Search Results

I assume "$p$-summable" means "absolutely $p$-summable", i.e., $\sum \|x_n\|^p < \infty$. Since all that matters about $x_n$ is its norm, the question reduces to the scalar case. Yes. Find a strictl …

No, this is false --- trivially, because if $M$ is the completion of $N$ then $M$ and $N$ have the same Arens-Eells space. But if you require $M$ and $N$ to both be complete there is a more interestin …

Okay, even assuming that $Y$ separates points the answer is still no. Take $X = l^1$ and let $Y$ be the set of elements $(a_n)$ of $l^\infty$ which satisfy $\lim a_n = 2a_1$. It's easy to see that $Y$ …

No, that's not true. Let $\mathbb{N}^* = \mathbb{N} \cup \{\infty\}$ and set $X = C(\mathbb{N}^*) \cong c$ and $X^* = l^1(\mathbb{N}^*) \cong l^1$. Take as the basic sequence of $X^*$ the vectors $e_n …

Yeah, this is true. Let ${\cal T}$ be an LCH topology on $X$ which makes the unit ball compact. Then every ${\cal T}$-continuous linear functional takes the unit ball to a compact subset of the scalar …

One way of disproving this conjecture is to construct a norm convergent sequence in E which is not order bounded. E.g. the sequence $(\frac{1}{n}e_n)$ in $l^1$.

Yes, see Proposition 8.1.3 of the book Mathematical Quantization. This describes a duality between Banach spaces and "dual unit balls", which are defined as compact, convex, balanced subsets of locall …

Let $C'= \bigcup_n nC$ be the cone generated by $C$ and let $X$ be the closed linear span of $C$. Theorem: A bounded linear functional $q$ with the stated property exists if and only if $C'$ is not d …

Yes, it is metrizable. Assume the scalar field is real; if it is complex, replace the functions $f_n$ with their real and imaginary parts. WLOG each $f_n$ maps $K$ into $[0,1]$. Amalgamate the $f_n$ i …

I think the general fact is that $\mathcal{B}_1$ consists of precisely the Banach spaces (isometrically isomorphic to) the spaces $C(X)$ for $X$ extremally disconnected and compact. In particular, eve …

Define $P: c_0 \to c_0$ by $P(a_0, a_1, a_2, \ldots) = (\sum \frac{a_n}{2^n}, 0, 0, \ldots)$. Then $P(a_0, 0, 0, \ldots) = (a_0, 0, 0, \ldots)$, so this is a projection onto the first coordinate. But …

Great question! What you need is Sandy Grabiner's approximation lemma: Lemma: Let $E$ and $F$ be Banach spaces, let $T \in B(E,F)$, and let $M > 0$ and $0 < r < 1$. Suppose that for each $y \in [F]_1 …

Find a sequence of disjoint balls $(B_n)$, all contained in the unit ball of $X$ and all having radius greater than some fixed $\epsilon$. Select continuous functions $f_n: B_n \to X$ however you like …

Is this the statement you want: any locally constant, compactly supported function from ${\bf Q}_p^n$ to ${\bf C}$ is uniformly approximated by linear combinations of products $f_1\cdots f_n$ where ea …

I haven't looked at the paper but your counterexample is mistaken. The basis projections generate not $B(l^p)$ but the algebra of multiplication operators, which is isometrically isomorphic to $l^\inf …