I'm reading KKT's condition in preparation for the final exam. Assume I have the following problem:
$$\begin{align*} \text{min} & \quad f(x) \\ \text{s.t} & \quad g_i(x) \le 0 \quad \forall i = \overline{1,l} \\ & \quad h_j(x) = 0 \quad \forall j = \overline{1,m} \end{align*}$$
where $f,g_i,h_j \in \mathcal C^1(\mathbb R^n,\mathbb R)$ for all $i$ and $j$.
One of the constraint qualifications mentioned in Wikipedia is Mangasarian-Fromovitz constraint qualification, which is
The gradients of the equality constraints are linearly independent at $x^{*}$ and there exists a vector $d \in \mathbb{R}^{n}$ such that $\langle \nabla g_{i}\left(x^{*}\right), d \rangle<0$ for all active inequality constraints and $\langle \nabla h_{j}\left(x^{*}\right), d\rangle=0$ for all equality constraints.
Assume that our problem does not contain equality constraints, i.e.
$$\begin{align*} \text{min} & \quad f(x) \\ \text{s.t} & \quad g_i(x) \le 0 \quad \forall i = \overline{1,l} \\ \end{align*}$$
I have two ways of interpreting this case which leads to opposite conclusions:
My professor said that in this case, $h_j \equiv 0$. It follows from $h_j \equiv 0$ that $\nabla h_j (x) \equiv 0$. The gradients of the equality constraints are consequently NOT linearly independent at $x^{*}$ because $\sum \lambda_j \nabla h_j (x) = 0$ for any sequence of $(\lambda_j)$. As such, the Mangasarian-Fromovitz constraint qualification is no way satisfied.
Because there is no equality constraints, we only check if there exists a vector $d \in \mathbb{R}^{n}$ such that $\langle \nabla g_{i}\left(x^{*}\right), d \rangle<0$ for all active inequality constraints.
Could you please explain which one is correct? Thank you so much for your help!
The second one is correct. It would be very weird to force an identical zero equality if there is no equality constraint (it would break indeed the CQ condition). In the case the set of equality gradients is empty then all equality gradients are linearly independent and satisfy whatever condition you like.