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.)

1. 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 \(|\sum_{n = 1}^{N + 1} b_n z^n| \le |\sum_{n = 1}^{N + 1} \bar{b_n} z^n| \le sup_{n \ge 1} |\frac{\sum_{i = 1}^n b_i}{n}| \sum_{n = 1}^{N + 1} z ^ n\)

2. 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.