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Assume we have this strictly convex quadratic programming:

$$f(x) = x^\top A x + b^\top x,$$ $$Ax \leq b$$ $$ 0 \leq x \leq 1$$

Where $A$ is symmetric and positive definite, and the feasible set is nonempty. Does strong duality and Slater's condition holds in this case.

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    $\begingroup$ you said the feasible set is nonempty, so I do not understand your first question (feasibility means that the optimal value is finite too) $\endgroup$ – LinAlg Jun 30 '18 at 12:25
  • $\begingroup$ you mean the second question, right, since the first one is about the duality? I removed the second question, how about the duality? $\endgroup$ – Faroq AL-Tam Jun 30 '18 at 12:30
  • $\begingroup$ Slater's condition requires a strictly feasible solution. If your linear equalities are such that (e.g.) there is a unique feasible solution then Slater's condition wouldn't be satisfied. Depending on your constraints, other constraint qualifications, such as the linear independence constraint qualification (LICQ), might hold. $\endgroup$ – Brian Borchers Jun 30 '18 at 14:19
  • $\begingroup$ @Brian Borchers All the constraints are linear, that is a constraint qualification. Slater's condition only pertains to nonlinear constraints, so you could say it is trivially satisfied in this case. $\endgroup$ – Mark L. Stone Jun 30 '18 at 15:33
  • $\begingroup$ @MarkL.Stone yes, but I was hoping that the OP could find this out for themselves... $\endgroup$ – Brian Borchers Jun 30 '18 at 15:56
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The feasible set is nonempty and compact. The objective is continuous. So by Weirstrass we have a finite optimal value. The constraints and the objective are all convex functions. Finally, since the feasible set is nonempty, we have a vector which satisfies the linear inequalities. Thus we have Slater's condition (linear inequalities do not require strict feasibility). So we also have strong duality.

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  • $\begingroup$ Thank you, Does this proof hold if we also have additional equality constraints $Cx = 1$. $\endgroup$ – Faroq AL-Tam Jun 30 '18 at 14:45
  • $\begingroup$ Yes. Equality constraints do not change the compactness of the feasible set nor the fact that Slater's condition holds. $\endgroup$ – Alex Shtof Jun 30 '18 at 21:30

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