# Tag Info

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 ${\mathbb Z}/p^k ... 1 There is a nice duality between$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)$. However people rarely think of duals/biduals of these spaces like that. ... 0 Actually, the reverse statement is (more or less) true. I assume that by$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} ...

1

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.

Top 50 recent answers are included