[Math] Valid KKT Constraint Qualification? Linear constraints not full rank, Jacobian of nonlinear constraints full rank and independ. of linear constraints

nonlinear optimization

For a nonlinear optimization problem having only linear constraints, by the Linearity Constraint Qualification, no further constraint qualification is required for the Karush-Kuhn-Tucker (KKT) conditions to hold. https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions . In particular, the Jacobian of active constraints need not be full rank.

The Linear Independence Constraint Qualification (LICQ) for KKT is stated as: the Jacobian of active constraints is full rank, i.e., the gradients of the active constraints are linearly independent.

Suppose the following:

1) Jacobian of active nonlinear constraints is full rank

2) Jacobian of active nonlinear constraints is independent of Jacobian of active linear constraints, i.e., rank(Jacobian of all active constraints) = rank(Jacobian of active nonlinear constraints) + rank(Jacobian of active linear constraints)

Is the combination of (1) and (2) a valid constraint qualification for KKT? Note that this would not require that the Jacobian of active linear constraints be full rank. If this is valid, it seems strange that I have never seen LICQ stated this way, because my candidate constraint qualification (of as yet unknown correctness) has a weaker hypothesis, and therefore would be a more powerful and widely applicable constraint qualification, than how I have seen it stated.

I am trying to determine whether I can "get away with" having some redundant (active) linear constraints on a non-pre-solved optimization model when I also have active nonlinear constraint(s). I can get away with it when there are no nonlinear constraints.

I have not found a counterexample, but will hazard a guess that if there is a counterexample, there exists a counterexample having one nonlinear constraint and two linear constraints.

This is not a homework or test problem, but maybe it should be: I.e., prove it true or false.

Thanks.

Bumping this since no one has answered. This seems like it shouldn't be that hard to either prove true or produce a counterexample. I'm more a user of Constraint Qualifications than I ever was a prover of them. But this should be a piece of cake for you math/geometry whizzes.

Best Answer

This is true. It's a particular case of the Constant Rank Constraint Qualification from Janin.

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