What can we say about the inner product of two Cauchy sequences?

Question

Let $(x_n)$, $(y_n)$ be two Cauchy sequences in an inner a real or complex product space $X$, and let the sequence $(\alpha_n)$ be given by $$ \alpha_n \colon= \ \langle x_n, y_n \rangle \ \ \ \mbox{ for } \ n = 1, 2, 3, \ldots. $$

Then is the sequence $(a_n)$ also Cauchy and therefore convergent? If so, how to show this rigorously?

If not, what counter-example can be given?

Answer 1

$$a_n-a_m =\langle x_n, y_n\rangle -\langle x_m, y_m\rangle$$

$$= \langle x_n, y_n\rangle - \langle x_m, y_n\rangle +\langle x_m, y_n\rangle -\langle x_m, y_m\rangle$$

$$= \langle x_n-x_m, y_n\rangle +\langle x_m, y_n-y_m\rangle$$

Hence

$$\left|a_n-a_m\right| \leq \left| \langle x_n-x_m, y_n\rangle +\langle x_m, y_n-y_m\rangle \right|$$

$$\left|a_n-a_m\right| \leq \| x_n-x_m\|\| y_n\| +\| x_m \| \|y_n-y_m\|$$

And as $(x_n)$ and $(y_n)$ are Cauchy sequences, this converge to 0