cesaro sum和tauber型定理
缓更
Definition.1: Given a sequence \(\{a_n\} (n \ge 1)\), the cesaro sum of a denote $\lim_{n \rightarrow \infty} \frac{\sum_{i = 1}^n a_i}{n} $,if this limit is convergent.
(Remark: I will also use \(\tilde{a}\) to denote the cesaro sum of a)
Lemma1:Suppose \(\{a_n\} (n \ge 1)\) is a sequence consisting of non-negative numbers, and forall \(z \in (0,1)\), \(\sum_{n \ge 1}a_n z^n\) is convergent.
prove that if \(\tilde{a}\) exists,\(\lim_{z \rightarrow 1^{-}} (1 - z) \sum_{n \ge 1} a_n z^n = \tilde{a}\)
proof:
Let \(f(z)\) denotes \(\sum_{n \ge 1} a_n z^n\)
so \((1 - z) f(z) = (1 - z)\sum_{n \ge 1} (a_n - \tilde{a}) z^n + \tilde{a}((1 - z)\sum_{n \ge 1}z^n )\)
\(= (1 - z) \sum_{n \ge 1}(a_n - \tilde{a}) z^n + \tilde{a}\)
Let \(b_n = a_n - \tilde{a}\)
It's easy to prove that \(\tilde{b} = 0\)
so we only need to prove that \(\lim_{z \rightarrow 1^{-}} (1 - z) \sum_{n \ge 1} b_n = 0\)
We make a adjustion step by step for the sequence b to complete this proof by mathmatical induction.
suppose the adjusted sequence is \(\bar{b}\)
for a Fixed N, suppose \(|\sum_{n = 1}^N b_n z^n| \le |\sum_{n = 1}^N \bar{b_n} z^n| \le sup_{n \ge 1} |\frac{\sum_{i = 1}^n b_i}{n}| \sum_{n = 1}^N z ^ n\).
then for (N + 1),we have that
(Assume that \(\sum_{n = 1}^{N + 1} b_n z^n\) > 0,since another condition is similar.)
- if \(b_{N + 1} \ge \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)
\(\forall i\),if (\(\bar{b_i} < \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)),we can reduce some value from \(b_{N + 1}\),while add it to \(\bar{b_i}\) until \(\bar{b_i} = \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)
or \(b_{N + 1} = \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)
let \(\bar{b}_{n + 1} = \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)
since \(z^n\) is decending,so these adjustions will make sure that \(|\sum_{n = 1}^{N + 1} b_n z^n| \le |\sum_{n = 1}^{N + 1} \bar{b_n} z^n| \le sup_{n \ge 1} |\frac{\sum_{i = 1}^n b_i}{n}| \sum_{n = 1}^{N + 1} z ^ n\)
- if \(b_{N + 1} < \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)
do nothing,and let \(\bar{b}_{n + 1} = b_{n + 1} < \frac{\sum_{n = 1}^{N+1}b_n}{N + 1}\)
and then it's obvious that \(|\sum_{n = 1}^{N + 1} b_n z^n| \le |\sum_{n = 1}^{N + 1} \bar{b_n} z^n| \le sup_{n \ge 1} |\frac{\sum_{i = 1}^n b_i}{n}| \sum_{n = 1}^{N + 1} z ^ n\)
So \((1 - z) |\sum_{n \ge 1} b_{n} z^n| <= (1 - z)|\sum_{n = 1} ^ N b_{n} z^n| + (1 - z)sup_{n \ge N + 1} |\frac{\sum_{i = 1}^n b_i}{n}| \sum_{n = 1}^{N + 1} z ^ n\)
taking the limit \(z \rightarrow 1^{-}\),the first term will tend to 0.
and the second term is less or equal:
\(sup_{n \ge N + 1} |\frac{\sum_{i = 1}^n b_i}{n}|\)
since the N is arbitrary,and the cesaro sum of b is 0. so the second term will also tend to 0.
By now,we have completed this proof.
Q.E.D
一些感想:想试着用英语写,不过写着写着发现我的英语表达能力是灾难级的.接下来可能不得不放弃吧,哭死.
另一个证明方法:
Lemma2: \(\forall \phi(x)\),如果 \(\phi(x)\) 在 \([0,1]\)上只有有限个不连续点,且有界,那么 \(\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx} \phi(e^{-nx}) = \int_{0}^{1}\phi(x)\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx}\)
考虑对于任意的多项式函数 \(\phi(x) = x^s\)
\(\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx} \phi(e^{-nx})
= \lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-n((s + 1)x)}\)
令 \(x := -(s + 1)x\)
则原式等于 \(\frac{1}{s + 1}\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx} = \int_{0}^{1}\phi(x)\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx}\)
由Weistrass-stone定理,任意连续函数都可以由多项式无限逼近
于是对于任意 \(\phi(x)\),我们有\(\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx} \phi(e^{-nx}) = \int_{0}^{1}\phi(x)\lim_{x \rightarrow} 0^{+} x \sum_{n \ge 0} a_n e^{-nx}\)(具体细节读者自行推导,很平凡)
事实上可以拓展到有有限个非连续点的有界函数,在中间一个足够小的 \(\epsilon\) 区间把两段连起来即可(证明也是很平凡的)
令 \(\phi(x) = 0\quad x \in [0,e^{-1})\)
\(\quad \qquad= \frac{1}{x} \quad x \in[e^{-1},1]\)
\(\lim_{x \rightarrow 0^{+}} x \sum_{n \ge 0} a_n e^{-nx}\phi(e^{-nx}) = \lim_{N \rightarrow \infty} \frac{\sum_{n = 0}^N a_n}{N}\)
我们希望证明 \(\lim_{x \rightarrow 0^{+}} x \sum_{n \ge 0} a_n e^{-nx}\phi(e^{-nx}) = \lim_{x \rightarrow 0^{+}} x \sum_{n \ge 0} a_n e^{-nx}\)
注意到 \(\int_{0}^1 \phi(x) = 1\)
套用上面的 Lemma2 即证
