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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 $$ ...


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. ...


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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} ...


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.


1

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 ...


0

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 ∧ ...


1

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.


0

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 ...


3

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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