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4

This is an overkill I guess. Let $\mu$ be a Radon measure on $X$, then $\mu(K) <\infty$. Also $\| f_n\|_{L^1}$ is uniformly bounded by $\mu(K)$ as $|f_n|\le 1$. By Lebesgue's dominated convergence theorem, $$\tag{1} \int_K f_n d\mu \to \int_X f d\mu.$$ By Riesz' representation theorem, all bounded linear functional on $C(K)$ is given by Radon measure, ...

3

As suggested in the comment, we show that the graph of $T$ is closed. Assume that $(x_n , T(x_n) ) \to (x, y)$. Then $x_n - x \to 0$. From the assumption, for all $f\in \mathcal F$, $$f(T(x_n -x)) \to 0 \Rightarrow f(T(x_n)) \to f(T(x)).$$ On the other hand, for all $f\in Y^*$ we have $$T(x_n) \to y \Rightarrow f(T(x_n)) \to f(y).$$ That is, $$f(T(x)) = ... 3 Not in general: Take V = \ell^1, and for each i\in \mathbb{N}, let \alpha_i denote the "evaluation" map$$ \alpha_i ((x_n)) := x_i $$Then clearly$$ \bigcap_i \ker(\alpha_i) = 0 $$However, consider the dual space pairing$$ \ell^{\infty} \to (\ell^1)^{\ast} $$then the \alpha_i correspond to the elements$$ e_i = (0,0,0, \ldots, 0, 1,0,\ldots) \in ...

3

That's a good proof. Here is another one. Assume that $a^m=0$. If you know that $$\sigma(a)=\{f(a):\ f \text{ is a multiplicative functional }\},$$ then for any such $f$ we have $f(a)^m=f(a^m)=f(0)=0$. So $f(a)=0$, and then $\sigma(a)=\{0\}$. Yet another proof, without machinery: if $a^m=0$, then $1-a$ is invertible: indeed, the inverse is ...

2

Not in general. Here is a counterexample. Let $X = C([-1,1])$. For $t \in [-1,1]$, let $\delta_t$ denote the point mass / evaluation functional $\delta_t(x) = x(t)$. Let $D = \{ \delta_q : q \in [-1,1] \cap \mathbb{Q}\}$. Then $D$ is countable and we have $\|x\| = \sup_{f \in D} |f(x)|$ for every $x \in X$. Let $$y(t) = \begin{cases} 4t, & -1 \le t ... 2 This is not in fact true in the case that you are interested in. For instance, consider the functional I:C([0,1])\to\mathbb{C} given by I(f)=\int f d\mu, where \mu is Lebesgue measure. It is easy to see that any finite linear combination of evaluation functionals has distance \geq 1 from I (because you can find an element of C([0,1]) of norm ... 2 Suppose that K\subset \ell_\infty is weakly compact. Then, by Mazur's lemma, so is the closed convex hull of K\cup \{0\}, call it B_K. Consider \ell_\infty as the dual space to \ell_1. Since the weak* topology on \ell_\infty is coarser than the weak topology, by compactness, they must be equal on B_K. Since B_K is weakly compact, X_K is ... 2 As explained by anonymous, this can be achieved by a cut-off argument. However, it can be proved in a more general way. Let X be a Banach space, J : X \to X^{**} the canonical embedding into the bidual. For a Banach space Y, we denote by B_Y the unit ball of Y. Now, we will show that J(B_X) is weak-* dense in B_{X^{**}}. We will use the ... 2 We will show that \sigma(A) = \{0\}, and that 0 belongs to the residual spectrum. As you have shown, 0 is not an eigenvalue, and as \operatorname{im} A \subseteq \{x \in C[0,1]: x(0) = 0\}, A does not have dense image. Hence 0 \in \sigma_r(A). To see that A - \lambda is invertible for \lambda \ne 0, let y \in C[0,1] be given. We have to ... 2 In the category \mathbf{Ban}_1 of Banach spaces with contractive linear operators we do have products (\bigoplus_\infty-sums) and coproducts (\bigoplus_1-sums). Even more we have an isomorphism$$ \left(\bigoplus_1 X_\alpha\right)^*\underset{\mathbf{Ban}_1}{\cong}\bigoplus_\infty X^*_\alpha $$Unfortunately, it is not true, that$$ ...

1

If $\|T^n\| < 1$, then $I-T^n$ is invertible with inverse $\sum_{k=0}^{\infty}(T^n)^k$. Then \begin{align} I-T^n & = (I-T)(I+T+T^2+\cdots+T^{n-1}) \\ & = (I+T+T^2+\cdots+T^{n-1})(I-T). \end{align} It follows that $I-T$ is invertible because it must be injective and surjective by the above. And that's what you need.

1

This is not true. For a counterexample, take $X = \ell^1$, $X^* = \ell^\infty$. Define $\Lambda_n \in X^{**}$ by $$\Lambda_n(x) = x_n$$ for $x \in \ell^\infty$. Then, it is easy to see that this sequence $\{\lambda_n\}$ satisfies your assumptions. Moreover, $$A = \{x \in \ell^\infty: \lim_{n \to \infty} x_n \text{ exists}\}.$$ This set is not weak-$*$ ...

1

I assume by $C:= conv^{\ast}(e_i)$ you mean the closed convex hull of the $\{e_i\}$. Let $\varphi \in (\ell^1)^{\ast}$ denote the linear functional $$\varphi((x_n)) := \sum_{n=1}^{\infty} x_n$$ Then for any $z \in \text{conv}(\{e_i\})$, we have $$\varphi(z) = 1$$ Since $C$ is norm closed, it follows that $\varphi \equiv 1$ on $C$. In particular, $0\notin ... 1 If$X=Y$in your question, we have \begin{equation*} S+T: X \rightarrow X. \end{equation*}$S+T$is a injective map, because$(S+T)(x) = 0 \Rightarrow x= 0 $. Now, we want to show that for all$y \in X$, exists$x \in X$such that$(S+T)(x)=y$. See that, \begin{equation*} (S+T)(x) = y \Leftrightarrow S(x) +T(x) =y \Leftrightarrow T^{-1}(y) - T^{-1}S(x) = x ... 1 I have two comments. I'm not sure that$\Phi$is surjective. However, since you don't want to prove that$X$is isometrically isomorphic to$\ell_\infty$(but to a subspace of$\ell_\infty$), you can replace "linear bijection" by "linear injection". Every separable set has a countable norming set (Lemma 6.7 here). And any normed space with a countable ... 1 Given$\mu\in \mathcal{M}(K)$, note that for each finite set$J\subset I$we have $$\sum_{j\in J}\|f_j\|\leqslant \|v\|.$$ Thus $$\sum_{i\in I}\|f_i\|\leqslant \|v\|.$$ Set $$\nu_0 = \sum_{i\in I}f_i {\rm d}\mu_i.$$ Note that$\nu-\nu_0$is singular with respect to all$\mu_i$so by maximality, it must be the zero measure. Consequently,$$\nu = ... 1 This is not an answer to your question, but merely an attempt to encourage you to formulate the question in a way that will encourage more attention and answers. You write: By definition of a dual space$Lip_0(X)^*$, every element is an evaluation function$\mu : Lip_0(X) \rightarrow \mathbb{F}$. This is not true. The norm-closure in$Lip_0(X)^{*}$... 1 I will elaborate a little bit on the hint by zhw. You can use the uniform boundedness principle to infer the boundedness of$\{f_n\}$in$X^*$(here, it is important that$X$is complete). Then, you can show the desired convergence by using$0 = f_n(x) - f_n(x)$and the triangle inequality in$\mathbb{R}\$.

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