# Proving the inequality $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$.

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?

• Your proof does not work because you have distinct subsequences of $(a_n),(b_n),(c_n)$ converging to $a^*, b^*, c^*$, respectively. Feb 28 at 7:00
• @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.
– Koro
Feb 28 at 7:04
• $\infty -\infty$ is undefined, and nothing can be said in that case: $(n) + (-n) \to 0$, $(2n) + (-n) \to + \infty$, $(n) + (-2n) \to -\infty$, $(n+1) + (-n) \to 1$,$\ldots$ Feb 28 at 7:07
• @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. 😊
– Koro
Feb 28 at 7:12
• 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. 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.

• 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?
– Koro
Feb 28 at 12:10
• @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. Feb 28 at 12:13
• 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.
– Koro
Feb 28 at 22:35
• 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.
– Koro
Feb 28 at 22:57
– 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, $$a_{n_{k_l}}+b_{n_{k_l}}\to (\alpha)+(c^*-\alpha)\le a^*+b^*\implies c^*\le a^*+b^*$$.
Proved.

• 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. Mar 1 at 4:01
• @MartinR: Thanks a lot 😊
– Koro
Mar 1 at 4:10