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Prove that $\sup\{x_n+y_n\}\leq \sup\{x_n\}+\sup\{y_n\}$, if both sups are finite. Furthermore, prove that $\limsup\{x_n+y_n\}\leq \limsup\{x_n\}+\limsup\{y_n\}$ if both limsups are finite.

Martin Sleziak
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user156594
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    The part about limsup is answered in several posts on this site, for example http://math.stackexchange.com/questions/408135/prove-limsup-limits-n-to-infty-a-nb-n-le-limsup-limits-n-to-infty and http://math.stackexchange.com/questions/69391/subadditivity-of-the-limit-superior – Martin Sleziak Sep 14 '14 at 09:57

2 Answers2

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For the first part of your question, let $\alpha := \sup_n x_n$ and $\beta := \sup_n y_n$. Then simply note that

$$ x_n + y_n \leq \alpha + \beta $$

for all $n \in \Bbb{N}$. As the supremum of a set is the least upper bound, we conclude

$$ \sup_n (x_n + y_n) \leq \alpha + \beta = \sup_n x_n + \sup_m y_m. $$

Note that it can happen for the inequality to be strict. Consider for example $x_n = (-1)^n$ and $y_n = (-1)^{n+1}$. Then $\sup_n x_n = 1 = \sup_n y_n$, but $x_n + y_n = 0$ for all $n$, hence $$\sup_n (x_n+y_n) = 0 < 2 = \sup_n x_n + \sup_m y_m.$$

For the second part, as Matrin Sleziak wrote, see Prove $\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n$.

Community
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PhoemueX
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  • is the following notation $\sup_n (\cdot)$ equivalent to $\lim_{n \to \infty} \sup (\cdot)$ ? – Naz Oct 13 '16 at 07:24
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    @isquared-KeepitReal: No, $\sup_n a_n$ means $\sup {a_n : n \in \Bbb{N}}$. Then there is also $\limsup_n a_n$, which means $\lim_{n \to \infty} \sup {a_k : k \geq n}$. I am not sure what the notation $\lim_{n\to\infty}\sup (\cdot)$ is supposed to mean, probably one of the two things I just mentioned. – PhoemueX Oct 13 '16 at 12:41
  • hmm.. There is a paper I am looking at that talks about $\beta$-mixing $\beta(m)$. Where $\beta(m) = \frac{1}{2} \sup_n \sup \sum_{i=1}^I \sum_{j=1}^J |P(A_i \cap B_j) - P(A_i)P(B_j)|$ and I could not see where that $\sup_n$ is coming from.. – Naz Oct 13 '16 at 12:52
  • @isquared-KeepitReal: Also, $m$ seems not to be used anywhere. Without more context, I can not tell you what is meant. – PhoemueX Oct 13 '16 at 15:29
  • Precisely. I also wondered how can beta depend on m if it is not used. I will dig out the paper tomorrow – Naz Oct 13 '16 at 20:19
  • here is the paper, it is on the third page: https://arxiv.org/pdf/1402.4501v3.pdf – Naz Oct 14 '16 at 07:31
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You may also try:

(1)If A is a sub set in B, then sup(A) <= sup(B). (2)For A, B are subsets in R, sup(A + B) = Sup(A) + Sup(B). $A + B = \{c = a + b\mid a \in A, b \in B\}$.

Then notice that $\{an + bn\}\subseteq \{an\} + \{bn\}.$

(2)The second inequality follows naturally:

$Am$ = { $an | n >= m$ }, $Bm$ = { $bn | b >= m$}, $Cm$ = { $cn = an + bn | n >= m$}.

Then, by the first inequality, Sup(Cm) <= Sup(Am) + Sup(Bm), for each m.

And, the sequence inequality holds as m -> $+00$.

Benjamin
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