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Hint. First construct a sequence of compact sets $K_n\subset\subset K_{n+1}$, such that $$\lim_{n\to\infty}\mu(K_n)=\infty.$$ Define $f_n\ge 0$ continuous to but equal to $1$ in $K_n$ and $0$ in $K_{n+1}^c$. Then look for $f=\sum a_nf_n$, for suitable $a_n>0$, $\sum a_n<\infty$, so that $\int_X f\,d\mu=\infty$ and $f\in C_0$.

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Yes, the primal optimal value is $\inf_x \sup_{\lambda \geq 0} L(x,\lambda)$ and the dual optimal value is $\sup_{\lambda \geq 0} \inf_x L(x,\lambda)$. A primal and dual optimal pair of variables gives you a saddle point of the Lagrangian. This is discussed on p. 238 (section 5.4.1) of Boyd and Vandenberghe, and is also discussed in other convex ...

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The result is given by diagram chasing. The map $d_*$ is the connecting homomorphism, so by chasing it we shall do the following: 1.Find inverse image in $\Omega^*_c(U) \oplus \Omega^*_c(V)$. 2.differentiate it. 3.Find the inverse image in $\Omega^*_c(U \cap V)$ And you can check proof of Prop 2.7 on page 26 for the explicit maps in each step. I should ...

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Let $C=R[t]$ with the usual coalgebra structure $(\varepsilon(t)=1$, $\Delta(t)=t \otimes t$). Then a $C$-comodule is the same as an $\mathbb{N}$-graded $R$-module. In fact, given a coaction $\alpha : M \to M[t]$, let $M_n = \{m \in M : \alpha(m)=m \cdot t^n\}$ and show $M = \oplus_n M_n$. Thus, in this case your question is: Does every graded $R$-module $M$ ...

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In view of the comment, I imagine that $C$ and $D$ are subsets of some Euclidean space $E$, that $C^*=\{x\in E:(\forall y\in C)\,x\cdot y\geq0\}$, and similarly for $D^*$. If so, and if $C\subseteq D$, then any $x\in D^*$ satisfies $x\cdot y\geq 0$ for all $y\in D$, and that includes all $y\in C$, so $x\in C^*$. Since $x$ was an arbitrary element of $D^*$, ...

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You can't map a torsion module into a torsion-free module, so it's isomorphic to $0$. For any map $\phi:\mathbb Z/ n\mathbb Z\to\mathbb Z$, $$\phi(1)=m\implies \phi(0)=n\cdot \phi(1)=nm=0\quad\forall n\in\mathbb Z\implies m=0.$$ For the set $\operatorname{Hom}_\mathbb Z(\mathbb Z,\mathbb Z/n\mathbb Z)$, maps are completely determined by where $1$ is sent. ...

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Duality: two objects, each property of the first object is linked to a property of the second object Symmetry: same object, one or several other points of view, properties are exactly the same Equivalency: different objects, properties are exactly the same Invariance: same object but transformed (and we say the object is some transformation-invariant), ...

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Apparently, $X$ is a dense subspace of $\bar X$, in order to define its dual. Clearly, if $\ell\in X^*$, then $\ell$ extends uniquely, by a standard density argument to an $\bar\ell\in \bar X^*$, and clearly $\|\bar\ell\|=\|\ell\|$. Inversely, if $\bar\ell\in\bar X^*$, then its restriction to $X$ is a bounded linear functional on $X$.

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