Definition: $\limsup x_n$= supremum of all subsequential limits of sequence $(x_n)=\sup E_x$, where $E_x$ is set of all subsequential limits of sequence $(x_n)$. Let $x^*=\limsup x_n$.

Given any two sequences $(a_n)$ and $(b_n)$, it is to be proven that: $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$, provided sum on right side is not of the form $\infty -\infty$. $\tag 1$
I tried to prove it as follows:
Let $c_n=a_n+b_n$ for all $n\in \mathbb N$

Case $1: a^*, b^*, c^* \in \mathbb R$
It follows that the sequences $(a_n),(b_n),(c_n)$ are bounded. And since set of all subsequential limits of a sequence is closed, it follows that $a^*\in E_a, b^*\in E_b, c^*\in E_c\implies$ There exist subsequences $(a_{n_k}), (b_{n_l})$ and $(c_{n_m})$ such that $a_{n_k}\to a^*, b_{n_l}\to b^*, c_{n_m}\to c^* $.
For any $\epsilon \gt 0, \exists s_\epsilon\in \mathbb N$ such that $s\ge s_\epsilon \implies |a_{n_s}-a^*|\lt \epsilon/4 \space, |b_{n_s}-b^*|\lt \epsilon/4 \implies |a_{n_s}+b_{n_s}-a^*-b^*|\lt \epsilon/2$.
Now suppose on the contrary that $c^*\gt a^*+b^*$. We have: for any $\epsilon \gt 0$
$\exists M_\epsilon \in \mathbb N: m\ge M_\epsilon\implies |c_{n_m}-c^*|\lt \epsilon/2.$
Hence for $r\in \mathbb N: r\ge \sup(M_\epsilon, s_\epsilon),$ we have:
$0\lt c^*-a^*-b^*=|c^*-(a_{n_r}+b_{n_r})+ (a_{n_r}+b_{n_r}-a^*-b^*)|\le |c^*-(a_{n_r}+b_{n_r})|+ |(a_{n_r}+b_{n_r}-a^*-b^*)|\lt\epsilon \implies c^*=a^*+b^* $, which is contradiction. Therefore, $c^*\le a^*+b^*$.

Case $2$: $a^*=\infty, b^*\in \mathbb R\cup \{\infty\},$ then the result is trivial.

Case $3$: $b^*=\infty, a^*\in \mathbb R\cup \{\infty\}$, then as in Case $3$, the result is true.

Case $4$: $a^*\in \mathbb R, b^*=-\infty .$
$E_b=\{-\infty\}$ and $ (a_n)$ is bounded by some $M\gt 0$. So, $(c_n)$ is not bounded below as $(b_n)$ is not bounded below.
So for some $M\in \mathbb N$, $n\ge M\implies c_n\lt a^*+b_n\implies c^*\le a^*+b^* $.

Case $5$: $b^*\in \mathbb R, a^*=-\infty .$
Same as case $4$.
So proved.

My question is what happens when right hand side of $(1)$ is $\infty -\infty $?

Consider for example: $(a_n)=(n), (b_n)=(-n)$, clearly $c_n=0\to \limsup c_n=0$. But can we claim that $0\gt \infty -\infty $ (I don't think so as $\infty -\infty $ is undefined.)?. If not, then $(1)$ always holds?

Please help. Thanks.

  • $\begingroup$ Your proof does not work because you have distinct subsequences of $(a_n),(b_n),(c_n)$ converging to $a^*, b^*, c^*$, respectively. $\endgroup$
    – Martin R
    Feb 28 at 7:00
  • $\begingroup$ @MartinR: I have skipped some steps in Case 1. And $c_n=a_n+b_n$ for all $n$. but it seems that you may be right. $\endgroup$
    – Koro
    Feb 28 at 7:04
  • $\begingroup$ $\infty -\infty$ is undefined, and nothing can be said in that case: $(n) + (-n) \to 0$, $(2n) + (-n) \to + \infty$, $(n) + (-2n) \to -\infty$, $(n+1) + (-n) \to 1$,$ \ldots$ $\endgroup$
    – Martin R
    Feb 28 at 7:07
  • $\begingroup$ @MartinR: Sir, I think this proof works because $c_n=a_n+b_n$ for every $n$. Indeed, sequences converging to $a^*,b^*,c^*$ are distinct but for large enough subscripts of $n$, they will be related as I have shown in my proof. Please help me understand if still my proof is wrong. Thanks for response. 😊 $\endgroup$
    – Koro
    Feb 28 at 7:12
  • $\begingroup$ Example: $a_n = (-1)^n$ and $b = (-1)^{n+1}$. Then $a_{2n} \to 1$ and $b_{2n+1} \to 1$ are the subsequences converging to the suprema, but these have no indices in common. $\endgroup$
    – Martin R
    Feb 28 at 7:14

Your proof is not correct. You have (possibly) distinct subsequences $(a_{n_k})$, $(b_{m_k})$ of the original sequences converging to $a^*$ and $b^*$, and these need not have any indices in common.

Also $\infty-\infty$ is undefined, and nothing can be said about $\limsup (a_n+b_n)$ if the right-hand side of $(1)$ is of that form. It can be $-\infty$, $+\infty$, or any finite real number.

Using your definition of $\limsup$ as the supremum of all subsequential limits, I would argue as follows:

If one of $\limsup a_n$ or $\limsup b_n$ is equal to $+\infty$ then the other cannot be $-\infty$. In that case the right-hand side of $(1)$ is $+\infty$ and the inequality holds.

Therefore is suffices to consider the case that both $\limsup a_n$ and $\limsup b_n$ are a real number or $-\infty$. In particular, both sequences are bounded above.

We can also assume that $c^* = \limsup(a_n+b_n)$ is not $-\infty$.

For every $\epsilon > 0$ there is a convergent subsequence $c_{n_k}$ of $(c_n)$ with $$ \lim_{k \to \infty} (a_{n_k} + b_{n_k}) = \lim_{k \to \infty} c_{n_k} > c^* - \epsilon \, . $$ Now $(a_{n_k})$ and $(b_{n_k})$ are both bounded sequences. Then $(a_{n_k})$ has a convergent subsequence $(a_{n_{k_l}})$ and $(b_{n_{k_l}})$ has a convergent subsequence $(b_{n_{k_{l_m}}})$. Then $$ c^* - \epsilon < \lim_{m \to \infty} (a_{n_{k_{l_m}}} + b_{n_{k_{l_m}}}) = \lim_{m \to \infty} a_{n_{k_{l_m}}} + \lim_{m \to \infty} b_{n_{k_{l_m}}} \le a^* + b^* \, . $$ This holds for all $\epsilon > 0$, which implies that $c^* \le a^* + b^*$.

Remark: The difference to your proof is that we started with a convergent subsequence $(c_{n_k})$ of $(c_n)$ and then chose “sub-sub-sequences” of $(a_n)$ and $(b_n)$ which are “simultaneously” convergent.

  • $\begingroup$ I think it is correct but what is wrong in my proof is not clear to me yet. In my case 1, I note that there is a subsequence $a_{n_k}$, which converges to $a^*$. Indices are $n_1<n_2<...,<n_k,... $. Similarly, $b_{m_l}$ is a convergent subsequence which converges to $b^*$. Indices are $m_1<m_2<...<...$. So this is what is problematic? $\endgroup$
    – Koro
    Feb 28 at 12:10
  • 1
    $\begingroup$ @Koro: Yes. For example, what if the indices for $a$ are $1, 3, 5, 7, \ldots$ and the indices for $b$ are $2, 4, 6, 8, \ldots$? Your proof claims that there are “common subsequences” $(a_{n_s})$ and $(b_{n_s})$ such that $a_{n_s}+b_{n_s} \to a^*+b^*$, but that need not be the case. $\endgroup$
    – Martin R
    Feb 28 at 12:13
  • $\begingroup$ In your answer, what if cgt. subsequence of $a_{n_k}$ has indices $n_{k_{l}}$ such that $4<8<12<...<$.Similarly, for cgt subsequence of $b_{n_k}$ we may have indices $n_{k_{s}}$ like this $2,6,18,...$. Now both indices are not common. $\endgroup$
    – Koro
    Feb 28 at 22:35
  • $\begingroup$ But I observe that it will not be the case as $(a_{n_k} + b_{n_k}) $ is cgt. And subsequence $a_{n_{k_l}}$ is cgt. therefore $ b_{n_{k_l}}= (a_{n_{k_l}} + b_{n_{k_l}})- (a_{n_{k_l}})$ is also cgt. $\endgroup$
    – Koro
    Feb 28 at 22:57
  • $\begingroup$ Can you please have a look at my answer below? Thanks in advance 😊 $\endgroup$
    – Koro
    Feb 28 at 23:15

I am posting an answer based on inputs by Mr. Martin R:

If exactly one of $a^*$ or $b^*$ is infinite, we are done. If both $a^*$, $b^*$ are infinity of same sign, then also we are done.

We consider the case, when $a^*, b^*$ and $c^*$ are finite. Since $c^*\in E_c, \exists$ subsequence $c_{n_k}\to c^*$. That is $c_{n_k}=a_{n_k}+b_{n_k} \to c^* $.

$(a_{n_k})$ is bounded and therefore has convergent subsequence $(a_{n_{k_l}}): (a_{n_{k_l}})\to \alpha$. Therefore, $b_{n_{k_l}}=(a_{n_{k_l}}+b_{n_{k_l}})-(a_{n_{k_l}})\to c^*-\alpha$.

Now, $a_{n_{k_l}}+b_{n_{k_l}}\to (\alpha)+(c^*-\alpha)\le a^*+b^*\implies c^*\le a^*+b^*$.

  • 1
    $\begingroup$ That is correct and works even for the case that $a^*$ or $b^*$ are $-\infty$. It is quite similar to my approach, only that you (correctly) observed that $(b_{n_{k_l}})$ is already convergent, so that it is not necessary to chose another subsequence. $\endgroup$
    – Martin R
    Mar 1 at 4:01
  • $\begingroup$ @MartinR: Thanks a lot 😊 $\endgroup$
    – Koro
    Mar 1 at 4:10

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.