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EdisonCruise
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Posts: 92
Joined: September 15th, 2012, 4:22 am

Why the minimum is taken here when its derivative > 0?

May 7th, 2019, 2:12 am

Below are sentensce taken from a paper.
"In fact, we are going to prove that
$$\forall t\in[0,T],\forall q\in\{-Q,...,Q\},\nu_q(t)\ge e^{-(\alpha Q^2-\eta)(T-t)} $$
If this was not true then there would exist \(\epsilon \gt0\) such that:
$$\min_{t,q}e^{-2\eta(T-t)}(v_q(t)-e^{-(\alpha*Q^2-\eta)(T-t)})+\epsilon (T-t) \lt 0$$
But this minimum is achieved at some point \(t^*,q^*\) with \(t^* \lt T\) and hence:
$$ \frac{d}{dt}e^{-2 \eta (T-t)}(\nu_q^*(t)-e^{-(\alpha Q^2- \eta)(T-t)})|_{t=t^*} \geq \epsilon$$
"
Note that Q takes discrete integer value above.
I don't understand in the last equation why the derivative is greater than a positive number. Instead, I think the minimum should be achieved when the derivative is zero. 
 
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Alan
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Joined: December 19th, 2001, 4:01 am
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Re: Why the minimum is taken here when its derivative > 0?

May 10th, 2019, 2:38 pm

I think it's ok:

[$]\min_t  \left\{f(t) + \epsilon \times (T-t) \right\} < 0[$], for some [$]\epsilon>0[$] [$]\Rightarrow[$] (with smoothness)

[$]f'(t^*) = \epsilon  \Rightarrow f'(t^*) > 0 \Rightarrow f'(t^*) \ge \delta[$] for some [$]\delta > 0[$].

Now rename [$]\delta \rightarrow \epsilon[$]. 
 
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Edophokles
Posts: 6
Joined: June 27th, 2017, 11:11 am

Re: Why the minimum is taken here when its derivative > 0?

May 11th, 2019, 5:29 am

Hi EdisonCruise, 

what is the meaning of [$]\nu^{*}[$] in the paper?
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