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I have the following normal cone inclusion

$$-(A x + b) \in \mathcal{N}_\mathcal{C}(x) \qquad (1)$$

where $\mathcal{N}_\mathcal{C}$ denotes the normal cone to the convex set $\mathcal{C}$ at the point $x \in \mathcal{C}$. Matrix $A$ is non-symmetric.

Normally if $A$ would be symmetric, the convex optimization problem is

$$\min \frac{1}{2} x^\top A x + b^\top x + I_{\mathcal{C}}(x) \qquad (2)$$ where $I_{\mathcal{C}}(x)$ is the indicator function of the set $\mathcal{C}$. The inclusion (1) is the necessary and sufficient optimality condition of the convex optimization problem (2).

However, what are the implications of $A$ beeing non-symmetric and how does that relate to an optimization function? I guess there does not exist an optimization function?

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  • $\begingroup$ It's not clear to me what the connection is here. In what way, precisely, is the convex optimization problem you have cited related to the normal cone inclusion above? There is context here you haven't shared with us. $\endgroup$ – Michael Grant Dec 1 '15 at 14:40
  • $\begingroup$ OK, I think I understand. If $x$ is on the boundary of $C$, then that first equation is the optimality condition that must be satisfied if that point is an optimum. If my understanding is correct, you are putting the cart before the horse. The optimization problem determines the inclusion, not the other way around. $\endgroup$ – Michael Grant Dec 1 '15 at 15:41
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    $\begingroup$ Not always, in mechanics, when modeling set-valued force laws: we have the inclusion and we want to know if there exists an optimization problem (hopefully convex) you are right, the optimality condition is the normal cone inclusion of the above mentioned convex optimization problem. Optimal point does not need to be on the boundary (of course its best visualized if it does so) but it can also be inside $\mathcal{C}$ then the normal cone is the Zero vector. $\endgroup$ – Gabriel Jan 29 '16 at 20:59

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