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 $\liminf_{n→∞}z_n=-1$?

Question

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 $b\in\mathbb R$,

4. $\liminf_{n\to\infty}(a_n+b_n)\ge\liminf_{n\to\infty}a_n+\liminf_{n\to\infty}b_n$;
5. $\liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}a_n+b$, if $b_n\xrightarrow{n\to\infty}b$; and
6. $\liminf_{n\to\infty}(a_nb_n)=b=\liminf_{n\to\infty}a_n$, 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?

Answer 1

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)).

Answer 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.

Answer 3

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.