# How can we compute $\liminf_{n→∞}(x_n+y_nz_n)$ for monotone $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ knowing that $\liminf_{n→∞}z_n=-1$?

Let $$(x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N},(z_n)_{n\in\mathbb N}\subseteq\mathbb R$$ with $$\liminf_{n\to\infty}z_n=-1\tag1.$$

I want to show that

1. if $$(x_n)_{n\in\mathbb N}$$ is decreasing with $$x_n\xrightarrow{n\to\infty}-\infty\tag2$$ and $$(y_n)_{n\in\mathbb N}$$ is increasing with $$y_n\xrightarrow{n\to\infty}\infty\tag3,$$ then $$\liminf_{n\to\infty}(x_n+y_nz_n)=-\infty\tag4;$$
2. if $$(x_n)_{n\in\mathbb N}$$ is decreasing with $$(2)$$ and $$(y_n)_{n\in\mathbb N}$$ is decreasing with $$y_n\xrightarrow{n\to\infty}0\tag5,$$ then $$(4)$$; and
3. if $$(x_n)_{n\in\mathbb N}$$ is increasing with $$x_n\xrightarrow{n\to\infty}0\tag6$$ and $$(y_n)_{n\in\mathbb N}$$ is decreasing with $$(5)$$, then $$\liminf_{n\to\infty}(x_n+y_nz_n)=0\tag7.$$

I'm a bit rusty in dealing with the limit inferior. My biggest problem is that we usually only know that for bounded $$(a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\subseteq\mathbb R$$ and $$b\in\mathbb R$$,

1. $$\liminf_{n\to\infty}(a_n+b_n)\ge\liminf_{n\to\infty}a_n+\liminf_{n\to\infty}b_n$$;
2. $$\liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}a_n+b$$, if $$b_n\xrightarrow{n\to\infty}b$$; and
3. $$\liminf_{n\to\infty}(a_nb_n)=b=\liminf_{n\to\infty}a_n$$, if $$b_n\xrightarrow{n\to\infty}b$$ and $$b\ge0$$.

So, these properties seem either not useful or not applicable. So, how can we show the desired claims?

## 3 Answers

The following two descriptions of the upper and lower limit might turn out to be more useful here:

• $$x \in \mathbb R \cup \{ \pm \infty \}$$ is a limit point of $$(x_n)$$ if and only if there exists a subsequence $$(x_{n_k})$$ with $$x_{n_k} \to x$$;

• $$\liminf x_n$$ is the infimum of all the limit points of $$(x_n)$$; in other words, there exists a subsequence $$(x_{n_k})$$ with $$x_{n_k} \to x$$ and if there exists $$y$$ and a subsequence $$(x_{m_l})$$ with $$x_{m_l} \to y$$ then $$x \le y$$;

• a similar characterization for $$\limsup$$;

• if you work with nets instead of sequences, then you'll have to use subnets instead of subsequences; everything else remains unchanged.

1) If $$\liminf z_n = -1$$, there exists a subsequence $$(z_{n_k})$$ with $$z_{n_k} \to -1$$. It follows immediately that the subsequence $$(x_{n_k} + y_{n_k} z_{n_k})$$ of the sequence $$(x_n + y_n z_n)$$ converges to $$-\infty + (-1) \cdot \infty = -\infty$$, showing that $$-\infty$$ is a limit point of $$(x_n + y_n z_n)$$. Since, by its very nature, $$-\infty$$ is the smallest possible element of $$\mathbb R \cup \{ \pm \infty \}$$, every other limit point of $$(x_n + y_n z_n)$$ must be greater than or equal to it, therefore $$-\infty$$ is the infimum of all the limit points of $$(x_n + y_n z_n)$$, therefore is its $$\liminf$$.

2) Exactly as above, with $$-\infty + (-1) \cdot \infty$$ getting replaced by $$-\infty + (-1) \cdot 0$$, which is still $$-\infty$$.

Notice that, so far, we haven't needed the monotonicity of $$(x_n)$$ and $$(y_n)$$, but only their limits. Notice also that we haven't used that $$-1$$ is the $$\liminf$$ of $$(z_n)$$, but only that it is a limit point.

3) Again, with the same construction as above, the subsequence $$(x_{n_k} + y_{n_k} z_{n_k})$$ tends to $$0 + (-1) \cdot 0 = 0$$, so $$0$$ is a limit point of $$(x_n + y_n z_n)$$. If $$r < 0$$ is any other limit point, then there exists a subsequence $$q_{m_l}$$ of $$(x_n + y_n z_n)$$ with $$q_{m_l} \to r$$. Consider then the subsequences $$(x_{m_l})$$, $$(y_{m_l})$$ and $$(z_{m_l})$$. Since $$(x_n)$$ is convergent (to $$0$$), so will be $$(x_{m_l})$$, therefore $$y_{m_l} z_{m_l} \to r$$.

If $$(y_n)$$ is constant $$0$$ from some $$N$$ onwards, then $$x_n + y_n z_n = z_n$$ from that $$N$$ onwards, and the result is trivial. If $$(y_n)$$ is not eventually $$0$$, then from $$(y_{m_l})$$ I may extract yet another subsequence $$(y_{m_{l_i}})$$ such that $$y_{m_{l_i}} \ne 0$$ for all $$i$$. Since $$(y_n)$$ decreases to $$0$$, it follows that $$y_n \ge 0$$, therefore $$y_{m_{l_i}} > 0$$.

Since $$y_{m_{l_i}} z_{m_{l_i}} \to r$$, we deduce that for every $$\varepsilon > 0$$ there exists $$I \in \mathbb N$$ such that for $$i \ge I$$ we have $$| y_{m_{l_i}} z_{m_{l_i}} - r | < \varepsilon$$. Taking $$\varepsilon = - \frac r 2$$, this implies that for $$i \ge I$$ we have $$y_{m_{l_i}} z_{m_{l_i}} - r < \frac r 2$$, which implies that $$z_{m_{l_i}} < \frac r {2 y_{m_{l_i}}} \to -\infty$$ (because $$r<0$$), which implies that $$zy_{m_{l_i}} \to -\infty$$, which means that $$-\infty$$ is a limit point of $$(z_n)$$. But this is a contradiction, because $$-1$$ is its $$\liminf$$, which is the infimum of the limit points. Therefore, $$r \ge 0$$, i.e. every other limit point of $$(x_n + y_n z_n)$$ is $$\ge 0$$, which (together with the already proven fact that $$0$$ is a limit point) makes $$0$$ its $$\liminf$$.

Notice that, unlike for (1) and (2), this time we have used the monotonicity of $$(y_n)$$ in an essential way, but (again) not the one of $$(x_n)$$. Notice also that we have used the fact that $$-1$$ is the $$\liminf$$ of $$(z_n)$$ in an essential way (unlike for (1) and (2)).

Here's my attempt at the first claim. I use the following definition for limit inferior $$\liminf_{n\to\infty} a_n = \sup\{\inf\{a_n : n \geq N\} : N \geq 0\}$$.

To show $$\liminf_{n\to\infty} (x_n+y_nz_n) = -\infty$$ it suffices to show that for every real number $$M < 0$$ and natural number $$N \geq 0$$ there exists an $$n_0 \geq N$$ such that $$x_{n_0} + y_{n_0}z_{n_0} \leq M$$. Let $$M < 0$$ and $$N \geq 0$$. Since $$\lim_{n\to\infty} y_n = \infty$$ there exists $$N' \geq 0$$ such that if $$n \geq N'$$ then $$y_n > 0$$. Since $$\lim_{n\to\infty} x_n = -\infty$$ there exists $$N'' \geq \max\{N',N\}$$ such that if $$n \geq N''$$ then $$x_n \leq M$$. Since $$\liminf_{n\to\infty} z_n = -1$$ we have $$\inf\{z_n : n \geq N''\} \leq -1$$. In particular there exists $$n_0 \geq N''$$ such that $$z_{n_0} < -1 + 1/2 < 0$$. Then $$n_0 \geq N'' \geq \max\{N',N\}$$ with $$x_{n_0} + y_{n_0}z_{n_0} \leq M - y_{n_0} < M$$ as required.

For 1 and 2, we just need to find some subsequence which converges to minus infinity. This is given by the appropriate subsequence of z(n) which converges to - 1.

For 3, we need to: a) Find a subsequence converging to zero. b) Show that every convergent subsequence has a non-negative limit. (a) is given by any convergent subsequence of z(n), for example the one converging to - 1. (b) follows from the given limits of x(n), y(n) and the fact that z(n) is bounded from below.