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I am mainly motivated by the question regarding how to evaluate the goodness of an inequality $$f(x) \leq g(x), \text{ }\forall \text{ }x.$$

The first approach is using numerical evaluations to compare $f(x)$ and $g(x)$ by sampling many different $x$ values.

Another approach is to use the concept of 'sharp inequalities'. There are several questions already regarding sharp inequalities, e.g., this question and this one. It is clear that the inequality is sharp if $\exists$ $x_0$ such that $f(x_0) =g(x_0)$. Some people call such a bound `best possible'.

However, what if we could find another function $f_2$ such that $f(x) \leq f_2(x) \leq g(x), \text{ }\forall \text{ }x$, and $\exists$ $x_1 \not= x_0$ such that $f(x_1) < f_2(x_1)$? In this case, both $f$ and $f_2$ are sharp lower bounds for $g$; however, it seems that $f_2$ is 'better' than $f$, as $f(x_1) < f_2(x_1)$.

To define the notion of 'being better' above, we clearly cannot use the concept of 'sharpness', as both bounds are sharp. Then what concept should we use here? Should we say that $f_2$ is a 'tighter' lower bound than '$f$'? How is the concept of 'being tight' or 'tighter' formally defined?

Are there other ways to analyze whether $f$ is a good lower bound for $g$?

It would be much appreciated if references are also provided.

Shi James
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    I wouldn't think there was a single answer to that sort of question, rather it depends on your purposes. For instance, one might say that $f_1$ was a better bound than $f_2$ if $\max |f_1(x)-g(x)|<\max |f_2(x)-g(x)|$, so a "worst case" test. Or you might want the average error to be lower. And so on. Or you might want to restrict the family of functions to which $f$ might belong (after all, $g(x)$ is the best lower bound for $g(x)$ if you have no restrictions). – lulu Nov 06 '23 at 23:29
  • Should say, the mere fact that the functions agree at a point isn't worth much. $f(x)=-|x|$ and $g(x)=x^{10}$ coincide at $x=0$ but $f$ is an absolutely terrible lower bound for $g$. – lulu Nov 06 '23 at 23:33
  • I agree. If I am only given $f_1$, for example, (no comparison $f_2$ is given or known), then is it true that the best thing we could do is to state that $f_1$ is a sharp bound? (or we could always compare $f_1$ and $g$ of course, but there is no other bound to compare to.) – Shi James Nov 06 '23 at 23:34
  • As I say, the mere fact that the two function meet at a point tells you very little. You'll want to specify more...presumably you have more information regarding the nature of $f$. Otherwise you could just take $f=g$ which, presumably, isn't helpful. – lulu Nov 06 '23 at 23:36
  • Again: the way this sort of thing typically goes is that you specify a family of functions (polynomials of low degree, say) and you specify a test (minimize the average error on some interval, for example) and then you have a more classical minimization problem. But the choice of function space and the choice of metric are usually up to the user. – lulu Nov 06 '23 at 23:39
  • Or, in those situations in which $g(x)$ is so terrible that you can't reasonably hope to solve any minimization problem involving it, the goal tends to be simpler: a lower bond is good if it involves a tractable function which you can prove to be a lower bound and which can't obviously be improved. Often times, that's the best you can hope for. – lulu Nov 06 '23 at 23:52

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I don't know if there is a formal framework for analyzing inequalities this way. We usually want to prove sharp inequalities, but a sharp inequality is not always the "best possible" inequality. As a silly example, $$ |a\cdot b|\le 1\qquad(a,b\in\mathbb R^n, |a|,|b|\le 1) $$ is a sharp inequality because it is attained for various values of $a$ and $b$ in the description, but it's not the best possible. There is a better inequality which is of course $$ |a\cdot b|\le |a|\cdot|b|, $$ and this inequality is also sharp. It's better than the first inequality because $|a|\cdot|b|\le 1$.

If we have a non-negative quantity $f(a_1,a_2,\dots)$ which depends on the parameters $a_1,a_2,\dots$, we generally want to find the best expression $g(a_1,a_2,\dots)\ge 0$ which satisfies $$ f(a_1,a_2,\dots) \le g(a_1,a_2,\dots)\qquad(\text{for all $a_1,a_2,\dots$)} $$ in the sense that for any other quantity $g'(a_1,a_2,\dots)$ which also satisfies the above inequality for all the inputs $a_1,a_2,\dots$, we have $g(a_1,a_2,\dots)\le g'(a_1,a_2,\dots)$.

To connect with the earlier example, of course $g(a,b) = |a|\cdot|b|$ "beats" $g'(a,b) = 1$ for $|a|,|b|\le 1$, although both are sharp upper bounds for $f(a,b) = |a\cdot b|$.

Alex Ortiz
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