请教魏尔斯特拉斯逼近定理的证明有一处不明白

abababa

Stone-Weierstrass Theorem.pdf (142.59 KB, Downloads: 780)

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如题，附件里是下载的证明，但有一处不明白，就是
For $0 \le x \le 1$ define
\[P_n(x)=\int_{-1}^{1}g(x+t)Q_n(t)dt\]

$P_n(x)$ is a polynomial of degree $\le 2n$:
\[\int_{-1}^{1}g(x+t)t^kdt=\int_{x-1}^{x+1}g(u)(u-x)^kdu=\sum_{i=0}^{k}(-1)^i\binom{k}{i}x^i\int_{x-1}^{x+1}g(u)u^{k-i}du\]

这里说$P_n(x)$是一个次数不超过$2n$的多项式，我不明白它为什么是多项式？求和里面虽然有$x^i$，但还要乘上积分的那项，而积分的那项积出来应该是一个$x$的函数，特别是根据后面的叙述，那个$g(u)$还是从$f(x)$得到的，这一项怎么保证是多项式呢？
把附件前几段的证明打在这里：
考察任意连续函数$g:\mathbb{R}\to\mathbb{R}$，并且满足$g(x)=0,x\not\in[0,1]$。对每个$n\ge 1$设
\[Q_n(x)=\frac{(1-x^2)^n}{\int_{-1}^{1}(1-x^2)^ndx}\]
则$Q_n(x)$是钟形的，关于$x=0$对称的，并且在$[-1,1]$上积分为$1$。
接下来附件里的证明，就是我所问的问题。

abababa

回复 1# abababa

pdf里的证明还是没弄懂。我在这部分做了其它处理，先用$g(x)$的支撑性把积分区间变一下。
因为当$x<0$或$x>1$时$g(x)=0$，因此$x+t<0$即$t<-x$，或$x+t>1$即$t>1-x$时$g(x+t)=0$，于是
\[\int_{-1}^{1}g(x+t)t^kdt=\int_{-x}^{1-x}g(x+t)t^kdt=\int_{0}^{1}g(u)(u-x)^kdu\]
这样积分限变为实数，就能二项展开了，展开后得到的也正是一个多项式。

hbghlyj

原地址 Sequences and Series of Functions.pdf

Sequences and Series of Functions (Rudin)

Stone-Weierstrass Theorem: Let $f : [a, b] →\Bbb R$ be continuous. Then there exists a sequence of polynomials $(P_n(x))$ such that converges uniformly to $f$ on $[a, b]$.

Proof: Consider any continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ that satisfies $g(x)=0$ for $x \notin[0,1]$. For each $n \geq 1$ set
$$
Q_n(x)=\frac{\left(1-x^2\right)^n}{\int_{-1}^1\left(1-x^2\right)^n d x} .
$$
The area under the curve $y=Q_n(x)$ over $[-1,1]$ is equal to 1 , and has the shape of a bell curve with most of it's area concentrated in a narrow band above $x=0$. The polynomials $Q_{10}(x), Q_{20}(x), Q_{30}(x)$ are plotted below:

For $0 \leq x \leq 1$ define
$$
P_n(x)=\int_{-1}^1 g(x+t) Q_n(t) d t .
$$
$P_n(x)$ is a polynomial of degree $\leq 2 n$ :
$$
\int_{-1}^1 g(x+t) t^k d t=\int_{x-1}^{x+1} g(u)(u-x)^k d u=\sum_{i=0}^k(-1)^i\left(\begin{array}{c}
k \\
i
\end{array}\right) x^i \int_{x-1}^{x+1} g(u) u^{k-i} d u .
$$
We have
$$
\left|P_n(x)-g(x)\right| \leq \int_{-1}^1|g(x+t)-g(x)| Q_n(t) d t .
$$
Let $M=\sup g(x)$. Since $g$ is uniformly continuous on $\mathbb{R}$, there exists $\delta_k>0$ such that $|t|<\delta_k$ implies $|g(x+t)-g(x)| \leq \frac{1}{k}$ for all $x$. This yields
$$
\left|P_n(x)-g(x)\right| \leq 2 M \int_{-1}^{-\delta_k} Q_n(t) d t+\frac{1}{k} \int_{-\delta_k}^{\delta_k} Q_n(t) d t+2 M \int_{\delta_k}^1 Q_n(t) d t .
$$
We have
$$
\frac{1}{k} \int_{-\delta_k}^{\delta_k} Q_n(t) d t \leq \frac{1}{k} .
$$
We also have
$$
\int_{-1}^1\left(1-x^2\right)^n d x \geq 2 \int_0^1(1-x)^n d x=\frac{2}{n+1},
$$
hence for $\delta_k \leq|x| \leq 1$ we have
$$
\left|Q_n(x)\right| \leq \frac{n+1}{2}\left(1-\delta_k^2\right)^n .
$$
Therefore
$$
|P_n(x)-g(x)| \leq 2 M(n+1)\left(1-\delta_k^2\right)^n+\frac{1}{k}
$$
for $0 \leq x \leq 1$, i.e.
$$
\left\|P_n-g\right\| \leq 2 M(n+1)\left(1-\delta_k^2\right)^n+\frac{1}{k} .
$$
Let $\epsilon>0$ be given. Choose $k$ so that $\frac{1}{k}<\frac{\epsilon}{2}$. For sufficiently large $n$, $2 M(n+1)\left(1-\delta_k^2\right)^n<\frac{\epsilon}{2}$, hence $\left\|P_n-g\right\|<\epsilon$. Therefore $P_n \rightarrow g$ uniformly on $[0,1]$.

Given $f:[a, b] \rightarrow \mathbb{R}, g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$ satisfies $g(a)=$ $g(b)=0$ and $h(x)=g((b-a) x+a)$ satisfies $h(0)=h(1)=0$. If $P_n(x) \rightarrow$ $h(x)$ uniformly on $[0,1]$, then $P_n(x) \rightarrow g((b-a) x+a)$ on $[0,1]$, therefore $P_n\left(\frac{x-a}{b-a}\right) \rightarrow g(x)$ uniformly on $[a, b]$, therefore
$$
P_n\left(\frac{x-a}{b-a}\right)+f(a)+\frac{f(b)-f(a)}{b-a}(x-a) \rightarrow f(x)
$$
uniformly on $[a, b]$.

hbghlyj

第2页的不等式$$\int_{-1}^1\left(1-x^2\right)^n\rmd x \geq 2 \int_0^1(1-x)^n\rmd x=\frac{2}{n+1},$$证明如下:
\begin{align*}\int_{-1}^1\left(1-x^2\right)^n\rmd x&=2\int_0^1\left(1-x^2\right)^n\rmd x
\\&\ge 2 \int_0^1(1-x)^n\rmd x&\because x>x^2\text{ for }0<x<1\\&=2 \int_0^1x^n\rmd x\\&=\frac{2}{n+1}\end{align*}

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这里与Rudin原书上不同.
Rudin, 2nd Edition第147页 (3rd Edition第159-160页) 使用的不等式是 $(1-x^2)^n\ge1-nx^2$

第159页下半	第160页上半