Convergence of series composed of sequences from $l^2$ space

Question

I need help with this question:

Let $(x_{j})_{j \in \mathbb{N}}$ and $(y_{j})_{j \in \mathbb{N}}$ $\in l^2$. Show that the serie $(x_{j})_{j \in \mathbb{N}} \cdot (y_{j})_{j \in \mathbb{N}} = \sum_{i=1}^{\infty}x_{i}y_{i}$ converge.

My attempt: Let $(x_{j})_{j \in \mathbb{N}}$ and $(y_{j})_{j \in \mathbb{N}}$ $\in l^2$. Therefore, $\sum_{j=1}^{\infty}|x_{j}|^{2} \lt \infty$ and $\sum_{j=1}^{\infty}|y_{j}|^{2} \lt \infty$, hence, $\sum_{j=1}^{\infty}|x_{j}y_{j}|^{2} \lt \infty$. (Can I do this?)

Let $c_{j} = \sum_{j=1}^{\infty}|x_{j}y_{j}|^{2}$. Since the summation is finite, $c_{j}$ converges. Therefore, $\forall \varepsilon \gt 0, \exists J^{*} \in \mathbb{N}: ||c_{j}-c|| \lt \varepsilon, \forall j \gt J^{*}$.

But note that $||c_{j}-c|| = ||\sum_{j=1}^{\infty}|x_{j}y_{j}|^{2}-c|| \lt \varepsilon$, i.e, $|x_{1}y_{1}|^{2} + |x_{2}y_{2}|^{2} + ... \lt \varepsilon$.

By the triangle inequality: $|(x_{1}y_{1})^2+(x_{2}y_{2})^2 + ...| \lt |x_{1}y_{1}|^2 + |x_{1}y_{1}|^2 + ... \lt \varepsilon$.

Therefore, $\big|\sum_{j=1}^{\infty}(x_{j}y_{j})^{2}\big| \leq \sum_{j=1}^{\infty}|x_{j}y_{j}|^{2} \lt \varepsilon, \forall j \geq J^{*}$, hence, $\sum_{j=1}^{\infty}(x_{j}y_{j})^{2}$ converges.

But how do I prove what is asked?

Answer 1

Take $p=q=2$, then $\frac{1}{p}+\frac{1}{q}=1$ so we can apply Hölder's inequality and the result is straighforward:

$\sum_{i=1}^{n}|x_{i}y_{i}| \leq \lim_{n \to \infty} \sum_{i=1}^{n}|x_{i}y_{i}|= \sum_{i=1}^{\infty}|x_{i}y_{i}| \leq ||x||_{2}||y||_{2} \lt \infty$

since $x,y \in l^{2}$.

Answer 2

Hint

Using Cauchy-Schwarz you have $\forall N \in \Bbb N$: $$\sum_{j=1}^N | x_j y_j| \leq \left(\sum_{j=1}^N |x_j|^2 \right)^\frac{1}{2}\left(\sum_{j=1}^N |y_j|^2 \right)^\frac{1}{2}$$