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Let $S(x)=c(x)+Q(x)K := \{c(x) + Q(x)z \mid z \in K \}$ where $K$ is a nonempty closed and convex subset and $c,Q$ continuous. Moreover, let $\phi(x,y) = \langle F(x), y-x \rangle + \frac{\alpha}{2} ||x-y||^2 $ with $F$ continuous.

Consider the following reformulation of an optimization problem \begin{align} \min_y \varphi(x,y) ~~ s.t. ~~ y\in S(x) \tag{1}\\ &\iff \min_y \varphi(x,y) ~~ s.t. ~~y \in c(x)+Q(x)K \\ &\iff \min_y \varphi(x,y) ~~s.t.~~ \exists z \in K: y=c(x)+Q(x)z \\ &\iff \min_{y,z} \varphi(x,y) ~~s.t. ~~ y=c(x)+Q(x)z, z \in K\\ &\iff \min_z \psi(x,z) ~~s.t. ~~ z \in K \tag{2} \end{align} where \begin{align} \psi(x,z) :&= \varphi(x,c(x) + Q(x)y) \notag \\ &= F(x)^T (c(x) - x) + \frac{\alpha}{2} z^T Q(x)^T Q(x)z + \frac{\alpha}{2} \|c(x) - x\|^2 \notag \\ &\quad + \big(F(x) + \alpha (c(x) - x)\big)^T Q(x)z \end{align}

If I assume that $S(x)$ is closed and convex and $\varphi$ is strongly convex, then I can conclude that the minimizer of the initial problem is unique. My question now is: Is the minimizer of problem (2) unique as well? I know that $\psi$ is strongly convex if and only if the $Q(x)$ has full rank. However, this is not necessary for problem (1) to have a unique solution. How do the problems connect?

sunflower234
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1 Answers1

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To answer the first question, problem (2) is not guaranteed to have a unique optimum for singular matrices $Q(\boldsymbol{x})$, as you have correctly observed yourself by the strong convexity argument. Contrary to the assumption you put forward, however, this is also necessary as long as you want to have the unique minimizer in (1). The reason for that is that you cannot guarantee the closedness of $S(\boldsymbol{x})$ in the case when $Q$ can be singular.

For a counterexample, let $S(\boldsymbol{x}) \subseteq \mathbb{R}^2$ and consider the linear transform from the answer linked above, that is, $$K=\{ (x_1, x_2) \mid x_2 \ge e^{x_1} \}, Q \equiv \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix};$$ last, assume for simplicity that $c(\boldsymbol{x}) \equiv \boldsymbol{0}$ and $F(\boldsymbol{x}) \equiv \boldsymbol{0}$. Observe that the feasible set, which also happens to be the same as $QK$, is then equal to $\{ 0 \} \times (0, +\infty)$, or, in other words, $$\boldsymbol{y} \in S(\boldsymbol{x}) \Leftrightarrow y_1 = 0, y_2 > 0.$$

Given that, you can now see that neither (1) nor (2) have unique, and, in fact, any minimizers as long as $x_2 < 0$. This is trivial to check for (2), as after substituting the parameters, simplifying the $Qz$ terms, and discarding constant terms, you get the problem $$\frac{\alpha}{2}q^2 - \alpha x_2 q \to \min, q > 0$$ with $q = (Qz)_2$. As for problem (1), doing the same there transforms it to $$\frac{\alpha}{2}\left( x_2 - y_2 \right)^2 \to \min, y_2 >0,$$ which is essentially the same problem but in a form that conveys the lack of minimizer more clearly:)

Konstantin Sidorov
  • 691