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We have a sequence of functions $K_n$, where

$$K_n=F_n+G_n+H_n$$

So, is $$\sup(K_n)\le \sup(F_n)+\sup(G_n)+\sup(H_n)?$$

$$\forall n\ge 1$$ Also, what will the infimum of the sum be like?

esssystephen
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  • Is there any reason for an inequality? I would think that both are equal. – Dominique Oct 30 '24 at 13:44
  • @Dominique, they're not always equal. take this for example – esssystephen Oct 30 '24 at 13:48
  • ... and what is "this"? (You might edit your question) – Dominique Oct 30 '24 at 13:49
  • @Dominique. yes, sorry. For confirmation, $Sup(A+B) \le Sup A + Sup B$?? So my question is if that applies to any number of functions to be added – esssystephen Oct 30 '24 at 13:53
  • @Dominique that is not even my main problem, I just need to know if I can apply it to my proof – esssystephen Oct 30 '24 at 13:54
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    Can you just give an example of two sets $A$ and $B$ where $\sup(A+B) < \sup(A) + \sup (B)$? I really don't see it. – Dominique Oct 30 '24 at 13:55
  • @esssystephen Please see problem 2 here – Soham Saha Oct 30 '24 at 13:58
  • @Dominique for the set $A$, $a\le\beta, \forall a \in A$ and $b\le \alpha, \forall b\in B$, where $\beta = Sup(A)$ and $ \alpha = Sup(B)$. Take the sum and you see the $C=${$a+b;a\in A,b\in B$} is bounded above by $\beta+\alpha$. – esssystephen Oct 30 '24 at 14:11
  • I still don't see it. Can you give an example with some numbers? In my opinion, the supremum of $C$ is simply $\alpha + \beta$. – Dominique Oct 30 '24 at 14:15
  • @SohamSaha I have gone through it. however, i don't get what happened with $\epsilon$ from claim 2 – esssystephen Oct 30 '24 at 14:17
  • @Dominique, take $A=(-1)^n$ and $B=(-1)^{n+1}$ where $n\ge 1$ – esssystephen Oct 30 '24 at 14:26
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    @esssystephen It sounds like you are saying $A$ is the set ${-1,1}$ and so is $B$. But you have in your head that they are sequences where only corresponding terms can be added. So you are mixed up with notation. – 2'5 9'2 Oct 30 '24 at 14:39
  • I still don't see it: $A={1, -1}$ and $B={-1, 1}$. Hence $A+B={-2,0,2}$, while you seem to think it only contains the value zero. Am I right? – Dominique Oct 30 '24 at 14:40
  • @Dominique i mentioned in an earlier comment that they are functions. I seem to have edited it out. – esssystephen Oct 30 '24 at 14:42
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    @Dominique yes, what you said is true for sets but not for sequences/functions. that's why I disagreed – esssystephen Oct 30 '24 at 14:44
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    Are they sequences of functions, or just sequences of real numbers? If they are indeed functions, are the suprema over $n$ or over the domain of the functions? Do they even have the same domain? – Theo Bendit Oct 30 '24 at 14:51
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    @Dominique I have edited the question now. so it is clear – esssystephen Oct 30 '24 at 14:52
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    @TheoBendit They are all sequences of functions and the suprema are over $n$ – esssystephen Oct 30 '24 at 14:56
  • Generally speaking, if $a_n$ and $b_n$ are real sequences, $\sup_n(a_n+b_n)\le\sup_na_n+\sup_nb_n$. This is proven here. This implies$$\sup_n(a_n+b_n+c_n)\le\sup_n(a_n+b_n)+\sup_nc_n\le\sup_na_n+\sup_nb_n+\sup_nc_n.$$This applies pointwise to each of your functions. – Theo Bendit Oct 31 '24 at 02:24

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