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In writing the optimization problem in epigraph form, $max$ is eliminated. In this way, it is assumed that $\alpha$ is fixed. But we must be careful since in this optimization problem we have the constraint $\forall \alpha: 0\leq \alpha \leq 1$, i.e. the optimization problem , \begin{align} \min_{0\leq \eta \leq e,M\succeq \eta\eta^T}\max_{0\leq \alpha \leq ...

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Note that $\alpha=0$ is feasible and gives $-\alpha^T (R\odot M)\alpha+\alpha^Tq=0$ $\quad\Rightarrow\quad$ $f(q,M)\ge 0$ $\quad\Rightarrow\quad$ $g(q,M)=f(q,M)+\frac12\|q\|^2\ge 0$ $\quad\Rightarrow\quad$ $\min\, g(q,M)\ge 0$. Now take $q=0$ $\quad\Rightarrow\quad$ $f(0,M)=0$ $\quad\Rightarrow\quad$ $g(0,M)=0$. Conclusion: the minimum is zero, attained at ...

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The core of a game is not self-dual, to see this, consider an average-convex game quantified by (lexi-order): v=[0,0,0,0,4,9,8,2,4,10,13,13,39/2,14,149/6] which gives another counter-example. The core of the game is given by the following extreme points: 65/6 0 13/6 71/6 9 0 71/6 4 ...

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Yes the proof is correct. The crucial point is to show that the matrix associate to $\text{}^tT$ is $\text{}^tA$ where $A$ is the matrix associate to $T$.

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For the sake of not leaving this question unanswered. You may take the following approach: Let $H_i = {\rm span\,}\{f_1,\dotsc,\widehat{f_i},\dotsc,f_n\}$ (where $\widehat{f_i}$ means omitting $f_i$). Then clearly $\dim H_i = n-1$. Now, consider $V_i:= \{x\in V\,|\, f(x) = 0 \text{ for all$f\in H_i$}\}$. Lemma: Let $V$ be a finite dimensional vector space ...

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This is basically an application of the Hyperplane separation theorem. Take a look at the proof of Gordan's theorem here, where the hyperplane theorem is used.

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Here's one way to look at it. The first condition can be written as $A^T y > 0$. Gordan's theorem says that either the range of $A^T$ intersects the positive orthant, or the null space of $A$ intersects the nonnegative orthant (at a point other than the origin). Because the null space of $A$ and the range of $A^T$ are orthogonal complements of ...

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If $X$ is a Banach space then a subspace is norm-closed if and only if it is weakly closed. (This is a major difference between weak and weak*). Say $S$ is a weakly dense countable set. Let $V$ be the norm-closed span of $S$. Then $V$ is weakly closed, hence $V=X$. So the linear combinations of elements of $S$ with rational coefficients are dense in $X$, ...

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I think that does not exist. Indeed, if $X$ is a normed space, then $X$ is weak separable if and only if $X$ is norm separable. Proof. "$\Leftarrow$": Let $X$ be norm separable. Let $A$ be countable and norm dense in $X$. Then, we use the fact that the weak topology is indeed weaker than the norm topology to see that \$\overline{A}^w\supset ...

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