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2

If $X$ is a Banach space then a subspace is norm-closed if and only if it is weakly closed. (This is a major difference between weak and weak*). Say $S$ is a weakly dense countable set. Let $V$ be the norm-closed span of $S$. Then $V$ is weakly closed, hence $V=X$. So the linear combinations of elements of $S$ with rational coefficients are dense in $X$, ...

2

I think that does not exist. Indeed, if $X$ is a normed space, then $X$ is weak separable if and only if $X$ is norm separable. Proof. "$\Leftarrow$": Let $X$ be norm separable. Let $A$ be countable and norm dense in $X$. Then, we use the fact that the weak topology is indeed weaker than the norm topology to see that $\overline{A}^w\supset ... 0 This is a well-known result from the theory of locally compact abelian groups. But I think that its proof should be complicated. For instance, in [Pon] this is a corollary from$\S$34.C, with a long full proof, which is based on irreducible matrix representations of compacts abelian groups, which are bases on integration with respect to invariant measures on ... 0 Salam brother Mahmud, The procedure is extremely similar to sec 2.10-7 in the Erwin's book (almost copy and paste with slight modifications for the new space$c_0$) and goes as follows. A Schauder basis (Sec. 2.3 in the Erwin book) for$c_0$is$(e_k)$where$e_k=(\delta_{kj})$has 1 in the kth place and zeros otherwise;$\begin{matrix} ...

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The statement (1) is the observation that $\delta(\cdot|K)$ is the convex conjugate of $h$ (and vice-versa). So (1) is always true under your assumptions. The function $h$ is usually called the support functional of $K$. The statement (2) states that there is no duality gap between the primal problem $$\min_{x'\in X} \|x-x'\| + \delta(x'|K)$$ and its ...

2

The dual of $C = C[a,b]$ is the Banach space $\mathcal {M}$ of finite Borel measures on $[a,b].$ Now $C$ is separable (polynomials with rational coefficients), but $\mathcal {M}$ is not (consider $\delta_{t}, t \in [a,b]$). There's a theorem somewhere that says that if $Y$ is nonseparable, then so is $Y^*.$ If we use this theorem, we're done. But we can show ...

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Regarding the differentiability of the dual function $g(\lambda)$, here is a sufficient condition (See Lemma 6.3.2, Theorem 6.3.3 and Theorem 6.3.4 of Bazaraa et al. 's Nonlinear Programming: Theory and Algorithms, 2006). Let the domain $X$ be a nonempty compact set in $R^n$, and $f$ and $h$ be continuous. For any $\lambda$, if the minimizer of $L(x, ... 1 What you have done is correct and you have shown$||f||_{c_0'} \le ||\hat f||_{\ell_1}$, where$\hat f$is the sequence given by$\hat f_i = f(e_i)$. On the other hand, let$g_n\in c_0\$ be defined so that (g_n)_i = \begin{cases} 1 & \text{ if } f(e_i) \ge 0, i\le n \\ -1 & \text{ if } f(e_i)<0 , i\le n\\ 0 & \text{ if } i ...

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