The optimality condition should have been \(\geq\) instead of \(>\) (according to Lemma 1.27 of the lecture notes). Looks like a typo. Proving equality also proves \(\geq\).

The optimality condition should have been \(\geq\) instead of \(>\) (according to Lemma 1.27 of the lecture notes). Looks like a typo. Proving equality also proves \(\geq\).

## Question exam 2019

Hello,

I have troubles understanding the logic of the correction of question 23 in exam 2019: we have that

\( y^* is an optimum \Leftrightarrow \nabla g(y^*)^T (y - y^*) >0 \)

So we verify if the RHS is verified for the proposed \( y^* \) in order to conclude the LHS: what I don't understand is that we check that

\( \nabla g(y^*)^T (y - y^*) \textbf{=}0 \) which is not equivalent to \( \nabla g(y^*)^T (y - y^*) >0 \)

Thanks in advance for any help!

Best,

Yann

## 1

The optimality condition should have been \(\geq\) instead of \(>\) (according to Lemma 1.27 of the lecture notes). Looks like a typo. Proving equality also proves \(\geq\).

I was also very confused by this. It seems to me that they are doing some sort of proof by induction, but it doesn't convince me.

The optimality condition should have been \(\geq\) instead of \(>\) (according to Lemma 1.27 of the lecture notes). Looks like a typo. Proving equality also proves \(\geq\).

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