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Is $\sum\limits_{n=1} ^\infty |x_n||y_n| ≤ \left(\sum\limits_{n=1} ^\infty |x_n|\right)\left(\sum\limits_{n=1} ^\infty |y_n|\right)$?

I want to use this in a separate exercise but I can't think if this is true let alone how to prove it and I can't find any resources to help.

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    $\begingroup$ Without even thinking about it I'd explore the Cauchy-Schwarz inequality. $\endgroup$ – Git Gud Nov 10 '14 at 14:54
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    $\begingroup$ Yep. Cauchy-Schwarz plus $\left(\sum |x_i|\right)^2 \left(\sum |y_i|\right)^2 \geq \sum x_i^2 \sum y_i^2$ and you are there. $\endgroup$ – Winther Nov 10 '14 at 15:05
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Apply Cauchy-Schwarz

$$(x\cdot y)^2 \leq ||x ||^2~||y||^2$$

to the vectors $x =(|x_1|,|x_2|,\ldots,|x_n|)$ and $y = (|y_1|,|y_2|,\ldots,|y_n|)$ to get

$$\left(\sum x_i y_i\right)^2 \leq \sum x_i^2\sum y_i^2$$

Now use that

$$\sum x_i^2 \sum y_i^2 \leq \left(\sum |x_i|\right)^2 \left(\sum |y_i|\right)^2$$

since $\left(\sum |x_i|\right)^2 = \sum x_i^2 + \sum_{i\not=j} |x_ix_j| \geq \sum x_i^2$ and the result follows.

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$$\sum\limits_{n=1} ^\infty |x_n||y_n|=|x_1||y_1|+\sum_{n=2}^\infty|x_n||y_n|\le |x_1|\sum_{n=1}^\infty|y_n|+\sum_{n=2}^\infty |x_n||y_n|\le$$

$$|x_1|\sum_{n=1}^\infty|y_n|+|x_2||y_2|+\sum_{n=3}^\infty |x_n||y_n|\le |x_1|\sum_{n=1}^\infty|y_n|+|x_2|\sum_{n=1}^\infty|y_n|+\sum_{n=3}^\infty |x_n||y_n|\le$$

$$\vdots$$

$$|x_1|\sum_{n=1}^\infty|y_n|+|x_2|\sum_{n=1}^\infty|y_n|+|x_3|\sum_{n=1}^\infty|y_n|+\cdots =$$

$$ \sum\limits_{n=1} ^\infty |x_n|\sum\limits_{n=1} ^\infty |y_n|$$

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How about something more elementary?

If $x_i=0$ for all $i$, it is easy. So assume not.

Let $M = \max\{|y_1|,|y_2|,\dots\}$. If $M=\infty$ it is easy, so assume not.

Certainly $\sum_{i=1}^\infty |y_i| \ge M$.

For all $i$, we have $|x_i y_i| \le |x_i|\; M$, so $$ \sum_{i=1}^\infty |x_i y_i| \le \sum_{i=1}^\infty \big(|x_i|M\big) = \left(\sum_{i=1}^\infty |x_i|\right)M \le \sum_{i=1}^\infty |x_i| \sum_{i=1}^\infty |y_i| $$

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  • $\begingroup$ Why $M=\infty$? Well, if, for example, $y_i=i$ for all $i$, then $M=\infty$. Why do the $x_i=0$ case first? Maybe not needed, but let's avoid having to think about $0 \cdot \infty = ?$. $\endgroup$ – GEdgar Nov 12 '14 at 1:44

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