous $g(x)$ with $g(0)=0$, here's how the proof dynamics shift:

• If you're just leaning on Weierstrass itself: Since $g(x)$ is continuous on $[-1,1]$, Weierstrass already guarantees it can be uniformly approximated by polynomials. So if your goal is to prove any $h(0)=0$ continuous function can be approximated by polynomials, you don't need $g(x)$ at all—Stone-Weierstrass directly applies to the full algebra of polynomials, which separates points and includes constants, hence is dense in $C([-1,1])$.
• If you try to mimic the $|x|$-based construction: The original intuition with $|x|$ is that some $h(0)=0$ functions can be written as $h(x) = k(x)|x|$ where $k(x)$ is continuous. But this only works if $\lim_{x \to 0} \frac{h(x)}{g(x)}$ exists and is finite, making $k(x) = \frac{h(x)}{g(x)}$ (with $k(0)$ defined as that limit) continuous. This isn't true for all pairs of $h(0)=0$ and $g(0)=0$:
  • Example: Take $g(x)=x^2$ and $h(x)=x$. $\frac{h(x)}{g(x)} = \frac{1}{x}$, which blows up as $x \to 0$, so you can't write $h(x)$ as a continuous $k(x)$ times $g(x)$.
  • Even for $g(x)=|x|$, there are counterexamples: $h(x)=x\sin\left(\frac{1}{x}\right)$ (with $h(0)=0$) gives $\frac{h(x)}{|x|} = \text{sign}(x)\sin\left(\frac{1}{x}\right)$, which oscillates wildly near 0 and has no limit, so $k(x)$ isn't continuous.
• If you're talking about the algebra generated by $g(x)$: If you want to use polynomials in $g(x)$ (i.e., functions like $a_0 + a_1g(x) + a_2g(x)^2 + ... + a_ng(x)^n$) to approximate continuous functions, this only works if $g(x)$ separates points on $[-1,1]$ (i.e., $x \neq y$ implies $g(x) \neq g(y)$). For example:
  • If $g(x)=x$ (strictly monotonic), the algebra is just all polynomials, so dense.
  • If $g(x)=|x|$ (not injective), the algebra only approximates even functions—you can't get odd functions like $f(x)=x$ since all polynomials in $|x|$ are even.

Rudin is referring to a concrete polynomial sequence built to approximate $|x|$. Here's the breakdown:

• He starts with the binomial expansion of $(1-t)^{1/2}$ for $t \in [0,1]$:
  $$(1-t)^{1/2} = \sum_{k=0}^\infty \binom{1/2}{k} (-1)^k t^k$$
  This series converges uniformly on $[0,1]$ (it's an alternating series with decreasing terms, so the remainder is bounded by the first omitted term).
• Let $P_n^*(t)$ be the first $n+1$ terms of this expansion—this is a polynomial in $t$.
• Now substitute $t = 1 - x^2$: since $x \in [-1,1]$, $t \in [0,1]$, and $(1 - (1-x²⁾⁾{1/2} = |x|$. So the polynomial $P_n(x) = P_n^*(1 - x^2)$ uniformly approximates $|x|$ on $[-1,1]$.

That's what Rudin means by $P_n(x)=P_n^*(x)...$—he's taking the polynomial approximating $(1-t)^{1/2}$, plugging in $1-x^2$, and getting a polynomial that approximates $|x|$.

内容的提问来源于stack exchange，提问作者kemb