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I am confused about what a "sharp" upper bound means. I have already seen this question asked elsewhere (linked) but it does not answer my question: What does it mean when a bound is sharp?

The main question that I have is:

Is a "sharp" upper bound unique?

It is easier to work with an example, so the following is a verbose version of the question:

Consider a set of scaled sinusoidal functions, $$ S = \{f_t~|~\forall x \in \mathbb{R}~~f_t(x) = t\sin(x),~~t \in [0, 1]\}. $$ I can establish the upper bound, $$ \forall f_t \in S,~~f_t(x) \leq 1~~\forall x. $$ Clearly, the upper bound is attained for the function $f_1$. Would this upper bound be called sharp? What about the following upper bound which is clearly better? $$ \forall f_t \in S,~~f_t(x) \leq t~~\forall x. $$ Would both these bounds be called sharp?

diff2
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    See https://math.stackexchange.com/questions/1985794/what-does-it-mean-when-a-bound-is-sharp. – edm Feb 27 '18 at 16:04
  • Thanks. I have already seen that question (I have also linked it above in my question). I don't think the linked question clarifies my doubt: specifically, with respect to the example given above, both upper bounds qualify as "sharp" according to the linked question as they are both attained. The second bound seems "sharper", though. So I am not sure if the first can be called "sharp". – diff2 Feb 27 '18 at 16:08
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    I wouldn't think the term is terribly precise. Informally it means that no simple modification of the bound improves it, in which case both of your examples would be sharp (again, informally. Without worrying too much about what "simple" means, say). But if something more than an informal sense is intended, I'd set out a definition specific to the context. – lulu Feb 27 '18 at 16:10
  • I guess my question boils down to this: "an upper bound being attained" is not equivalent to "an upper bound that can not be improved upon". In the linked question these two concepts are treated as equivalent; my example shows that they may not be. Both bounds are attained, but the second bound is the one that can not be improved upon. Agreed? – diff2 Feb 27 '18 at 16:11
  • To stress: You can't expect that "sharp" means "can't be improved". Unless your inequality is tautological, as in $f(x)≤f(x)$, you can always improve it. – lulu Feb 27 '18 at 16:12
  • In context it may be clear that you are only looking at bounds of a certain type. Like "constant" or whatever. Within such a well defined context it might make sense to seek the optimal bound (or it may not). Context and definition are critical. – lulu Feb 27 '18 at 16:14
  • @lulu OK, this is what my doubt was, thanks. In the answers to the other question that I linked, they treat these two concepts as equivalent. I am of the same view as you, and would consider both bounds as sharp, but I wanted to gauge what the general opinion is. – diff2 Feb 27 '18 at 16:14

1 Answers1

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I would say that a "sharp" upper bound is unique, yes. But hidden in this phrase is some subtlety: what do you mean by an "upper bound"? If we have a collection of things that we allow ourselves to call "upper bounds", we can ask whether one of those is sharp at all, but if we don't define "upper bound" clearly, it might not be obvious which objects count and which don't.

In your example, you clearly want to extend the notion of "upper bound" from a single function $f$ to a family of functions $f_t$ parametrised by some $t$. You have (at least!) two options:

  • a "global" upper bound, i.e. a single upper bound for the entire family of functions, i.e. some constant $a$ such that $f_t(x) \leq a$ for all $t$ and $x$
  • a "local" upper bound, i.e. a family of upper bounds for the functions, i.e. a family of constants $a_t$ such that $f_t(x) \leq a_t$ for all $t$ and $x$.

Now, all global upper bounds are local upper bounds, but not vice-versa; and something that is sharp as a global upper bound (i.e. is "best possible" among all global upper bounds) might not be sharp as a local upper bound.

  • Your first bound is sharp as a global upper bound, but not sharp as a local upper bound.
  • Your second bound isn't a global upper bound at all, but is a local upper bound and is sharp as such.

(I've just made up the terminology "local" and "global" here, by the way.)

Billy
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    Thanks. I don't know who downvoted you, but this made some sense to me. I suppose the downvote means that someone thinks we are still missing something, so I will keep the question open to see what other opinions people might present here. – diff2 Feb 27 '18 at 16:22
  • I won't lose sleep over it! – Billy Feb 27 '18 at 16:26