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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 $$... 2 The Pruefer group {\mathbb Z}_{p^{\infty}} is the direct limit of the sequence$${\mathbb Z}/p{\mathbb Z}\hookrightarrow {\mathbb Z}/p^2{\mathbb Z}\hookrightarrow ...$$Applying (-)^{\wedge} = \text{Hom}_{\text{cnt}}(-,{\mathbb S}^1) shows that \left({\mathbb Z}_{p^\infty}\right)^{\wedge} is the inverse limit of the Pontryagin duals of {\mathbb Z}/p^k ... 0 There is a nice duality between C(K) and L_1(\mu)-spaces. The double dual of C(K) is of the form C(L) for some huge compact space L. (Actually it is also isometric to L_\infty(\nu) for some huge measure \nu.) The second dual of L_1(\mu) is also of the form L_1(\nu). However people rarely think of duals/biduals of these spaces like that. ... 0 Actually, the reverse statement is (more or less) true. I assume that by X \subset Y, you mean that the inclusion$$ \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 ...

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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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Nice question! In fact there are no other examples. Let $G$ be a connected locally compact abelian group whose Pontryagin dual $G^{\vee}$ is also connected. Because $G^{\vee}$ is connected, it can have no discrete quotients; taking Pontryagin duals, $G$ can have no compact subgroups. By the Gleason-Yamabe theorem, it follows that $G$ has an open subgroup ...

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It seems that there is a little bit of terminological confusion on this issue. See Steven Givant & Paul Halmos, Introduction to Boolean algebras (2009), Ch.4 : The Principle of Duality, page 4 : Every Boolean polynomial has a dual: it is defined to be the polynomial that results from interchanging $0$ and $1$, and at the same time interchanging ∧ ...

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Hint: If $E=\{0\}$, then it is obvious by computation that $E^*=\{0\}$. If $E\ne \{0\}$, then you can construct at least one non-trivial linear functional on it.

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There are two possible answers to your question. First I will assume that you don't distinguish between different sizes of infinity. Let $L \colon V \to W$ of infinite rank be given, and let $n$ be a natural number. It is possible to find a finite-dimensional subspace $V_1$ of $V$ and a finite dimensional quotient $W_1$ of $W$ such that the composite ...

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I asked a (near-) duplicate of this question on mathoverflow, where it was answered; indeed, it turns out that the triangle is optimal. The proof of this is a simple and very pretty application of the Poisson summation formula.

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I had to prove this for an assignment the other day and I came up with a fairly clean proof that I thought I should share: We define a' and b' as follows: $$a'\triangleq\underset{a\in A}{argmin}\left\{ \underset{b\in B}{\max}f(a,b)\right\} \\ b'\triangleq\underset{b\in B}{argmax}\left\{ \underset{a\in A}{\min}f(a,b)\right\}$$ a' is the point in A where ...

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Your first statement is too vague. Let's make it more specific: in category theory, we can dualize categorical statements by reversing the directions of all of the arrows. This always produces a second true statement, as long as you're careful to take the dual correctly. Here's an example of taking the dual incorrectly: it's true that finite limits commute ...

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