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As for 3. the answer is no: P. Terenzi, Successioni regolari negli spazi di Banach, Milan J. Math., 57 (1) (1987), 275–285. E. Glakousakis and S. K. Mercourakis, Examples of infinite dimensional Banach spaces without infinite equilateral sets, preprint, 2015, arXiv:1502.02500, to appear in Serdica Math. J., 42 (2016). see also Introduction in T. ...

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In the plane we can take the vertices of an equilateral triangle. In $n=3$ the vertices of a tetrahedron. In general, the vertices of a regular $n$-simplex. The construction is described in Wikipedia like this: Begin with a point A. Mark point B at a distance r from it, and join to form a line segment. Mark point C in a second, orthogonal, ...

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The answer to your question is included in the third line of the proof of the theorem you are interested in! By a classical result of Aharoni (see Theorem 7.11, p. 176 in 1), we know that there is a 3-Lipschitz-homeomorphism between $C(K)$ and some subset of $c_0$. This proof can be summarized in four steps as follows: By Sobczyk's theorem, any ...

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That proof starts with the sentence that $c_0$ is $2$-complemented in $C(K)$. And indeed $c_0$ is isometrically embedded into $C(K)$. This holds as $K$ is an infinite compact metric space, so has a convergent sequence (with all different terms) $x_n$ with limit $x$ (unequal to all $x_n$). Call $S = \{x_n: n \in \mathbb{N}\} \cup \{x\}$. Then $C(S)$ embeds ...

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Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, $X$ is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case. References: P. Bohnenblust, A characterization of complex Hilbert spaces, ...

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In this case you have $$\|p_q(x)\| = \max_{s\ne q} \|x_s\| = \|x_r\| = \|x_q\|.$$ Hence, $$1-\frac{\|p_q(x)\|}{\|x_q\|} = 0$$ and similarly for $r$.

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Hint: use $$\|x(t+h)\|-\|x(t)\|\le\|x(t+h)-x(t)\|$$ and the definition of derivative.

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Assuming that the derivative on the left hand side exists, it is the limit of $$\frac{||x(t+d)||-||x(t)||}{d}$$ The numerator is less than $||x(t+d)-x(t)||$ so the absolute value of the fraction is bounded by $||\frac{x(t+d)-x(t)}{d}||$. which converges to the norm of the derivative of x.

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Even more is true. Every copy of $c_0$ in $C(K)$ for $K$ compact, metric is complemented by a projection of norm at most 2. Indeed, $C(K)$ is in this case separable (as $K$ is second-countable we may use the Stone–Weierstrass theorem to get the claim) and then we may apply Sobczyk's theorem.

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Let $t_n \in K$ be a convergent sequence of distinct points with limit $t_\infty$ also distinct from all $t_n$, consider $C_0(K)=\lbrace f\in C(K): f(t_\infty)=0\rbrace$ and consider $P:C_0(K)\to c_0$, $f\mapsto (f(t_n))_{n\in\mathbb N}$. Conversely, choose peak functions $\varphi_n \in C(K)$ with disjoint supports and $\varphi_n(t_k)=\delta_{n,k}$ and ...

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b) Convergence in $C^0[0,1]$ is uniform convergence, and uniform convergence on $[0,1]$ implies convergence of integrals. (And it preserves the property $f(0)=0$ too.) Hence $F$ is closed. c) There is no function $f$ in the closed unit ball of $E$ that is at distance $1$ from $F$. Indeed, fix any $f$ in the closed unit ball. From $\sup_{[0,1]} |f|\le 1$ ...

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If you want to avoid the measure theoretic framework, there is a quick and easy way to build a useful integral for Banach-valued functions. It's called the regulated integral / Cauchy integral. A good reference is Dieudonne's Foundations of Modern Analysis, but I think you can also find it in Bourbaki. In any case, it avoids measure theory and gives an ...

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Convolution really comes out of gathering like powers of things such as powers of exponentials or of powers of a complex variable. \begin{align} \sum_{n=-\infty}^{\infty}a_n e^{in\theta}\sum_{n=-\infty}^{\infty}b_ne^{in\theta}&=\sum_{n=-\infty}^{\infty}\left(\sum_{j+k=n}a_j b_k\right)e^{in\theta} \\ & = ...

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It's an easy thing to show that $\sigma(T) \subseteq \{ \lambda : |\lambda| \le \|T\| \}$ because the following inverse series converges in operator norm for $\|T\| < |\lambda|$ \begin{align} (T-\lambda I)^{-1} &= \frac{1}{\lambda}(\frac{1}{\lambda}T-I)^{-1} \\ & = -\frac{1}{\lambda}(I-\frac{1}{\lambda}T)^{-1} \\ ...

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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 ...

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For $(X,\mathcal{T},\mu)$ a measure space and $E$ a (complex or real) Banach space, you can define the integral of a function $f$ defined from $X$ to $E$. For this, you can use the space $\mathcal {E}(X, \mathcal{T};E)$ of simple functions defined from $X$ to $E$. A function $f : X \to E$ defined almost everywhere is said to be $\mu$-integrable (or just ...

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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-$*$ ...

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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, $0\notin ... 0 Let$X$be a Banach space and let$\{x_d\colon d\in D\}$be a dense subset of the unit ball of$X$. Consider the space$\ell_1(D)$of all absolutely summable sequences on$D$. We define a linear map$T\colon \ell_1(D) \to X$by $$T\Big((\lambda_d)_{d\in D}\Big) = \sum_{d\in D}\lambda_d x_d\qquad ((\lambda_d)_{d\in D} \in \ell_1(D)).$$ This is a well-defined ... 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 ... 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$\mathbb{R}$. 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$Lip_0(X)^{*}$... 2 Theorem. Let$X$be an infinite dimensional Banach space. Then$\,\mathrm{dim}\,X\ge 2^{\aleph_0}$. Sketch of Proof. Based on M.G. McKay's proof. Let$\{w_n : n\in\mathbb N\}\subset X$be a linearly independent set. Step A. Using Hahn-Banach, we shall construct another linearly independent set$\{v_n : n\in\mathbb N\}\subset X$, and a set of linear ... 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 ... 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, ... 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 $$... 3 As suggested in the comment, we show that the graph of T is closed. Assume that (x_n , T(x_n) ) \to (x, y). Then x_n - x \to 0. From the assumption, for all f\in \mathcal F,$$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)) = ... 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 ... 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 $$e_i = (0,0,0, \ldots, 0, 1,0,\ldots) \in ... 1 Given \mu\in \mathcal{M}(K), note that for each finite set J\subset I we have$$\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 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 ... 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 ... 0 Here is an alternative proof of the fact that r(T)=0 ( I add it because it does not seem to appear in the related posts and I find it quite nice). First show by induction that |A^n x(t)|\leq \frac{t^n}{n!}\|x\|_\infty for all t\in[0,1]: Basis n=0: |x(t)|\leq \|x\|_\infty for all t\in[0,1] by definition of the supremum norm. Inductive step: For ... 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 ... 0 It is not enough to just know that \dim U=\dim V. For instance, suppose U=H and V is the orthogonal complement of a single nonzero vector v\in H. As long as H is infinite-dimensional, U and V will have the same dimension, but clearly no unitary T:H\to H can map U to V, since then T would not be surjective. More generally, one can see ... 0 You need exactly \dim U=\dim V and \dim U^\perp=\dim V^\perp , in the sense that they have bases with the same cardinality. Suppose that \{e_j\}_{j\in J} is an orthonormal basis of U and that \{f_j\}_{j\in J} is an orthonormal basis of V. Extend both basis to orthonormal bases of H, \{e_j\}_{j\in K}, \{f_j\}_{j\in K}, where K\supset ... 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 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 ... 0 Let's first notice that X_0 is a closed subspace of X. Hence we can define the quotient norm by\Vert \overline{f} \Vert = \inf_{g \in X_0} \Vert f- g \Vert$$... with an obvious abuse of notation. The application$$ \begin{array}{l|rcl} \phi : & X/X_0 & \longrightarrow & \mathbb C\\ & \overline{f} & \longmapsto & f(0) ... 0 Observe thatA^n(x) = T^n x + \beta_n$, where$\beta_n = \sum_{k=1}^{n-1}T^k \eta + \eta$. Thus, $$\|A^n(x) - A^n(y)\| = \|T^n(x - y)\| \leq \|T^n\|\|x - y\|$$ which implies that$A^n$is a contraction. From here you can use the result that if$A^n$is a contraction, then$A$has a unique fixed point. For example, see the proof of this result here. 1 You can extend$L$uniquely (by weak* continuity) to the weak*-closure of$M$. And the preannihilator of$M$coincides with the preannihilator of its weak* closure. 1 Let$0\ne c\in l^p$. By Hahn-Banach, there exists a linear$T:l^p\rightarrow\mathbb{R}$such that$Tc = 1$and$|Tv|\le\frac{\|v\|_{l^p}}{\|c\|_{l^p}}$for all$v\in l^p$. Let$A:l^p\rightarrow l^p$be defined by $$Av = (Tv,0,0,\dots)$$ It follows that$\|Av\|_{l^p} = |Tv|\le\frac{\|v\|_{l^p}}{\|c\|_{l^p}}$, and hence$\|A\|_{l^p\rightarrow ...

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For any sequence $y\in c_0$ we have $$\|x-y\|_\infty = \sup_n|x_n-y_n|\ge \limsup_{n\to\infty} |x_n-y_n|=\limsup_{n\to\infty} |x_n|$$ This gives a lower bound on the distance. To get the matching upper bound, let $y_n=x_n$ when $n\le N$, and $y_n=0$ otherwise. This is an element of $c_0$, and $$\|x-y\|_\infty = \sup_{n>N}|x_n| ... 1 Actually equality holds because the norm |\cdot | is a continuous function. To show that it is continuous, simply note that it is Lipschitz since for all x,y$$||x|-|y|| \le |x-y|$$2 Fixing x, y\in X and consider$$g(t) = \frac{f(x+ ty) - f(x)}{t}.$$Then your question reduces to an easy question in real analysis. You need only that |\cdot| is a continuous function. 3 The separability of X is irrelevant here. From the Banach-Alaoglu theorem, we know that for all \varphi \in X^{\ast} and r \geqslant 0 the norm-closed ball K_r(\varphi) := \{ \psi \in X^{\ast} : \lVert \varphi - \psi\rVert \leqslant r\} is weak^{\ast}-compact. Then if S is any weak^{\ast}-closed nonempty subset of X^{\ast}, the family ... 0 Yes this is true and it can be proven along the following lines: find a sequence \{\psi_n\} \subset M, such that \|\psi_n - \varphi\| \to \inf\{\|\varphi - m\| : m \in M\} extract a weak-* convergent subsequence with limit \psi conclude that \psi has the desired properties 1 Let H be an infinite-dimensional separable Hilbert space. Let \{E_{kj}\} be the canonical set of matrix units. Let$$ X=\sum_{k=1}^\infty \frac1k\,E_{2k,2k+1}, \ \ \ E=\sum_{k=1}^\infty E_{2k,2k} $$and let P=E+X, Q=I-E+X^*. Note that X^2=0, EX=X, and XE=0, so$$ P^2=P, Q^2=Q. $$Also,$$ PQ=E(I-E)+EX^*+X(I-E)+XX^*=X+XX^*,  ...

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If $x\neq y$, let $Y\subset X$ be the span of $x-y$ and let $f_0:Y\to \mathbb{C}$ be the linear functional sending $x-y$ to $\|x-y\|$. By Hahn-Banach, $f_0$ extends to an $f\in X^*$ of norm $1$. By construction, $f(x-y)=f_0(x-y)\neq 0$, so $f(x)\neq f(y)$.

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Without getting into any particularly deep analysis, here are my thoughts: (let's assume, like you say, for simplicity we run over all convex $f$s that are continuous and defined on all of $X$) First I would start with thinking about what it would mean for $f(S)$ to not be an interval: this means there is some $x\in\mathbb{R}$ that "separates" $f(S)$ (i.e., ...

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