Prove Schwarz's & Triangle Ineqs for Inf Seqs: Abs Conv

In summary, the Schwarz's inequality can be used to show that if two infinite sequences $\sum\limits_{n = -\infty}^{\infty}|a_n|^2$ and $\sum\limits_{n = -\infty}^{\infty}|b_n|^2$ are both finite, then the series $\sum\limits_{n = -\infty}^{\infty} a_nb_n$ converges absolutely. This can be proven by first showing that $|a_n|^2 + |b_n|^2 \geq 2|a_nb_n|$ and then using this to show that $\sum\limits_{n = -\infty}^{\infty}|
  • #1
Dustinsfl
2,281
5
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?
 
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  • #2
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.
 
  • #3
girdav said:
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 $|a_nb_n|^2$?
 
  • #4
Where?
 
  • #5
girdav said:
Where?

In your post.
 
  • #6
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?
 
  • #7
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
$$
\alpha + \beta \geq 2\sum\limits_{n = -\infty}^{\infty}|a_nb_n|\iff \sum\limits_{n = -\infty}^{\infty}|a_nb_n|\leq\frac{\alpha + \beta}{2} < \infty.
$$
 
  • #8
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:
  • #9
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 $n$, $a_n=\lambda b_n$.
 

FAQ: Prove Schwarz's & Triangle Ineqs for Inf Seqs: Abs Conv

What is Schwarz's inequality for infinite sequences?

Schwarz's inequality for infinite sequences states that for two infinite sequences, A and B, the inner product of A and B is less than or equal to the product of the norms of A and B. This can be written as |∑n=1AnBn| ≤ (∑n=1|An|2)1/2(∑n=1|Bn|2)1/2.

What is the Triangle inequality for infinite sequences?

The Triangle inequality for infinite sequences states that for three infinite sequences, A, B, and C, the norm of the sum of A and B is less than or equal to the sum of the norms of A and B. This can be written as (∑n=1|An+Bn|2)1/2 ≤ (∑n=1|An|2)1/2 + (∑n=1|Bn|2)1/2.

What does it mean for a sequence to be absolutely convergent?

A sequence is absolutely convergent if the sum of the absolute values of its terms is finite. In other words, if ∑n=1|An| is a finite value, then the sequence is absolutely convergent.

How do you prove Schwarz's inequality for infinite sequences?

To prove Schwarz's inequality for infinite sequences, one can use the Cauchy-Schwarz inequality and the fact that the inner product is a bilinear form. By expanding the inner product and applying the Cauchy-Schwarz inequality, one can show that the inner product is less than or equal to the product of the norms of the sequences.

Can the Triangle inequality for infinite sequences be extended to more than three sequences?

Yes, the Triangle inequality can be extended to any finite number of sequences. However, the Triangle inequality for infinite sequences is a special case that is often used in mathematical analysis and is particularly useful for proving other theorems involving infinite sequences.

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