# Prove $\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty \iff \lim_{n\to\infty}|x_n| =\lim_{n\to\infty}|y_n| =\lim_{n\to\infty}x_ny_n =+\infty$

Let $$x_n$$ and $$y_n$$ denote sequences such that: $$\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty$$ Prove: $$\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty \iff \lim_{n\to\infty}|x_n| =\lim_{n\to\infty}|y_n| =\lim_{n\to\infty}x_ny_n =+\infty$$

I've started with the first case ($$\implies$$): $$\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty \\ \stackrel{\text{def}}{\iff} \forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies ||x_n + y_n| - |x_n - y_n|| \ge \epsilon$$

We want to show that $$|x_n| \ge \epsilon$$ and $$|y_n|\ge \epsilon$$. By triangular inequality: $$|x_n + y_n| + |x_n - y_n| \ge ||x_n + y_n| - |x_n - y_n|| \ge \epsilon$$

At the same time: $$|x_n + y_n| + |x_n - y_n| \ge |(x_n + y_n) + (x_n - y_n)| = 2|x_n|$$ Or: $$|x_n + y_n| + |y_n - x_n| \ge |(x_n + y_n) + (y_n - x_n)| = 2|y_n|$$ Here is where the first questionable case comes in. It seems that: $$|x_n + y_n| + |x_n - y_n| \ge 2|x_n| \ge ||x_n + y_n| - |x_n - y_n|| \ge \epsilon\\ \text{and}\\ |x_n + y_n| + |y_n - x_n| \ge 2|y_n| \ge ||x_n + y_n| - |x_n - y_n|| \ge \epsilon \tag1$$

But I'm not sure how to justify that. I've checked various cases like $$x<0 \land y<0$$ and 3 others, for all it seems to hold. So based on this: $$\forall \epsilon >0 \ \exists N\in\Bbb N: \forall n> N \implies 2|x_n| \ge \epsilon \\ \forall \epsilon >0 \ \exists N\in\Bbb N: \forall n> N \implies 2|y_n| \ge \epsilon$$

Which shows: $$\lim_{n\to\infty}|x_n| = \lim_{n\to\infty}|y_n| = +\infty$$

In this part I'm interested in how to justify $$(1)$$ and where from it follows that: $$\lim_{n\to\infty}x_ny_n = +\infty$$

Second case $$(\impliedby)$$. This case I've no idea where to start from. We basically given three things: $$\forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies |x_n| \ge \epsilon\\ \forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies |y_n| \ge \epsilon \\ \forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies |x_ny_n| \ge \epsilon$$

The problem is I don't see where to go from this.

Could you please verify the overall reasoning and help with ($$\impliedby$$) and question from ($$\implies$$), I still often have troubles with constructing those proofs since i'm a self-learner and have no one to refer to. Thank you!

$$(\Longrightarrow)$$ Note that by triangle inequality, we have $$2|x_n|=|(x_n+y_n)+(x_n-y_n)|\ge|x_n+y_n|-|x_n-y_n|$$ and $$2|y_n|=|(x_n+y_n)-(x_n-y_n)|\ge|x_n+y_n|-|x_n-y_n|.$$ By taking $$n\to\infty$$, we get $$\lim_{n\to\infty}|x_n|=\lim_{n\to\infty}|y_n|=\infty.$$ Also note that $$|x_n+y_n|\ge |x_n-y_n|$$ holds eventually. Hence, by squaring both sides, we have $$x_ny_n\ge 0$$. It follows that $$\lim_{n\to\infty}x_n y_n=\lim_{n\to\infty}|x_n| |y_n|=\infty.$$
$$(\Longleftarrow)$$ To show the converse, note that $$\lim_{n\to\infty}x_n y_n=\infty$$ implies that $$|x_n+y_n|-|x_n-y_n|\ge 0$$ eventually. And since it is a positive sequence, we have $$\lim_{n\to\infty}|x_n+y_n|-|x_n-y_n|=\infty \Longleftrightarrow \lim_{n\to\infty}\left[|x_n+y_n|-|x_n-y_n|\right]^2=\infty.$$ Now, we have $$\left[|x_n+y_n|-|x_n-y_n|\right]^2= 2(|x_n|^2+|y_n|^2) -2|x_n^2-y_n^2|=:L.$$ Observe that if $$x_n^2\ge y_n^2$$, then it holds $$L=4|y_n|^2$$ and otherwise, $$L=4|x_n|^2$$. This gives $$L=4 \min\{|x_n|^2,|y_n|^2\}$$ and $$\lim_{n\to\infty}\left[|x_n+y_n|-|x_n-y_n|\right]^2= 4\lim_{n\to\infty}\min\{|x_n|^2,|y_n|^2\}=\infty$$ by the assumption that $$\lim_{n\to\infty}|x_n|=\lim_{n\to\infty}|y_n|=\infty$$. This proves the converse claim.