How can I prove this general case of the inequality?

Question

I think I need some help with this problem.

According to p.544 of Feller's An Introduction To Probability Theory. Vol II if $|a_k|\leq c_k$, $|b_k|\leq c_k$ then we have $|\prod_{i=1}^na_i -\prod_{i=1}^n b_i|\leq\sum_{k=1}^n (a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i)$

I could only prove $|a^n-b^n|=\bigg|\sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i})\bigg|=\bigg|\sum_{i=0}^{n-1}a^{i+1}b^{n-1-i}-\sum_{i=0}^{n-1}a^{i}b^{n-i}\bigg|\\=\bigg|(a-b)\sum_{i=0}^{n-1}a^ib^{n-1-i}\bigg|\leq n|a-b|c^{n-1}$

but I don't know how I can do the more general case.

What does $\sum_{k=1}^n c_1\cdot ... \cdot c_{k-1}a_k-b_kc_{k+1}\cdot ...\cdot c_n$ mean? Do you mean $\prod_{k=1}^n c_ka_k-\prod_{k=1}^nc_kb_k$? — R. Burton, May 22 '19 at 19:10
The inequality is used on p. 544 in An Introduction to Probability Theory and Its Applications Vol2 from Feller — Testname420, May 22 '19 at 19:41
Possible counterexample?: $\forall k\in\Bbb{N}.a_k<0\land b_k>0$; $|\prod_{i=1}^na_i -\prod_{i=1}^n b_i|>0$ and $\sum_{k=1}^n (a_k\prod_{i=1}^{k-1}c_k-b_k\prod_{i=k+1}^n c_i)<0$, leading to a contradiction. — R. Burton, May 22 '19 at 20:04
Maybe? It might just be a typo. If it's meant to say $|\prod_{i=1}^na_i -\prod_{i=1}^n b_i|\leq\sum_{k=1}^n |(a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i)|$, then I might get it, but I'm not sure. — R. Burton, May 22 '19 at 20:18
In that case, I think it might be the same as this: https://math.stackexchange.com/questions/2560702/prove-by-induction-that-bigg-vert-prod-k-1n-a-k-prod-k-1n-b-k-b?rq=1. On account of the products only serve to scale the terms of the sum. — R. Burton, May 22 '19 at 20:28
Good find but i am still too dumb to get the desired result :( — Testname420, May 22 '19 at 21:22

score 0 · Accepted Answer · answered May 23 '19 at 14:45

Assuming that the line "$|\alpha_1\cdots\alpha_n-\beta_1\cdots\beta_n|\leq\sum_{k=1}^n\gamma_1\cdots\gamma_{k-1}\alpha_k-\beta_k\gamma_{k+1}\cdots\gamma_n$" is indeed a mistake, and that it is meant to read $|\alpha_1\cdots\alpha_n-\beta_1\cdots\beta_n|\leq\sum_{k=1}^n|\gamma_1\cdots\gamma_{k-1}\alpha_k-\beta_k\gamma_{k+1}\cdots\gamma_n|$, we have...

$$\forall k,n\in\Bbb{N}.|a_k|\leq c_k\land |b_k|\leq c_k\implies\left\vert\prod_{i=1}^na_i -\prod_{i=1}^n b_i\right\vert\leq\sum_{k=1}^n \left\vert a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i\right\vert$$

Now, from this post, we know that...

$$\forall\mathbf{a},\mathbf{b}\in\Bbb{C}^n:\left(\forall k\leq n\in\Bbb{N}.|a_k|\leq 1\land |b_k|\leq 1\right).\left\vert\prod_{i=1}^na_i -\prod_{i=1}^n b_i\right\vert\leq\sum_{k=1}^n \left\vert a_k-b_k\right\vert$$

That $\forall k\leq n\in\Bbb{N}.|a_k|\leq 1\land |b_k|\leq 1$ is implied by context (if not, then disregard this answer), so the question becomes "how does Feller's sum differ?"

From $|\cdot|:\Bbb{C}\to[0,\infty)$, we know that $\forall k\in\Bbb{N}.|a_k|,|b_k|\geq 0$, whence...

$$\forall k\in\Bbb{N}.0\leq|a_k|,|b_k|\land|a_k|,|b_k|\leq c_k\implies 0\leq c_k$$

Thus, $\prod_{k\in K}c_k\geq0$ for all valid indexing sets $K$. Saying anything more than this will require more information on what $\mathbf{c}$ is. Suffice to say that if...

$$\exists x\in[1,\infty).\prod_{i=1}^{k-1}c_i=\prod_{i=k+1}^n c_i=x$$

...then...

$$\sum_{k=1}^n \left\vert a_k-b_k\right\vert\leq\sum_{k=1}^n \left\vert a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i\right\vert=x\sum_{k=1}^n \left\vert a_k-b_k\right\vert$$

Thanks you for your hard work – Testname420 May 24 '19 at 15:57 — Testname420, May 24 '19 at 15:57

How can I prove this general case of the inequality?

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