Schauder Bases Approximation and Banach Space Pathologies — Examples

For $N\ge 1$, $P_N$ retains coordinates $0,\dots ,N-1$, and $P_0=0$. [given, L1, L2] Thus in $c_0$, [L1] (with its truncation index $N-1$) gives $P_{N}x\to x$ in supremum norm. In ${\mathrm{\ell }}^p$, [L2] gives $\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_p^p={\sum }_{m\ge N}\mathrm{\mid }{x}_m{\mathrm{\mid }}^p\to 0$; for $p=1$ this is also [L1]. The coefficients are necessarily the coordinates, so the expansions are unique.

Deleting coordinates cannot increase either the supremum norm or the [given, L3, step 1.1] $p$-norm, hence $\mathrm{\parallel }{P}_N\mathrm{\parallel }\le 1$. In a nonzero space $P_N{e}_1=e_1$ for $N\ge 1$, so $\mathrm{\parallel }{P}_N\mathrm{\parallel }=1$ and [L3] gives basis constant one.

Every finite linear combination of standard unit vectors has finite [given, L2] support. A supremum-norm limit of finite-support sequences lies in $c_0$: for a given tolerance, approximate uniformly by one finite-support sequence and use its finite support to bound the tail.

The constant-one sequence belongs to ${\mathrm{\ell }}^{\mathrm{\infty }}$ but not to $c_0$. [given, L1, L2, step 1.1] Therefore it is not the norm limit of standard-unit-vector partial sums, in violation of [L1]. This explicit witness refutes the statement.

For $x=(x_k{)}_{k\ge 0}\in c_0$ set $a_n=x_{n-1}-x_n$ for $n\ge 1$. [given] The $k$th coordinate of ${\sum }_{n=1}^N{a}_n{s}_n$ is $x_k-x_N$ when $0\le k<N$ and zero otherwise. Hence the error has supremum at most $\max \{\mathrm{\mid }{x}_N\mathrm{\mid },{\sup }_{k\ge N}\mathrm{\mid }{x}_k\mathrm{\mid }\}\to 0$. Conversely the coordinate identities force $a_n=x_{n-1}-x_n$, so $(s_n)$ is a Schauder basis.

Take $x_k=(-1{)}^k\mathrm{/}(k+1)$ for $k\ge 0$. Then $a_n=(-1{)}^{n-1}(1\mathrm{/}n+1\mathrm{/}(n+1))$. The subseries over the odd indices has first coordinate

Put $x=(x_k{)}_{k\ge 0}$ with $x_k=(-1{)}^k\mathrm{/}(k+1)$, and for $n\ge 1$ put

The odd-indexed terms have positive first coordinate $a_n$, whose sum [given, step 1.1] diverges, while the even-indexed terms have negative first coordinate and the sum of their absolute first coordinates diverges. Also $\mathrm{\parallel }{y}_n{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\mid }{a}_n\mathrm{\mid }\to 0$. Starting at zero, take consecutive unused odd terms until the first coordinate exceeds $1$, then consecutive unused even terms until it is below $0$, and repeat. Each stage ends after finitely many terms because the corresponding signed tail diverges.

Infinitely many stages of each parity occur, and each stage consumes at [given, L2, step 2.1] least one term in that parity's original order. Consequently every odd and every even term is eventually used exactly once, so the procedure defines a permutation of ${\mathbb{N}}_{\ge 1}$. The first coordinates of its partial sums exceed $1$ and fall below $0$ infinitely often, so the rearranged vector series diverges. Step 1.1 gives convergence in the original order, thereby refuting the statement and, consistently with [L2], witnessing failure of unconditional convergence.

Use [A1] exactly through [L1] and define $\nu$ by [L2]. Positivity of $L$ [given, A1, L1, L2] makes $\nu$ positive, and normalization gives $\nu (\mathbb{N})=L(\mathbf{1})=1$.

Shift invariance makes all singleton masses equal, say to $c\ge 0$. [given, L1, L2, step 1.1] Finite additivity gives $Nc\le \nu (\mathbb{N})=1$ for every positive integer $N$, hence $c=0$. If $\nu$ were countably additive, the disjoint singleton decomposition of $\mathbb{N}$ would give $\nu (\mathbb{N})={\sum }_{n}0=0$, a contradiction.

AC in [A1] supplies the ultrafilter lemma, DC, Hahn--Banach, and Countable [given, A1, L1, L2] Choice needed by [L1] and [L2]. Thus [L1] already refutes reflexivity of ${\mathrm{\ell }}^1$, over both scalar fields.

By [L3], $({\mathrm{\ell }}^1{)}^{\ast }={\mathrm{\ell }}^{\mathrm{\infty }}$. If ${\mathrm{\ell }}^{\mathrm{\infty }}$ were reflexive, [given, L3, L2, step 1.1] the reverse implication in [L2] would make ${\mathrm{\ell }}^1$ reflexive, contradicting step 1.1. Hence ${\mathrm{\ell }}^{\mathrm{\infty }}$ is not reflexive.

Independently, [L4] exhibits the bidual surplus: under the identification [given, L4, A1, step 2.1] $({\mathrm{\ell }}^1{)}^{\ast \ast }=({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }=ba$, the canonical ${\mathrm{\ell }}^1$ image is only the proper subspace of countably additive charges. This is a concrete failed- surjectivity witness consistent with steps 1.1-2.1.