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By admin 数学代写 2022年2月23日

$6.10$ Weierstrass Approximation
It turns out that if $f$ is a continuous real valued function defined on an interval, $[a, b]$ then there exists a sequence of polynomials, $\left{p_{n}\right}$ such that the sequence converges uniformly to $f$ on $[a, b]$. I will first show this is true for the interval $[0,1]$ and then verify it is true on any closed and bounded interval. First here is a little lemma which is interesting for its own sake in probability. It is actually an estimate for the variance of a binomial distribution.
Lemma 6.10.1 The following estimate holds for $x \in[0,1]$ and $m \geq 2$.
$$
\sum_{k=0}^{m}\left(\begin{array}{c}
m \
k
\end{array}\right)(k-m x)^{2} x^{k}(1-x)^{m-k} \leqslant \frac{1}{4} m
$$
Proof: Using the binomial theorem and expanding $(k-m x)^{2}$,
$$
\begin{gathered}
\sum_{k=0}^{m}\left(\begin{array}{c}
m \
k
\end{array}\right)(k-m x)^{2} x^{k}(1-x)^{m-k} \
=\sum_{k=0}^{m}\left(\begin{array}{c}
m \
k
\end{array}\right) k^{2} x^{k}(1-x)^{m-k}-2 m x \sum_{k=0}^{m}\left(\begin{array}{c}
m \
k
\end{array}\right) k x^{k}(1-x)^{m-k}+m^{2} x^{2} \
=\sum_{k=0}^{m}\left(\begin{array}{c}
m \
k
\end{array}\right) k(k-1) x^{k}(1-x)^{m-k}+(k-2 m k x) \sum_{k=0}^{m}\left(\begin{array}{c}
m \
k
\end{array}\right) x^{k}(1-x)^{m-k}+m^{2} x^{2} \
=\sum_{k=2}^{m} \frac{m !}{(k-2) !(m-k) !} x^{k}(1-x)^{m-k}+ \
\quad(1-2 m x) \sum_{k=1}^{m} \frac{m !}{(k-1) !(m-k) !} x^{k}(1-x)^{m-k}+m^{2} x^{2} \
\quad(1-2 m x) \sum_{k=0}^{m-1} \frac{m !}{k !((m-1)-k) !} x^{k+1}(1-x)^{(m-1)-k}+m^{2} x^{2} \
\sum_{k=0}^{m-2} \frac{m !}{k !(m-2-k) !} x^{k+2}(1-x)^{(m-2)-k}+ \
x-1) x^{2}+(1-2 m x) m x+m^{2} x^{2}=m\left(x-x^{2}\right) \leq \frac{m}{4}
\end{gathered}
$$
Now let $f$ be a continuous function defined on $[0,1]$. Let $p_{n}$ be the polynomial defined by
$$
p_{n}(x) \equiv \sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right) f\left(\frac{k}{n}\right) x^{k}(1-x)^{n-k}
$$
Theorem $6.10 .2$ The sequence of polynomials in 6.5 converges uniformly to $f$ on $[0,1]$. These polynomials are called the Bernstein polynomials.
Proof: By the binomial theorem,
$$
f(x)=f(x) \sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right) x^{k}(1-x)^{n-k}=\sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right) f(x) x^{k}(1-x)^{n-k}
$$
and so by the triangle inequality
$$
\left|f(x)-p_{n}(x)\right| \leq \sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right)\left|f\left(\frac{k}{n}\right)-f(x)\right| x^{k}(1-x)^{n-k}
$$
At this point you break the sum into two pieces, those values of $k$ such that $k / n$ is close to $x$ and those values for $k$ such that $k / n$ is not so close to $x$. Thus
$$
\begin{aligned}
\left|f(x)-p_{n}(x)\right| \leq & \sum_{|x-(k / n)|<\delta}\left(\begin{array}{l} n \ k \end{array}\right)\left|f\left(\frac{k}{n}\right)-f(x)\right| x^{k}(1-x)^{n-k} \ &+\sum_{|x-(k / n)| \geq \delta}\left(\begin{array}{l} n \ k \end{array}\right)\left|f\left(\frac{k}{n}\right)-f(x)\right| x^{k}(1-x)^{n-k} \end{aligned} $$ where $\delta$ is a positive number chosen in an auspicious manner about to be described. Since $f$ is continuous on $[0,1]$, it follows from Theorems $4.8 .2$ and $6.7 .2$ that $f$ is uniformly continuous. Therefore, letting $\varepsilon>0$, there exists $\delta>0$ such that if $|x-y|<$ $\delta$, then $|f(x)-f(y)|<\varepsilon / 2$. This is the auspicious choice for $\delta .$ Also, by Lemma $6.3 .2$ $|f(x)|$ for $x \in[0,1]$ is bounded by some number $M$. Thus $6.6$ implies that for $x \in[0,1]$,
$$
\begin{aligned}
\left|f(x)-p_{n}(x)\right| \leq & \sum_{|x-(k / n)|<\delta}\left(\begin{array}{l}
n \
k
\end{array}\right) \frac{\varepsilon}{2} x^{k}(1-x)^{n-k} \
&+2 M \sum_{|n x-k| \geq n \delta}\left(\begin{array}{l}
n \
k
\end{array}\right) x^{k}(1-x)^{n-k} \
\leq & \frac{\varepsilon}{2}+2 M \sum_{|n x-k| \geq n \delta}\left(\begin{array}{l}
n \
k
\end{array}\right) \frac{(k-n x)^{2}}{n^{2} \delta^{2}} x^{k}(1-x)^{n-k} \
\leq & \frac{\varepsilon}{2}+\frac{2 M}{n^{2} \delta^{2}} \sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right)(k-n x)^{2} x^{k}(1-x)^{n-k}
\end{aligned}
$$
Now by Lemma $6.10 .1$ there is an estimate for the last sum. Using this estimate yields for all $x \in[0,1]$,
$$
\left|f(x)-p_{n}(x)\right| \leq \frac{\varepsilon}{2}+\frac{2 M}{n^{2} \delta^{2}} \frac{n}{4}=\frac{\varepsilon}{2}+\frac{M}{2 n \delta^{2}}
$$
Therefore, whenever $n$ is sufficiently large that $\frac{M}{2 n \delta^{2}}<\frac{\varepsilon}{2}$, it follows that for all $n$ this large and $x \in[0,1]$,
$$
\left|f(x)-p_{n}(x)\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .
$$
Now this theorem has been done, it is easy to extend to continuous functions defined on $[a, b]$. This yields the celebrated Weierstrass approximation theorem. Also note that this would hold just as well if the functions had values in $\mathbb{C}$ or even $\mathbb{C}^{n}$ provided you had a norm defined on $\mathbb{C}^{n}$. In fact, it would hold if the functions have values in any normed space, a vector space which has a norm. These Bernstein polynomials are very remarkable.

Theorem $6.10 .3$ Suppose $f$ is a continuous function defined on $[a, b]$. Then there exists a sequence of polynomials, $\left{p_{n}\right}$ which converges uniformly to $f$ on $[a, b]$.
Proof: For $t \in[0,1]$, let $h(t)=a+(b-a) t$. Thus $h$ maps $[0,1]$ one to one and onto $[a, b]$. Thus $f \circ h$ is a continuous function defined on $[0,1]$. It follows there exists a sequence of polynomials $\left{p_{n}\right}$ defined on $[0,1]$ which converges uniformly to $f \circ h$ on $[0,1]$. Thus for every $\varepsilon>0$ there exists $N_{\varepsilon}$ such that if $n \geq N_{\varepsilon}$, then for all $t \in[0,1],\left|f \circ h(t)-p_{n}(t)\right|<\varepsilon$. However, $h$ is onto and one to one and so for all $x \in[a, b]$,
$$
\left|f(x)-p_{n}\left(h^{-1}(x)\right)\right|<\varepsilon .
$$
Now note that the function $x \rightarrow p_{n}\left(h^{-1}(x)\right)$ is a polynomial because $h^{-1}(x)=\frac{x-a}{b-a}$. More specifically, if $p_{n}(t)=\sum_{k=0}^{m} a_{k} t^{k}$ it follows
$$
p_{n}\left(h^{-1}(x)\right)=\sum_{k=0}^{m} a_{k}\left(\frac{x-a}{b-a}\right)^{k}
$$
which is clearly another polynomial.

$6.10$ Weierstrass 近似
事实证明，如果 $f$ 是定义在区间 $[a, b]$ 上的连续实值函数，则存在一个多项式序列 $\left{p_{n}\right}$ 使得该序列在 $[a, b]$ 上一致地收敛到 $f$。我将首先证明这对于区间 $[0,1]$ 是正确的，然后在任何闭合和有界区间上验证它是正确的。首先，这里有一个小引理，它本身就概率而言很有趣。它实际上是对二项分布方差的估计。
引理 6.10.1 以下估计适用于 $x \in[0,1]$ 和 $m \geq 2$。
$$
\sum_{k=0}^{m}\left(\begin{array}{c}
米\
ķ
\end{array}\right)(k-m x)^{2} x^{k}(1-x)^{m-k} \leqslant \frac{1}{4} m
$$
证明：使用二项式定理并展开$(k-m x)^{2}$，
$$
\开始{聚集}
\sum_{k=0}^{m}\left(\begin{array}{c}
米\
ķ
\end{数组}\right)(k-m x)^{2} x^{k}(1-x)^{m-k} \
=\sum_{k=0}^{m}\left(\begin{数组}{c}
米\
ķ
\end{array}\right) k^{2} x^{k}(1-x)^{mk}-2 mx \sum_{k=0}^{m}\left(\begin{array}{ C}
米\
ķ
\end{数组}\right) k x^{k}(1-x)^{m-k}+m^{2} x^{2} \
=\sum_{k=0}^{m}\left(\begin{数组}{c}
米\
ķ
\end{数组}\right) k(k-1) x^{k}(1-x)^{mk}+(k-2 mkx) \sum_{k=0}^{m}\left(\开始{数组}{c}
米\
ķ
\end{数组}\right) x^{k}(1-x)^{m-k}+m^{2} x^{2} \
=\sum_{k=2}^{m} \frac{m !}{(k-2) !(m-k) !} x^{k}(1-x)^{m-k}+ \
\quad(1-2 mx) \sum_{k=1}^{m} \frac{m !}{(k-1) !(mk) !} x^{k}(1-x)^{mk }+m^{2} x^{2} \
\quad(1-2 mx) \sum_{k=0}^{m-1} \frac{m !}{k !((m-1)-k) !} x^{k+1}(1 -x)^{(m-1)-k}+m^{2} x^{2} \
\sum_{k=0}^{m-2} \frac{m !}{k !(m-2-k) !} x^{k+2}(1-x)^{(m-2) -k}+ \
x-1) x^{2}+(1-2 mx) m x+m^{2} x^{2}=m\left(xx^{2}\right) \leq \frac{m}{ 4}
\结束{聚集}
$$
现在让 $f$ 是定义在 $[0,1]$ 上的连续函数。令 $p_{n}$ 为定义的多项式
$$
p_{n}(x) \equiv \sum_{k=0}^{n}\left(\begin{array}{l}
n \
ķ
\end{数组}\right) f\left(\frac{k}{n}\right) x^{k}(1-x)^{n-k}
$$
定理 $6.10 .2$ 6.5 中的多项式序列在 $[0,1]$ 上一致地收敛到 $f$。这些多项式称为伯恩斯坦多项式。
证明：由二项式定理，
$$
f(x)=f(x) \sum_{k=0}^{n}\left(\begin{array}{l}
n \
ķ
\end{array}\right) x^{k}(1-x)^{n-k}=\sum_{k=0}^{n}\left(\begin{array}{l}
n \
ķ
\end{数组}\right) f(x) x^{k}(1-x)^{n-k}
$$
所以由三角不等式
$$
\left|f(x)-p_{n}(x)\right| \leq \sum_{k=0}^{n}\left(\begin{array}{l}
n \
ķ
\end{数组}\right)\left|f\left(\frac{k}{n}\right)-f(x)\right| x^{k}(1-x)^{n-k}
$$
此时，您将总和分成两部分，$k$ 的值使得 $k / n$ 接近 $x$ 和 $k$ 的值使得 $k / n$ 不那么接近 $ x$。因此
$$
\开始{对齐}
\left|f(x)-p_{n}(x)\right| \leq & \sum_{|x-(k / n)|<\delta}\left(\begin{array}{l} n \ ķ \end{数组}\right)\left|f\left(\frac{k}{n}\right)-f(x)\right| x^{k}(1-x)^{n-k} \ &+\sum_{|x-(k / n)| \geq \delta}\left(\begin{数组}{l} n \ ķ \end{数组}\right)\left|f\left(\frac{k}{n}\right)-f(x)\right| x^{k}(1-x)^{n-k} \end{对齐} $$ 其中$\delta$ 是一个正数，以即将被描述的吉祥方式选择。由于 $f$ 在 $[0,1]$ 上是连续的，因此从定理 $4.8 .2$ 和 $6.7 .2$ 可以得出 $f$ 是一致连续的。因此，令 $\varepsilon>0$，存在 $\delta>0$ 使得如果 $|xy|<$ $\delta$，则 $|f(x)-f(y)|<\varepsilon / 2美元。这是 $\delta .$ 的吉祥选择。此外，根据引理 $6.3 .2$ $|f(x)|$ for $x \in[0,1]$ 由某个数字 $M$ 界定。因此 $6.6$ 意味着对于 $x \in[0,1]$，
$$
\开始{对齐}
\left|f(x)-p_{n}(x)\right| \leq & \sum_{|x-(k / n)|<\delta}\left(\begin{array}{l}
n \
ķ
\end{数组}\right) \frac{\varepsilon}{2} x^{k}(1-x)^{n-k} \
&+2 M \sum_{|n x-k| \geq n \delta}\left(\begin{数组}{l}
n \
ķ
\end{数组}\right) x^{k}(1-x)^{n-k} \
\leq & \frac{\varepsilon}{2}+2 M \sum_{|n x-k| \geq n \delta}\left(\begin{数组}{l}
n \
ķ
\end{array}\right) \frac{(k-n x)^{2}}{n^{2} \delta^{2}} x^{k}(1-x)^{n-k} \
\leq & \frac{\varepsilon}{2}+\frac{2 M}{n^{2} \delta^{2}} \sum_{k=0}^{n}\left(\begin{array {l}
n \
ķ
\end{数组}\right)(k-n x)^{2} x^{k}(1-x)^{n-k}
\end{对齐}
$$
现在通过引理 $6.10 .1$ 可以估算出最后一笔金额。使用这个估计产生所有 $x \in[0,1]$，
$$
\left|f(x)-p_{n}(x)\right| \leq \frac{\varepsilon}{2}+\frac{2 M}{n^{2} \delta^{2}} \frac{n}{4}=\frac{\varepsilon}{2}+ \frac{M}{2 n \delta^{2}}
$$
因此，只要 $n$ 足够大到 $\frac{M}{2 n \delta^{2}}<\frac{\varepsilon}{2}$，那么对于所有这么大的 $n$ 和x \in[0,1]$,
$$
\left|f(x)-p_{n}(x)\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon 。
$$
现在这个定理已经完成，很容易扩展到定义在 $[a, b]$ 上的连续函数。这产生了著名的魏尔斯特拉斯逼近定理。另请注意，如果函数在 $\mathbb{C}$ 甚至 $\mathbb{C}^{n}$ 中具有值，只要您在 $\mathbb{C}^{ n}$。事实上，如果函数有值，它就会成立