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I'm working through some analysis textbooks on my own, so I don't want the full answer. I'm only looking for a hint on this problem.

I'm trying to prove the subadditivity of the limit supremum, i.e.

$$ \limsup_{n \to \infty} (x_n + y_n) \le \limsup_{n \to \infty} (x_n) + \limsup_{n \to \infty} (y_n) $$

My book and the notes that accompany it say I should be able to do this without resorting to properties that relate the infimum to the supremum, so I'm assuming my approach shouldn't follow this question (which doesn't actually address this particular statement directly).

Here's my attempt at a proof that uses contradiction:

  1. Assume that $\limsup_{n \to \infty} (x_n + y_n) > \limsup_{n \to \infty} (x_n) + \limsup_{n \to \infty} (y_n)$

  2. Consider the sequences $(x_n) = (-1)^n$ and $(y_n) = (-1)^{n+1}$. I know that $\forall n, \sup(x_n) = \sup(y_n) = 1$, but $\sup(x_n + y_n) = 0$.

  3. This is a contradiction because I can't have $0 > 1$, so the proof is complete.

Is this a valid proof for this?

If this isn't a valid proof, is there a simple hint that may put me on the right track? I'm a bit stuck on this one.

EDIT:

Here's what I can think of so far, but I'm really stuck on this.

  1. I know that the supremums of subsequences form a monotonically decreasing sequence, and because that sequence converges, it's bounded.

  2. I know that $\limsup_{n \to \infty} (x_n)$ and $\limsup_{n \to \infty} (y_n)$ aren't necessarily upper bounds on the entire sequence; they're just upper bounds on some subsequence in the limit.

I'm really at a loss where to go from here. Can anyone give me some nudge in the right direction? After staring at this proof for days and seeing a lot of resources that say "it's trivial" which is pretty disheartening.

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M T
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2 Answers2

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First note that the claim is not quite true as stated: you have to exclude the case that $\limsup_{n \to \infty} x_n = +\infty$ and $\limsup_{n \to \infty} y_n = -\infty$ or vice versa, in which case the right side is undefined.

Suggestion:

  1. Let $a_m = \sup_{n \ge m} (x_n + y_n)$, let $b_m = \sup_{n \ge m} x_n$, $c_m = \sup_{n \ge m} y_n$. We want to show $\lim_{m \to \infty} a_m \le \lim_{m \to \infty} b_m + \lim_{m \to \infty} c_m$.

  2. Note that for any $n \ge m$ we have $x_n \le b_m$ and $y_n \le c_m$. Conclude that for any $n \ge m$ we have $x_n + y_n \le b_m + c_m$.

  3. Show that $a_m \le b_m + c_m$.

  4. By standard properties of limits, $\lim_{m \to \infty} a_m \le \lim_{m \to \infty} (b_m + c_m)$.

  5. By standard properties of limits, $\lim_{m \to \infty} (b_m + c_m) = \lim_{m \to \infty} b_m + \lim_{m \to \infty} c_m$. (Note that we have to assume, as I mentioned at the top, that the limit on the right side is not of the form $\infty - \infty$.)

Nate Eldredge
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    I sort of follow you until step 3, but here's my thought on #3. Say that this isn't true, and $a_m > b_m + c_m$. I keep wanting to bring this back to the sign of the elements, though, and say something like if $a_m > b_m + c_m$, then $\sup_{n \ge m} (x_n + y_n) > \sup_{n \ge m} x_n + \sup_{n \ge m} y_n$, so at some index $i \ge m$, then the element $x_i + y_i$ exceeds the upper bounds of both sequences put together. That seems intuitively correct, but I don't know how to make it rigorous. – M T Jan 15 '15 at 23:07
  • Obviously step #3 is the meat here, because once we have that everything follows quickly from standard properties of limits. – M T Jan 15 '15 at 23:08
  • Wait, no, I think I have #3 now that we're just talking about the supremums of sets. Since #3 doesn't involve the limits, we're just talking about the supremum of the sets $X = {x_n | n \ge m}$ and $Y = {y_n | n \ge m}$. If $u = \sup X$ and $v = \sup Y$, then $x \le u, \forall x \in X$ and $y \le Y, \forall y \in Y$, so by addition, $x + y \le u + v, \forall x \in X, y \in Y$. Is that it? – M T Jan 15 '15 at 23:18
  • @Michael: What you've written is actually the logic needed for #2, not #3. For #3, consider the following: in #2 we showed that $b_m + c_m$ is an upper bound for the set ${x_n + y_n \mid n \ge m}$. And $a_m$ is by definition the least upper bound of this set... – Nate Eldredge Jan 15 '15 at 23:48
  • Right, I forgot that last part, but I figured it out. $a_m$ is the least upper bound of the set, so since we found something that's an upper bound of the set, we know $a_m$ has to be $\le$ than it. Then the rest follows from the properties of limits. – M T Jan 15 '15 at 23:55
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You have only given one example of sequences where the opposite statement (sub-additivity fails) is false, by contradiction. A full proof by contradiction would have to consider arbitrary sequences where sub-additivity fails and then somehow derive a contradiction.

user2566092
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  • I'm an idiot and obviously I'm been working on this for too long because I didn't even realize that my proof is ridiculous. Thank you. Are these sequences at least an example of sequences where the inequality is strict? Or did I mix something up? – M T Jan 15 '15 at 21:13
  • Yes, your example gives a strict inequality and IMO is a good example. – user2566092 Jan 15 '15 at 21:19
  • Is it straightforward to prove this using contradiction in general, or am I barking up the wrong tree by trying? – M T Jan 15 '15 at 21:31
  • @Michael I think a direct proof is the most straightforward, using properties of lim sup but I may be wrong. Of course you could always turn that into a proof by contradiction by saying "Assume not" and then give the direct proof, giving a contradiction. – user2566092 Jan 15 '15 at 21:36
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    Ok, I'll work on this more. I think the only property I know so far is that the partial supremums form a monotonically decreasing sequence, but I'm not sure where to go from there yet. I really have no idea how to attempt this, and it's pretty frustrating. I don't know that the limit supremum is an upper bound on the entire sequence, so I'm completely stuck. Is there a specific property I should be targeting? I'm sure this is a simple proof, but obviously math isn't my strong suite. – M T Jan 15 '15 at 21:54