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

Posted:

**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.

"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.