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1

The problem in (1) is convex (since $\mathbf{A}$ is positive definite). It is also strictly feasible, since any x such that $\|x\| < 1$ is a (strictly) feasible solution. By Slater's condition strong duality holds and the optimal value of (1) will be equal to the optimum of its dual. The Lagrangian of (1) is $$ L(x, \lambda) = x^{T}\mathbf{A}x + ...


1

Since $\mathbb{Z}(p^\infty)$ is torsion, the image of a character $\mathbb{Z}(p^\infty)\to\mathbb{T}=\mathbb{R}/\mathbb{Z}$ is contained in the torsion part of $\mathbb{T}$, which is $\mathbb{Q}/\mathbb{Z}$. Since $\mathbb{Z}_{p^\infty}$ is a $p$-group, the image is further contained in the $p$-component, which is exactly $\mathbb{Z}_{p^\infty}$. So $$ ...


3

The Pruefer group ${\mathbb Z}_{p^{\infty}}$ is the direct limit of the sequence $${\mathbb Z}/p{\mathbb Z}\hookrightarrow {\mathbb Z}/p^2{\mathbb Z}\hookrightarrow ...$$ Applying $(-)^{\wedge} = \text{Hom}_{\text{cnt}}(-,{\mathbb S}^1)$ shows that $\left({\mathbb Z}_{p^\infty}\right)^{\wedge}$ is the inverse limit of the Pontryagin duals of ${\mathbb Z}/p^k ...


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There is a nice duality between $C(K)$ and $L_1(\mu)$-spaces. The double dual of $C(K)$ is of the form $C(L)$ for some huge compact space $L$. (Actually it is also isometric to $L_\infty(\nu)$ for some huge measure $\nu$.) The second dual of $L_1(\mu)$ is also of the form $L_1(\nu)$. However people rarely think of duals/biduals of these spaces like that. ...


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Actually, the reverse statement is (more or less) true. I assume that by $X \subset Y$, you mean that the inclusion $$ \iota : X \to Y, x \mapsto x $$ is a well-defined, bounded linear map. Then, for each $\varphi \in Y^\ast$, we get that $\varphi|_X = \varphi \circ \iota \in X^{\ast}$. If also $X \subset Y$ is dense, then the "inclusion" map $$ \Gamma ...


3

Let $\lambda_{\min}$ be the minimum eigenvalue of $Q$. The dual function is \begin{align} g(u) &= \inf_x L(x,u) \\ &= \begin{cases} -\infty \quad \text{if } u < -\lambda_{\min} , \\ -\frac12 u \quad \text{otherwise}. \end{cases} \end{align} The dual problem is \begin{align} \operatorname*{maximize}_{u} & \quad -\frac12 u \\ \text{subject to} ...


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Hint: Your problem is partially addressed in this question Note that this is one the rare non-convex problems that has a closed form solution and has zero duality gap.



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