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absolute convergence

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$\sum\limits_{n = -\infty}^{\infty} a_nb_n$ converges absolutely.

To show this, wouldn't I need to know that the |a_n| is bounded not the sum of squares?
 

girdav

Member
Feb 1, 2012
96
To see the series is absolutely convergent doesn't need Schwarz inequality, just write that $2|a_nb_n|\leq |a_n|^2+|b_n|^2$.

But we can use Schwarz inequality to get an estimation of the sum, using it first for finite sums, then taking the limit.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
To see the series is absolutely convergent doesn't need Schwarz inequality, just write that $2|a_nb_n|\leq |a_n|^2+|b_n|^2$.

But we can use Schwarz inequality to get an estimation of the sum, using it first for finite sums, then taking the limit.
Should $|a_nb_n|^2$?
 

girdav

Member
Feb 1, 2012
96
Where?
 

dwsmith

Well-known member
Feb 1, 2012
1,673

girdav

Member
Feb 1, 2012
96
Indeed, $|a_nb_n|^2$ will be summable, but here we show more (that $|a_nb_n|$ is summable). I think the inequality I wrote as stated. Did you try to show it and use it?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
We have that $|a_n - b_n|^2 \leq |a_n|^2 + |b_n|^2 - 2|a_nb_n| \geq 0$, i.e. $|a_n|^2 + |b_n|^2 \geq 2|a_nb_n|$.
Let $\sum\limits_{n = -\infty}^{\infty}|a_n|^2 = \alpha < \infty$ and $\sum\limits_{n = -\infty}^{\infty}|b_n|^2 = \beta < \infty$.
Then
$$
\alpha + \beta \geq 2\sum\limits_{n = -\infty}^{\infty}|a_nb_n|\iff \sum\limits_{n = -\infty}^{\infty}|a_nb_n|\leq\frac{\alpha + \beta}{2} < \infty.
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Now I am trying to show this for the same problem.
$$
\left|\sum_{n = -\infty}^{\infty}a_nb_n\right|^2\leq \left(\sum_{n = -\infty}^{\infty}|a_n|^2\right)\left(\sum_{n = -\infty}^{\infty}|b_n|^2\right)
$$


$$
\left|\sum_{n = -\infty}^{\infty}a_nb_n\right|^2\leq\sum_{n = -\infty}^{\infty}|a_nb_n|^2
$$

Is it this
$$
\sum_{n = -\infty}^{\infty}|a_nb_n|^2 = \sum_{n = -\infty}^{\infty}|a_n|^2\sum_{n = -\infty}^{\infty}|b_n|^2
$$
or
$$
\sum_{n = -\infty}^{\infty}|a_nb_n|^2 \leq \sum_{n = -\infty}^{\infty}|a_n|^2\sum_{n = -\infty}^{\infty}|b_n|^2
$$
 
Last edited:

girdav

Member
Feb 1, 2012
96
The latter (actually, a nice exercise is to show that we have the former if and only if we can find a constant $\lambda$ such that for each $n$, $a_n=\lambda b_n$.