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EdisonCruise
Topic Author
Posts: 97
Joined: September 15th, 2012, 4:22 am

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

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.

Alan
Posts: 9783
Joined: December 19th, 2001, 4:01 am
Location: California
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Re: Why the minimum is taken here when its derivative > 0?

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 $\delta \rightarrow \epsilon$.

Edophokles
Posts: 6
Joined: June 27th, 2017, 11:11 am

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

Hi EdisonCruise,

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