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Problem: 1. Let $(a_n)$ and $(b_n)$ be bounded sequences of real numbers. Prove that $$\limsup _{n\to \infty}(a_n+b_n)\leq \limsup _{n\to \infty}a_n + \limsup _{n\to \infty}b_n,$$ and that equality holds when $(a_n)$ or $(b_n)$ is convergent.

Weak Proof:

Let $(a_n)$ and $(b_n)$be bounded sequences of real numbers. Then the limit superior of $(a_n)$ and $(b_n)$ are defined $$ \limsup_{n\to\infty}a_n=\lim_{n\to\infty}\sup_{k\ge n}a_k $$ $$ \limsup_{n\to\infty}b_n=\lim_{n\to\infty}\sup_{k\ge n}b_k. $$ Note that $\{(j,k): j,k \geq n, j=k\} \subset \{(j,k): j,k \geq n\}$. Then $$ \sup_{k\ge n}\,(a_k+b_k)=\sup_{\substack{j\ge n\\k\ge n\\j=k}}\,(a_j+b_k) $$ and $$ \sup_{k\ge n}a_k+\sup_{k\ge n}b_k=\sup_{\substack{j\ge n\\k\ge n}}\,(a_j+b_k). $$ Since $\sup_{\substack{j\ge n\\k\ge n\\j=k}}\,(a_j+b_k)$ is over a smaller set than $\sup_{\substack{j\ge n\\k\ge n}}\,(a_j+b_k)$, we have $$ \sup_{k\ge n}\,(a_k+b_k)\le\sup_{k\ge n}a_k+\sup_{k\ge n}b_k. $$ Taking the limit of both sides we get $$\limsup _{n\to \infty}(a_n+b_n)\leq \limsup _{n\to \infty}a_n + \limsup _{n\to \infty}b_n.$$

Thanks for help. I also need to prove equality when either sequence is convergent.

JohanLiebert
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  • I think your proof is pretty good. – Pocho la pantera Oct 09 '13 at 15:22
  • @Pocholapantera Knowing my professor, this proof is sure to get marked down. Are you sure this is good enough? – JohanLiebert Oct 09 '13 at 15:30
  • Yes, I do (but I am not your professor). Just a comment: At the end of the proof, when you take limit of both sides, you are using that if the limits exist, then the limit of the sum is the sum of the limits. Something you are asked to prove. So, you must prove that. – Pocho la pantera Oct 09 '13 at 15:42
  • If $a<b$then $\lim a< \lim b$, right? The limit exists of those sup functions because they are descreasing – JohanLiebert Oct 09 '13 at 15:53
  • Your proof is fine. The limits exist because the sequences are non-decreasing, and addition is continuous, hence you can interchange addition and limits. – copper.hat Oct 09 '13 at 15:54

1 Answers1

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Suppose $a_n \to a$. Then all you need to show is $\limsup_n (a_n+b_n) \ge a + \limsup_n b_n$

Choose a subsequence $b_{n_k}$ such that $b_{n_k} \to \overline{b} = \limsup_n b_n$. Clearly $a_{n_k} \to a$ since $a_n \to a$.

Then $a_{n_k}+b_{n_k} \to a+ \overline{b}$, and so $\limsup_n (a_n+b_n) \ge \lim_k (a_{n_k}+b_{n_k}) = a+ \overline{b} = \limsup_n a_n + \limsup_n b_n$.

copper.hat
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