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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.

7 votes

Almost isometric embeddability implies isometric embeddability

The OP's intuition that $X:=(\oplus \ell^2_{p_n})_2$ with $p_n$ tending to $1$ (and $1<p_n < \infty$) provides an example is correct. Indeed, $X$, being an $\ell_2$ sum of strictly convex spaces, is …
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4 votes
Accepted

A question on compact operators with domain $l_{p}$

Yes. If $R:\ell_q \to X$ and $TR$ is not compact then there is a normalized block basis $u_n$ of the unit vector basis for $\ell_q$ s.t. $TRu_n$ is bounded away from zero. Let $V:\ell_p \to \ell_q$ …
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0 votes
Accepted

A question on unconditionally $p$-summable sequences

Yes. Choose $n_1<n_2<\dots$ s.t. for all $m$ $$ \sup_{x^{*}\in B_{X^{*}}}\sum_{k=n_m}^\infty |\langle x^*, x_k \rangle|^p < \epsilon^p/4^m. $$ Set $\lambda_k = 1$ for $k<n_1$ and $\lambda_k = 2^{-k} …
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1 vote
Accepted

On the separability of operator range

No. Consider the identity operator on $\ell_r(S)$ with $p<r<\infty$ and $S$ uncountable. Or consider any weakly compact operator with non separable range into a $C(K)$ space or an $L_1$ space and us …
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1 vote
Accepted

Weakly $p$-summable sequences in $L_{r}$

No, because $L_r$ has cotype 2 for $r\le 2$. In fact, every bounded linear operator from $\ell_q$ into $L_r$ is compact when $r \le 2 < q$.
Bill Johnson's user avatar
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1 vote
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On the normalized block basic sequences in $c_{0}\widehat{\otimes}_{\pi} c_{0}$

No. Consider $z_{2n+1} - z_{2n}$.
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6 votes

Equivalent definitions of $\mathscr{L}_p$-spaces

The equivalence of the two definitions of $\mathscr{L}_p$-space involves an elementary perturbation argument that is probably contained in the original article of Lindenstrauss and Pelczynski. That k …
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2 votes
Accepted

A quantity measuring weak non-compactness

Suppose that $A$ is a bounded subset of $X$ and $X$ is a subspace of $Y$. Then \begin{equation} wk_Y(A) \le wk_X(A) \le 2 wk_Y(A),\ \ \ (\#) \end{equation} and $2$ is the best constant in the right …
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2 votes
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Classification of subsymmetric basic sequences

In the theorem you quote, part of the definition of subsymmetric includes the hypothesis that the sequence is unconditionally basic. If you do not include this in the definition of subsymmetric, then …
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2 votes

The reflexivity of the space generated by a convex, balanced and compact set

Even if $K$ is just weakly compact, $X_K$ is isometrically isomorphic to a dual space with the weak$^*$ topology agreeing with the weak topology from $X$ on $K$. You can get a weakly compact convex s …
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1 vote
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A formally weaker form of the extendable local reflexivity for Banach spaces

The answer is yes by a small perturbation argument. For a small $\epsilon$, enlarge $F$ so that $F$ $1+\epsilon$-norms $E$. Get $T$ from the condition. Choose any Auerbach basis $u_k$ for $E$ …
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4 votes

A question in Banach space

Given a linear operator $A$ from a Banach space $E$ to a Banach space $F$, the quantity $\gamma_2(A)$ is the infimum (actually, it is a minimum) of $\|X \| \cdot \|Y\|$, where the minimum is over all …
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2 votes

A question on p-summing operators

Not unless $p=1$, in which case you interpret $\ell_q$ to be $c_0$. The most interesting case is $p=2$. Tomczak's theorem says that you can compute $\pi_2(T)$ for $T: \ell_2^n \to X$ up to the con …
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2 votes

Rearrangments of Fourier series

In every Banach space that has a Schauder basis there is a Schauder basis that is not unconditional (meaning that there is a rearrangement of the basis that is not a basis). See Pełczyński, A.; Sing …
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3 votes

Is the Banach space of continuously differential functions strongly regular?

$C^k(M)$ is isomorphic to $C(M)$ and hence, by Milutin's Theorem, to $C[0,1]$. The first statement is more or less obvious since the norm on $C^k(M)$ is equivalent on a $k$-codimensional subspace to …
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