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I read in this post that $\sup(f(x)+g(x))\le\sup f(x)+\sup g(x)$ (whenever all these quantities are defined) whereas I read somewhere that $\sup (A+B)=\sup A + \sup B$ where $A, B$ are real-valued, bounded sets.

Can someone please explain why this discrepancy exists (if it does at all)? Where is there an inequality in one case and an equality in the other?

Edit: Let $S=\{a+b: a\in A, b\in B\}$. Then $\sup(A+B)=\sup S$.

Ricky_Nelson
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  • How do you define $A+B$? – Jacky Chong Feb 19 '21 at 03:46
  • Then consider the simple example: $f(x) = x$ and $g(x) = -x$ on $[0, 1]$. See if you can convince yourself that there is no discrepancy. – Jacky Chong Feb 19 '21 at 03:52
  • @JackyChong So, there shouldn't be an inequality in the case of the 2 functions, right? – Ricky_Nelson Feb 19 '21 at 03:55
  • I didn't say that. – Jacky Chong Feb 19 '21 at 03:55
  • $\sup (f(x)+g(x))$ is much more rigid that $\sup(A+B)$. – Jacky Chong Feb 19 '21 at 03:56
  • @JackyChong If I take your example of $f(x), g(x)$, then $\sup (f(x)+g(x)) = \sup{0} = 0$, $\sup f(x) = 1$, and $\sup g(x) = 0$. I am still missing your point. – Ricky_Nelson Feb 19 '21 at 03:58
  • My point is: when you define $A+B$, you are looking at all possible addition of two numbers, one from set $A$ and one from set $B$. But it is not the case for $\sup(f(x)+g(x))$. So, while $\sup(A+B) = \sup A+\sup B$, it is very possible that $\sup(f(x)+g(x)) \ne \sup f(x) + \sup g(x)$. – Jacky Chong Feb 19 '21 at 04:02
  • @JackyChong So, the key to understanding this is that when we're looking at the supremum of the sum of two functions, we are looking at a "transformed" or new function, in some sense, that happened to be obtained by adding two functions. Makes perfect sense now, thanks! – Ricky_Nelson Feb 19 '21 at 04:23

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