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I am following a paper proof which starts with the following constraint:

$$\forall v \in [a,b], \forall \tilde{v} \in [a,b]$$ $$f_{i}(v) \geq f_{i}(\tilde{v}) + (v-\tilde{v})g_{i}(\tilde{v})$$

In the proof, the author writes let $h = \tilde{v} - v$ therefore we have:

$$ f_{i}(v) \geq f_{i}(h+v) + h*g_{i}(h+v)$$

I understand that this is when $\forall v \in [a,b]$ However, what interval will $h$ be defined in? Would it be that all $\forall h \in [a,b]$?

If $v=a$ and $\tilde{v} = a$ then $h \not\in [a,b]$?

Andrés E. Caicedo
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A K
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1 Answers1

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You can show that $h$ will verify $$-|a-b| \leq h \leq |a-b|$$

So indeed, often $h\not\in [a,b]$

Tryss
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  • Thank you for your help. Put then if $\tilde{v}=h+v$, how can it be that $\tilde{v} \in [a,b]$? For example, letting $[a,b] = [0,1]$, $h \in [-1,1]$ and as $\tilde{v} = h+v$, we get $\tilde{v} \in [-1,2]$? This contradicts the fact that $\tilde{v} \in [0,1]$ – A K Mar 10 '15 at 01:24
  • The value of h depend of the value of v : you can't have h=-1 if v is different from 1, and you can't have h=1 if v is different from 1 – Tryss Mar 10 '15 at 03:30