2
$\begingroup$

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?

$\endgroup$
1
+50
$\begingroup$

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)).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.