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Consider $\delta_{1/n}-\delta_0$; this defines a weak$^*$ null sequences which is not weakly null; e.g., $\langle \delta_{1/n}-\delta_0, \chi_{\{0\}} \rangle \not\to 0$. So if $\Omega$ contains a nont …

The norm of (even a continuous convex function on) a separable Banach space is Gâteaux differentiable on a dense $G_\delta$-set (Mazur), but the canonical norm on the nonseparable Banach space $\ell_1 …

The answer is no. Bourgain and Pisier have given a counterexample (A construction of $\mathcal{L}_\infty$-spaces and related Banach spaces. Bol. Soc. Bras. Mat. 14, No. 2, 109-123 (1983). See Zbl 0586 …

This is the question of duality of injective and projective tensor products of Banach spaces, and the natural question would be whether the dual of $K(X)$ can be represented by the functionals in $N(X …

$Y^{\bot\bot}$ need not be what you call Hahn-Banach smooth [see below] in $X^{\bot\bot}$: Take $X=L_1[0,1]$ and a smooth point of the unit sphere, e.g. $x=1$. Let $Y$ be the linear span of $x$, which …

I find the following criterion useful: A sequence $(x_n)$ is Cauchy iff for all subsequences $(x_{n_{k+1}}-x_{n_k})$ tends to $0$. This works for the norm topology, the weak topology and the weak$^*$ …

It was proved by E. Behrends in Studia Math. 55, 71-85 (1976) that apart from $E=\mathbb{R}^2$ with the sup norm, which is isometric to $E$ with the $\ell_1$-norm, a Banach space $E$ admits a decompos …

Actually, a somewhat weaker condition is sufficient, namely that $x_0$ is a strong extreme point, meaning that $\|z_n\|\to0$ whenever $\|x_0\pm z_n\|\to1$. This is the same as saying that for each $\v …

It is known that $C(K)$, for infinite $K$, contains a copy of $c_0$, hence it does not have nontrivial type (meaning $>1$) or nontrivial cotype (meaning $<\infty$).

The separable spaces that do not have your property for bounded $(x_n)$ (namely, there is a sequence in the unit sphere satisfying your limit condition) are characterised in the paper ``Thickness of t …

I think there are various reasons from abstract tensor norm theory why this is impossible, but I'll try to give a concrete example. First, a $C^*$-norm on an algebra is uniquely determined; therefore, …

Yes, it does. This is essentially an unpublished result due to Hermann Pfitzner, see III.3.6 and III.3.7 in $M$-ideals in Banach spaces and Banach algebras by P. Harmand, W. Werner and myself (Zbl 078 …

For $a=(a_n) \in \ell_\infty$ let $m(a) = (\sup a_n + \inf a_n)/2$. Then $a\mapsto P(a)=m(a) 1$ is the projection you're looking for. Dirk

Cameron's answer indicates why $Z(X)$ rather than $X$ contains an isomorphic copy of $c_0$. Using E. Behrends's function module representation theory, one actually obtains an isometric copy of $c_0$ i …

With the usual definition of a $p$-nuclear operator (see comment above), $\nu_p(P_\tau)\le1$: Let $x_i^*(f)= \int_{A_i} f / \mu(A_i)^{1/q}$ and $y_i= \chi_{A_i}/ \mu_(A_i)^{1/p}$. Then $P_\tau= \sum x …