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Consider the strongly monotonic function $w:\mathbb{R}\rightarrow [a\;b]$, with $a,b\in\mathbb R$, that is, there exist a scalar $\eta(x,h)$ that verifies $$ \frac{w(x+h)-w(x)}{h}\geq \eta(x,h)>0, \quad \forall x,h\in\mathbb{R}. $$

I was wondering if it is possible to write a simplified expression for the lower bound of $$ \mathcal T(x):=[w(\bar x+x+h_1)-w( \bar x)]\cdot[w(\bar x+x+h_2)-w( \bar x)], \quad \forall x,\bar x, h_1,h_2\in\mathbb{R}. $$ What if one defines $f(x):=w(\bar x+x+h_1)-w(\bar x)$ so to rewrite $$ \mathcal T(x)=f(x)\cdot f(x+h_2-h_1) $$ ? Is it possible to say something based on strong monotonicity? Otherwise, which further assumption do I need on $w$ to get this lower bound?

Daniele
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    There is no $\eta>0$ that works for all $x,h$. Take $w(x)=\operatorname{arctan}(x)$. Then $w'(x)=\frac{1}{1+x^2}$ which gets arbitrarily small. – Kurt G. May 23 '22 at 16:58
  • So do you mean that for $\eta$ to be NOT arbitrarily small the co-domain of $w$ should be the whole $\mathbb R$? @KurtG. – Daniele May 23 '22 at 17:10
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    Please clarify what $\mathcal T(x) := \ldots,,\forall x,\bar x, h_1,h_2$ is supposed to mean. Also there seems to be an error in the premises. The definition implies that $w(x) \geq \eta h + w(0)$ for $h\geq 0$ and $w(x)\leq \eta h + w(0)$ otherwise. Thus $w\to\pm\infty$ at $x\to\pm\infty$, which contradicts $w:\mathbb R\to[a,b]$. – Lazy May 23 '22 at 17:12
  • @Daniele . I only mean that it is confusing what you assume about $\eta$. The first two lines of this question (before you start wondering) should be unambiguously formulated. Do you assume that $\forall x\forall h \exists\eta >0$ or $\exists\eta>0\forall x\forall h$ ? – Kurt G. May 23 '22 at 17:15
  • @Lazy I would like to establish a lower bound for the term $\mathcal T(x)$, for a fixed $\bar x$ holds true for any $h_1,h_2,x\in\mathbb R^n$. Regarding the contradiction you mention, does this mean that for $\eta$ to be NOT arbitrarily small the co-domain of w should be the whole $\mathbb R$? Txs! – Daniele May 23 '22 at 17:17
  • @KurtG. I am just saying that there exists a positive $\eta$ for any $x,h$. As you said, this can be arbitrarily small. – Daniele May 23 '22 at 17:20
  • @Daniele If $\eta$ might depend on $x$ and $h$ then this is simply strict monotony, as then $\eta(x,h)$ can simple be chosen as $(w(x+h)-w(x))/h$ to even achieve equality. If on the other hand $\eta$ is fixed for all $x,h$ then for any interval $[x,y]$ you’d have $w(y)-w(x) \geq \eta (y-x)$, which gives you a contraint on how large the codomain has to be at the very least depending on the size of the domain. – Lazy May 23 '22 at 18:27

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