# proof of Slater's condition and strong duality when $\mathrm{relint}\:\mathcal D\ne\mathrm{int}\:\mathcal D$

Background. Consider a convex program $(P)$, \begin{aligned} \textrm{minimize } \quad & f_0(x) \\ \textrm{subject to } \quad &f_i(x) \le 0, \quad i=1,2,...,m \\ & Ax=b, \end{aligned}

and its dual program $(Q)$,

\begin{aligned} \textrm{maximize} \quad & g(\lambda, \nu)\\ \textrm{subject to} \quad & \lambda \succeq 0 \end{aligned}

where $g(\lambda, \nu)=\underset{x\in \mathcal D}{\inf} \left(f_0(x)+\sum_{i=1}^m \lambda_i f_i(x)+\nu^T (Ax-b)\right)$, and $\mathcal D\triangleq \left(\underset{i}{\cap} \: \mathrm{dom}\:f_i\right) \subset \mathbb{R}^n$. The Slater's condition implies strong duality, i.e. $p^*=d^*$, where $p^*$ and $d^*$ are the optimal value of $(P)$ and $(Q)$, respectively. (The Slater's condition is: There exists an $x\in \mathrm{relint}\:\mathcal D$ such that $Ax=b$ and $f_i(x)<0, \:i=1,2,...,m$.)

Boyd & Vandenberghe's book "Convex Optimization" proves the strong duality for a simplified case, i.e. when $\mathrm{relint}\:\mathcal D=\mathrm{int}\:\mathcal D$. Let's call this BV's Lemma.

Question. However, I had some difficulty extending it to the general case, when $\mathrm{relint}\:\mathcal D\ne\mathrm{int}\:\mathcal D$, which I suppose can happen only when the affine dimension of $\mathcal D$ is less than $n$. Does someone know how to prove this, or where I can find a proof? I'll also outline my attempt below, and would appreciate it if someone can point out flaws, if any, in the arguments or comment on/confirm my proof:

My attempted proof. If the affine dimension of $\mathcal D$ is less than $n$, then the affine hull of $\mathcal D$ can be expressed by $\{x\in \mathbb R^n:Bx=c\}$. So I suppose we can add $Bx=c$ to the equality constraints of the convex program, enlarge $\mathcal D$ so that its affine dimension becomes $n$, and obtain an equivalent convex program $(\hat P)$, but now with $\mathrm{relint}\:\mathcal{\hat D}=\mathrm{int}\:\mathcal{\hat D}$, where $\mathcal{\hat D}$ is the enlarged domain of $(\hat P)$. Then by BV's Lemma, strong duality holds for $(\hat P)$. As a result, letting $\hat p^*$ and $\hat d^*$ be the optimal value of $(\hat P)$ and its dual program, respectively, we have $p^*=\hat p^*=\hat d^*$.

Clearly, $p^*\ge d^*$ due to weak duality, and hence $\hat d^*\ge d^*$. But it appears that we also have $d^* \ge \hat d^*$. To see this, note that the dual function of $(\hat P)$ is:

$$\hat g(\lambda, \nu_A, \nu_B)=\underset{x\in \mathcal{\hat D}}{\inf} \left(f_0(x)+\sum_{i=1}^m \lambda_i f_i(x)+\nu_A^T (Ax-b)+\nu_B^T(Bx-c)\right),$$

Therefore, $g(\lambda, \nu) \ge \hat g(\lambda, \nu, \nu_B)$, since $\mathcal D \subset \mathcal{\hat D}$ and $Bx=c$ for any $x\in \mathcal D$. As a result,

$$d^*=\underset{\lambda \succeq 0}{\sup} g(\lambda, \nu) \ge \underset{\lambda \succeq 0}{\sup} \hat g(\lambda, \nu_A, \nu_B)=\hat d^*,$$

and hence $p^*=d^*$. QED.

Am I mistaken somewhere?

• See Theorem 28.2 in Convex Analysis by Rockafellar. Your approach is flawed since you obtain a different dual when you add $Bx=c$. – LinAlg Dec 1 '16 at 16:16
• @LinAlg Thanks a lot for the reference and comments. Yes, the dual program is different, but it appears that its optimal value (of the new dual) is no greater than the original? So if the strong duality holds for the new program, it must hold for the original program? – syeh_106 Dec 1 '16 at 16:30
• The optimal value if the modified primal is exactly the same as that of the original primal indeed. There are many optimization problems with the same objective value as the original primal. Strong duality however says something about a primal-dual pair. So you must look at the dual of the modified primal. If that dual is equivalent to the dual of the original primal your proof is finished. Otherwise, you haven't proven anything. – LinAlg Dec 1 '16 at 16:32
• @LinAlg Thanks again for the feedback, but I guess I still missed something... Let $p^*, \hat p^*$ be the optimal values of the original and modified programs, respectively, and let $d^*, \hat d^*$ be the optimal values of the original and modified dual programs, respectively. Then $\hat p^*= p^*\ge d^*\ge \hat d^*$. If $\hat p^*=\hat d^*$, wouldn't this imply $p^*=d^*$? – syeh_106 Dec 1 '16 at 16:43
• Why $d^* \geq \hat{d}^*$? – LinAlg Dec 1 '16 at 18:11