abs converge 2

dwsmith

Well-known member
I am not getting anywhere with this problem.

Prove the Schwarz's and the triangle inequalities for infinite sequences:
If
$$\sum_{n = -\infty}^{\infty}|a_n|^2 < \infty\quad\text{and}\quad \sum_{n = -\infty}^{\infty}|b_n|^2 < \infty$$
then
$\displaystyle\left(\sum_{n = -\infty}^{\infty}|a_n + b_n|^2\right)^{1/2}\leq \left(\sum_{n = -\infty}^{\infty}|a_n|^2\right)^{1/2}\left(\sum_{n = -\infty}^{\infty}|b_n|^2\right)^{1/2}$.

Sudharaka

Well-known member
MHB Math Helper
I am not getting anywhere with this problem.

Prove the Schwarz's and the triangle inequalities for infinite sequences:
If
$$\sum_{n = -\infty}^{\infty}|a_n|^2 < \infty\quad\text{and}\quad \sum_{n = -\infty}^{\infty}|b_n|^2 < \infty$$
then
$\displaystyle\left(\sum_{n = -\infty}^{\infty}|a_n + b_n|^2\right)^{1/2}\leq \left(\sum_{n = -\infty}^{\infty}|a_n|^2\right)^{1/2}\left(\sum_{n = -\infty}^{\infty}|b_n|^2\right)^{1/2}$.
Hi dwsmith,

You can generalize the triangle inequality for a finite sequence of terms using mathematical induction.

$\left|\sum_{k=1}^{n}a_k\right|\leq\sum_{k=1}^{n}|a_k|$

If $$\displaystyle\sum_{k=1}^{\infty}|a_k|$$ is convergent,

$\left|\sum_{k=1}^{n}a_k\right|\leq \sum_{k=1}^{n}|a_k|\leq \sum_{k=1}^{\infty}|a_k|$

Now $$(z_n)=\displaystyle\left|\sum_{k=1}^{n}a_k\right|$$ is a convergent sequence. By Theorem 3.2.6 (Page 66) of Introduction to Real Analysis by Robert G. Bartle we get,

3.2.6 Theorem: If $$X=(x_{n})$$ is a convergent sequence and if $$a\leq x_{n}\leq b$$ for all $$n\in\mathbb{N}$$, then $$a\leq \lim(x_{n})\leq b$$.
$\left|\sum_{k=1}^{\infty}a_k\right|\leq \sum_{k=1}^{\infty}|a_k|$

There are various approaches in proving the Cauchy-Schwarz inequality some of which are given >>here<<.

Kind Regards,
Sudharaka.