Welcome to our community

Be a part of something great, join today!

abs converge 2

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
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.