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I have tried to read many proofs of this but I'm not sure I get it, so please bare with me.

Show that $\lim_{n \rightarrow \infty} \sup (a_n+b_n) \leq \lim_{n \rightarrow \infty} \sup (a_n)+lim_{n \rightarrow \infty} \sup (b_n).$

I care about the case when $(a_n + b_n)$ is bounded.

Take an arbitrary subsequence $({a_n}_k + {b_n}_k)$, it must converge to a limit $l$ since $(a_n + b_n)$ is bounded.

Take the subsequences $({{a_n}_k}_j)$ and $({{b_n}_k}_j)$, since $({a_n}_k + {b_n}_k)$ converges to $l$, $({{a_n}_k}_j)$ + $({{b_n}_k}_j)$ must also converge to $l$.

So we have $$\lim_{n \rightarrow \infty}({a_n}_k + {b_n}_k)=lim_{n \rightarrow \infty}({{a_n}_k}_j)+\lim_{n \rightarrow \infty}({{b_n}_k}_j) \leq \lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n)$$ which is true by the definition of supremum. (Is there some intuition here?)

Well, any subsequence $(a_n+b_n)$ that converges is bounded above by $\lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n) $, so certainly $\lim_{n \rightarrow \infty} \sup (a_n+b_n)$ is bounded above, or in other words $$\lim_{n \rightarrow \infty} \sup (a_n+b_n) \leq \lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n).$$

Questions:

  1. Is this a correct proof?

  2. The inequality seems obvious by itself, because the suprema of two sequences will always be an upper bound of those two sequences added together. The only other thing that can happen is that those suprema are an upper bound for $(a_n+b_n)$, but not the least upper bound - if this is correct, is this a case of 'so easy it's hard' or am I missing an essential piece of the puzzle? None of the proofs I've seen online have made any sense to me, so I'm guessing it's the latter.

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1. The argument is not correct. Take $a_n=(-1)^n$ and $b_n=(-1)^{n+1}$. You have $a_n+b_n=0$ which converges, and yet neither $a_n$ nor $b_n$ converges. –  1015 Mar 8 '13 at 20:45
    
This is similar to (1) and (2). –  robjohn Mar 8 '13 at 20:47
    
@julien Could I then take another subsequence $({{{b_n}_k}_j}_p)$ and construct it so that $({{{b_n}_k}_j}_p) = (a+b)_{nkj}-(a_{nkj})$? In other words, can I choose my subsequences more carefully to have them converge to $l$? (and does that make the proof correct?) –  john.abraham Mar 8 '13 at 21:07
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No, this can't work. See robjohn's answer. –  1015 Mar 8 '13 at 21:16
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1 Answer 1

up vote 1 down vote accepted

The simplest way that I have found to think about the sub-additivity of $\limsup$, is to use the definition $$ \limsup_{n\to\infty}a_n=\lim_{n\to\infty}\sup_{k\ge n}a_k\tag{1} $$ The limit on the right exists since $\sup\limits_{k\ge n}a_k$ is a decreasing function of $n$.

Note that $$ \sup_{k\ge n}\,(a_k+b_k)=\sup_{\substack{j\ge n\\k\ge n\\j=k}}\,(a_j+b_k)\tag{2} $$ and $$ \sup_{k\ge n}a_k+\sup_{k\ge n}b_k=\sup_{\substack{j\ge n\\k\ge n}}\,(a_j+b_k)\tag{3} $$ Since the $\sup$ on the right hand side of $(2)$ is over a smaller set, it will be smaller than that on the right hand side of $(3)$. That is, $$ \sup_{k\ge n}\,(a_k+b_k)\le\sup_{k\ge n}a_k+\sup_{k\ge n}b_k\tag{4} $$ Taking the limit of both sides of $(4)$ yields $$ \limsup_{n\to\infty}\,(a_n+b_n)\le\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n\tag{5} $$

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I get that the left hand side of (3) is a larger set than (2); could you please explain how the right hand side of (3) conveys that? –  john.abraham Mar 8 '13 at 21:59
    
@Eriko: Comparing the left hand sides of $(2)$ and $(3)$ is what $(2)-(4)$ are all about. This is done by looking at the right hand sides. Note that the right hand sides of both $(2)$ and $(3)$ are exactly the same except for the restriction to $j=k$ in $(2)$. This means that we are taking a supremum over a larger set of $j,k$ in $(3)$ than in $(2)$. This is because $$\{(j,k):j,k\ge n,j=k\}\subset\{(j,k):j,k\ge n\}$$ –  robjohn Mar 8 '13 at 23:34
    
Is it right to say that $(a_j + b_k)$ where $(j,k)\geq n,j=k$ is a subset of $(a_j + b_k)$ where $(j,k)\geq n$? –  john.abraham Mar 9 '13 at 10:54
    
@Eriko: yes. If the indices are a subset, then the values are also a subset. –  robjohn Mar 9 '13 at 14:48
    
Great, I get it now. Thank you! –  john.abraham Mar 9 '13 at 15:01
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