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None of the limit superior inequality proofs in MSE did the trick for me, this is the inequality which I am trying to prove:

$$\lim \sup(a_n+b_n)\leq \lim \sup(a_n)+\lim\sup(b_n)$$

for $a_n$, $b_n$ bounded.

My reasoning is as such: if $a_n$ and $b_n$ are bounded, then surely $a_n + b_n$ is bounded as well. That means we can find subsequences $a_{n_k}$ and $b_{n_k}$ such that the sum of their limits is equal to the limit superior of the sum:

$$\lim\sup(a_n+b_n)=\lim(a_{n_k})+\lim(b_{n_k})$$

Then, I claim that $a_{n_k}$ and $b_{n_k}$ are not necessarily the subsequences that converge to the bigger of the sublimits of $a_n$ and $b_n$, and so we have:

$$\lim(a_{n_k})+\lim(b_{n_k})\leq\lim\sup(a_n)+\lim\sup(b_n)$$

Which concludes the proof. Is this correct? If yes, how can I make it more rigorous?

user401936
  • 1,071
  • You might wanna have a look at post of mine here on Math SE on this very result. See how you like it. Here is the link. https://math.stackexchange.com/questions/1757778/theorem-3-19-in-baby-rudin-the-upper-and-lower-limits-of-a-majorised-sequence-c – Saaqib Mahmood May 07 '18 at 17:07
  • This is not the same proof, I already worked out theorem 3.19 in Rudin but I fail to see how 3.19 directly implies the inequality above. – user401936 May 07 '18 at 17:11
  • Oh sorry! How about the post of mine with the following link? https://math.stackexchange.com/questions/2071290/prob-5-chap-3-in-baby-rudin-lim-sup-n-to-infty-lefta-nb-n-right-leq-l – Saaqib Mahmood May 07 '18 at 18:19

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