Schauder Bases Approximation and Banach Space Pathologies

The displayed formula is a norm on $E$: definiteness follows because its [given, L2] value zero forces every partial sum, hence every coefficient since $e_n\mathrm{\ne }0$, to vanish. Linearity of $E$ and the remaining norm axioms follow termwise from the norm axioms in $X$.

Let $(a^{(k)})$ be Cauchy in $E$. For fixed $n$, apply the $n$th coordinate [given, L1, step 1.1] map on $\mathrm{span}\{e_1,\dots ,e_n\}$ to ${\sum }_{j\le n}(a_j^{(k)}-a_j^{(\mathrm{\ell })})e_j$. By [L1], $(a_n^{(k)}{)}_k$ is Cauchy, so it has a scalar limit $a_n$.

Given $\epsilon >0$, choose $k_0$ so $\mathrm{\parallel }{a}^{(k)}-a^{(\mathrm{\ell })}{\mathrm{\parallel }}_E<\epsilon$ for $k,\mathrm{\ell }\ge k_0$. Fix $k\ge k_0$ and $N$, and let $\mathrm{\ell }\to \mathrm{\infty }$ in the finite sum. Coordinatewise convergence and continuity of finite sums give

Fix $k\ge k_0$. Since $a^{(k)}\in E$, its series has Cauchy tails. For [given, step 3.1] $M>N$, step 3.1 applied to the two partial sums bounds the corresponding finite block for $a-a^{(k)}$ by $2\epsilon$. Hence the partial sums for $a$ are Cauchy in the Banach space $X$, so $a\in E$. Step 3.1 then yields $\mathrm{\parallel }{a}^{(k)}-a{\mathrm{\parallel }}_E\le \epsilon$; thus $E$ is complete.

For $a\in E$, norm continuity gives [given, L2, step 4.1] $\mathrm{\parallel }Sa\mathrm{\parallel }={\lim }_N\mathrm{\parallel }{\sum }_{n\le N}{a}_n{e}_n\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_E$, so $S$ is bounded. Surjectivity and injectivity are respectively existence and uniqueness in [L2].

Apply [L2] to [L1]. The only choice use is [A1], through that bounded-inverse [given, L2, L1, A1] theorem. Thus $S^{-1}:X\to E$ is bounded.

Truncation $Q_N:E\to E$, $(a_n)\mapsto (a_1,\dots ,a_N,0,\dots )$, satisfies $\mathrm{\parallel }{Q}_{N}a{\mathrm{\parallel }}_E\le \mathrm{\parallel }a{\mathrm{\parallel }}_E$, because every partial sum of $Q_{N}a$ is a partial sum of $a$. Since $P_N=S{Q}_N{S}^{-1}$ and $\mathrm{\parallel }S\mathrm{\parallel }\le 1$,

For $n\ge 1$,

In the real case let $D$ consist of all finite linear combinations of the [given] basis vectors with rational coefficients. In the complex case use coefficients in $\mathbb{Q}+i\mathbb{Q}$. This is a countable union of countable finite products and hence is countable.

Given $x$, first use [L1] to choose a basis partial sum within [given, L1, step 1.1] $\epsilon \mathrm{/}2$ of $x$. Approximate its finitely many scalar coefficients by rational, respectively Gaussian-rational, scalars closely enough that the resulting finite combination changes by less than $\epsilon \mathrm{/}2$. It belongs to $D$ and is within $\epsilon$ of $x$. Thus $D$ is dense.

Suppose (3) fails, and enumerate finite subsets of ${\mathbb{N}}_{\ge 1}$ by their finite codes. [given] Recursively build a listing as follows. At stage $k$, first append the least positive integer not yet listed, let $M$ be the greatest integer listed so far, and then take the least coded finite set $F_k\subseteq \{M+1,M+2,\dots \}$ with $\mathrm{\parallel }{\sum }_{n\in F_k}{x}_n\mathrm{\parallel }\ge \epsilon$ and append its members in increasing order. The negation of (3) supplies such an $F_k$ after every finite stage, and least codes make the recursion unique.

No integer is listed twice, because every block $F_k$ lies beyond all [given, L2, step 1.1] earlier entries. Every positive integer is eventually listed, since each stage appends the current least omitted integer. Thus the listing is a permutation of ${\mathbb{N}}_{\ge 1}$. Each $F_k$ is a consecutive block whose increment has norm at least $\epsilon$, so the rearranged partial sums are not Cauchy. By [L2], (1) therefore implies (3).

Assume (3). Given $\epsilon$, choose $N$ for $\epsilon \mathrm{/}2$. If finite [given, L1, step 2.1] $E,F$ both contain $\{1,\dots ,N-1\}$, their sums differ by two disjoint finite tail sums and hence by less than $\epsilon$. In particular, the ordinary partial sums are Cauchy, so [L1] gives a limit $s\in X$. Applying (3) once more to a finite $F$ containing a sufficiently long initial segment shows $\mathrm{\parallel }{\sum }_{n\in F}{x}_n-s\mathrm{\parallel }<\epsilon$. Thus the finite-subset net converges to $s$, and (3) implies (2).

Every permutation's initial index sets are cofinal among finite subsets: [given, L2, step 3.1] each fixed finite set is eventually included. Hence (2) makes every rearranged partial-sum sequence converge to the net limit. The ordinary initial segments are also cofinal, so this limit is the fixed-order sum. Thus (2) implies (1).

Under (3), any finite tail of any subseries is a finite tail set of the [given, L1, step 1.1, step 4.1] original series. It satisfies the Cauchy criterion, so [L1] proves (4). Conversely, if (3) failed, the union of the disjoint blocks $F_k$ from step 1.1, listed increasingly, would define a subseries having successive block increments of norm at least $\epsilon$, hence not Cauchy. Thus (3) and (4) are equivalent.

Assume (3), let $\mathrm{\mid }{\lambda }_n\mathrm{\mid }\le M$, and take a finite tail set $F$. A finite layer-cake decomposition shows that for $0\le t_n\le 1$, ${\sum }_{n\in F}{t}_n{x}_n$ is a convex combination of subset sums of $F$. Writing a real multiplier as its positive part minus its negative part, and a complex multiplier as the same decomposition of real and imaginary parts, gives

Steps 2.1--6.1 give both directions among all five conditions. The [given, step 4.1, step 6.1] common-sum identification is the conclusion of step 4.1.

Passing to limits in the linear identities for [L2] shows that $T$ is [given, L2, L1] linear. By [L1], $\mathrm{\parallel }Tx\mathrm{\parallel }={\lim }_n\mathrm{\parallel }{T}_{n}x\mathrm{\parallel }\le M_0\mathrm{\parallel }x\mathrm{\parallel }$, so $T$ is bounded and $\mathrm{\parallel }T\mathrm{\parallel }\le M_0$.

Put $M:=M_0+\mathrm{\parallel }T\mathrm{\parallel }$. If $M=0$, every $T_n$ and $T$ is zero and the result [given, step 1.1] is immediate. Suppose $M>0$, fix compact $C$ and $\epsilon >0$, and choose a finite $\epsilon \mathrm{/}(3M)$-net $x_1,\dots ,x_r$ in $C$.

Pointwise convergence gives $n_0$ such that $\mathrm{\parallel }(T_n-T)x_j\mathrm{\parallel }<\epsilon \mathrm{/}3$ for all $j$ and $n\ge n_0$. For $x\in C$ choose $j$ with $\mathrm{\parallel }x-x_j\mathrm{\parallel }<\epsilon \mathrm{/}(3M)$. Then [L1] gives

Each $P_N$ has range in $\mathrm{span}\{e_1,\dots ,e_N\}$ and hence [given, L1, A1, L2] has finite rank. By [L1], using [A1] exactly through the coordinate-boundedness theorem, $\mathrm{\parallel }{P}_N\mathrm{\parallel }\le K$; by [L2], $P_{N}x\to x$ for every $x\in X$.

Apply [L3] to $(P_N)$ and the identity. On each compact $C$, [given, L3, L4, step 1.1] ${\sup }_{x\in C}\mathrm{\parallel }{P}_{N}x-x\mathrm{\parallel }\to 0$. Together with step 1.1, [L4] says precisely that $X$ has $K$-BAP. Since BAP implies AP by [L4], the consequence follows.

Let $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ and $\epsilon >0$. In the real case partition the [given, L1] bounded interval containing all $x_n$ into finitely many half-open intervals of length below $\epsilon$, and replace every $x_n$ by a fixed endpoint of its cell. The resulting sequence $s$ has finite range and $\mathrm{\parallel }x-s{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$.

In the complex case partition a square containing all $x_n$ into finitely [given, L1, step 1.1] many squares of side below $\epsilon \mathrm{/}\sqrt{2}$ and replace by one corner of the containing cell. Again $s$ has finite range and $\mathrm{\parallel }x-s{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$. This proves density in both scalar fields.

Two partition representations of $s$ have the common refinement [given, L1] $(A_j\cap B_k{)}_{j,k}$. Finite additivity replaces each original summand by its sum over the refinement; because both coefficients equal $s_n$ on a nonempty cell, the two refined sums agree. Thus $I_{\nu }^0$ is well-defined and linear.

For a partition representation,

Use the fixed dyadic grid to define a finite-range quantization $q_k(x)$ [given, L2, step 2.1] with $\mathrm{\parallel }{q}_k(x)-x{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^{-k}$ (coordinatewise in a square over $\mathbb{C}$). Step 2.1 makes $(I_{\nu }^0(q_k(x)){)}_k$ Cauchy; [L2] supplies its limit. The same estimate shows independence of any approximating finite-range sequence, linearity, uniqueness, and the bound $\mathrm{\parallel }{I}_{\nu }\mathrm{\parallel }\le \mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})$.

Given $\epsilon >0$, choose a finite partition $(A_j)$ with [given, step 3.1] ${\sum }_j\mathrm{\mid }\nu (A_j)\mathrm{\mid }>\mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})-\epsilon$. Put $c_j=1$ when $\nu (A_j)=0$ and otherwise $c_j=\overline{\nu (A_j)}\mathrm{/}\mathrm{\mid }\nu (A_j)\mathrm{\mid }$ (the same sign formula over $\mathbb{R}$). Then $\mathrm{\parallel }{\sum }_j{c}_j{\mathbf{1}}_{A_j}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and its integral is ${\sum }_j\mathrm{\mid }\nu (A_j)\mathrm{\mid }$. Thus $\mathrm{\parallel }{I}_{\nu }\mathrm{\parallel }\ge \mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})-\epsilon$; letting $\epsilon \downarrow 0$ proves equality. Linearity in $\nu$ is immediate from the formula.

Let $\phi \in ({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }$ and put ${\nu }_{\phi }(A)=\phi ({\mathbf{1}}_A)$. Linearity and ${\mathbf{1}}_{A\bigsqcup B}={\mathbf{1}}_A+{\mathbf{1}}_B$ give finite additivity. For a finite partition $(A_j)$ choose scalar phases $c_j$ with $c_j{\nu }_{\phi }(A_j)=\mathrm{\mid }{\nu }_{\phi }(A_j)\mathrm{\mid }$. Then $\mathrm{\parallel }{\sum }_j{c}_j{\mathbf{1}}_{A_j}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and [L2] gives

By construction, $I_{{\nu }_{\phi }}({\mathbf{1}}_A)=\phi ({\mathbf{1}}_A)$. [given, L1, step 1.1] Linearity gives equality on finite-range sequences, and density plus boundedness gives $I_{{\nu }_{\phi }}=\phi$ on ${\mathrm{\ell }}^{\mathrm{\infty }}$.

Conversely, $\mathrm{\Phi }(I_{\nu })(A)=I_{\nu }({\mathbf{1}}_A)=\nu (A)$, so the two maps are [given, L1, step 2.1] inverse. Finally [L1] gives $\mathrm{\parallel }{I}_{\nu }\mathrm{\parallel }=\mathrm{\parallel }\nu {\mathrm{\parallel }}_{ba}$, proving isometry and completing both surjectivity and injectivity.

For $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ define [given, L2, L3] $p(x)={\mathrm{lim\hspace{0.17em}sup}}_N\frac{1}{N+1}{\sum }_{n=0}^N{x}_n$. This is finite because $x$ is bounded. Linearity of finite averages, positive homogeneity of limsup, and [L2] show that $p$ is sublinear in the sense of [L3].

Let $c$ be the subspace of ordinarily convergent sequences and let [given, L1, A1, step 1.1] $f(x)={\lim }_n{x}_n$ on $c$. Cesaro means preserve an ordinary limit, so $f(x)=p(x)$ on $c$; in particular $f\le p$. Apply [L1]. The exact non-finite choice use is [A1] in Hahn--Banach, producing a real-linear extension $L$ with $L(x)\le p(x)$ for every $x$.

If $x_n\ge 0$, then $p(-x)\le 0$, hence [given, step 2.1] $-L(x)=L(-x)\le p(-x)\le 0$; thus $L$ is positive. Since $\mathbf{1}\in c$, $L(\mathbf{1})=1$. Positivity applied to $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathbf{1}\pm x$ gives $\mathrm{\mid }L(x)\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, while $L(\mathbf{1})=1$ gives the reverse norm bound. Therefore $\mathrm{\parallel }L\mathrm{\parallel }=1$.

The Cesaro average of $x-Sx$ equals [given, step 2.1, step 1.1, step 3.1] $(x_0-x_{N+1})\mathrm{/}(N+1)$ and tends to zero; the same is true of $Sx-x$. Therefore $p(x-Sx)=p(Sx-x)=0$. Domination gives $L(x-Sx)\le 0$ and $-L(x-Sx)=L(Sx-x)\le 0$, so $L(x-Sx)=0$ and $L(Sx)=L(x)$. On the original subspace $c$, step 2.1 already says that $L$ extends the ordinary limit.

Let $\nu$ be countably additive and put $a_n=\nu (\{n\})$. For every $N$, [given] the singleton partition of $\{0,\dots ,N\}$ gives ${\sum }_{n=0}^N\mathrm{\mid }{a}_n\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})$. Thus $(a_n)\in {\mathrm{\ell }}^1$. Countable additivity applied to $A={⨆}_{n\in A}\{n\}$ gives the displayed formula.

Conversely, if $(a_n)\in {\mathrm{\ell }}^1$, absolute convergence makes [given, step 1.1] ${\nu }_a(A)={\sum }_{n\in A}{a}_n$ independent of enumeration and countably additive; also $\mathrm{\mid }{\nu }_a\mathrm{\mid }(\mathbb{N})={\sum }_n\mathrm{\mid }{a}_n\mathrm{\mid }<\mathrm{\infty }$. This proves the equivalence and linearity of the subspace.

Use [A1] exactly through [L1], and let $L$ be the resulting mean. By [L2], [given, A1, L1, L2, step 2.1] $\nu (A):=L({\mathbf{1}}_A)$ is a charge. Shift invariance makes all singleton masses equal because $S{\mathbf{1}}_{\{n+1\}}={\mathbf{1}}_{\{n\}}$. Moreover $S{\mathbf{1}}_{\{0\}}=0$, so shift invariance and linearity give $\nu (\{0\})=L({\mathbf{1}}_{\{0\}})=L(0)=0$. Hence $\nu (\{n\})=0$ for every $n$ but $\nu (\mathbb{N})=L(\mathbf{1})=1$, so it is not countably additive. The subspace is proper.

For fixed $p$, $q_p(x)$ is $2^{-1\mathrm{/}2}$ times the Euclidean norm of the [given, L1] finite vector of cyclic successive differences. Euclidean Minkowski gives $q_p(x+y)\le q_p(x)+q_p(y)$ and homogeneity is immediate. Taking suprema shows that $J$ is a vector subspace and gives the triangle inequality and homogeneity for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_J$.

For $i<j$, the two-point tuple $(i,j)$ gives [given, L1, step 1.1] $q_{(i,j)}(x)=\mathrm{\mid }{x}_i-x_j\mathrm{\mid }$. Since $x_j\to 0$ for $x\in c_0$, letting $j\to \mathrm{\infty }$ yields $\mathrm{\mid }{x}_i\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_J$. Taking the supremum proves the first displayed inequality and definiteness, including at $x=0$.

For $x\in {\mathrm{\ell }}^2$ and $p=(p_1<\cdots <p_k)$, use [given, L1, step 2.1] $\mathrm{\mid }a-b{\mathrm{\mid }}^2\le 2\mathrm{\mid }a{\mathrm{\mid }}^2+2\mathrm{\mid }b{\mathrm{\mid }}^2$ in the cyclic sum. Every selected coordinate occurs twice before the factor $1\mathrm{/}2$, so $q_p(x{)}^2\le 2{\sum }_{j=1}^k\mathrm{\mid }{x}_{p_j}{\mathrm{\mid }}^2\le 2\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$. Taking the supremum gives ${\mathrm{\ell }}^2\subseteq J$ and the second inequality.

Let $(x^{(m)})$ be Cauchy in $J$. By [L1] it is Cauchy in $c_0$, so [L2] [given, L1, L2] gives $x^{(m)}\to x\in c_0$ uniformly. For each fixed tuple $p$, $q_p(x)={\lim }_m{q}_p(x^{(m)})$, whence ${\sup }_p{q}_p(x)<\mathrm{\infty }$. Letting $m\to \mathrm{\infty }$ in $q_p(x^{(n)}-x^{(m)})<\epsilon$ uniformly in $p$ yields $\mathrm{\parallel }{x}^{(n)}-x{\mathrm{\parallel }}_J\le \epsilon$. Thus $J$ is complete.

For $p=(p_1<\cdots <p_k)$ in the positive labeling of [L0], define the auxiliary endpoint variation

Finite-support sequences are dense. Indeed, for nonzero $x$ and $\epsilon >0$, choose $\delta >0$ with $4\delta R(x)<{\epsilon }^2$. Choose a tuple $p$ with $r_p(x)>R(x)-\delta$, append a sufficiently remote final index $N=p_k$ so this remains true, and ensure ${\sup }_{i\ge N}\mathrm{\mid }{x}_i\mathrm{\mid }<\delta$. Put $\xi ={\mathrm{\Pi }}_{N}x$. For any tuple $q$ meeting the tail, delete its indices at most $N$ and call the remaining tuple $q^{\mathrm{\prime }}$. Concatenating $p$ and $q^{\mathrm{\prime }}$ in the definition of $R(x)$ gives

Fix a tuple $q$. If it lies wholly before or after $N$, the two [given, step 2.1, step 3.1] contractive estimates for ${\mathrm{\Pi }}_{N}x$ and $x-{\mathrm{\Pi }}_{N}x$ are immediate. If it crosses $N$, delete respectively the tail or the head. Expanding the one new jump to zero shows that the resulting $q$-variation equals an auxiliary endpoint variation $r_{q^{\mathrm{\prime }}}(x)$, hence is at most $R(x)\le \mathrm{\parallel }x{\mathrm{\parallel }}_J$ by step 2.1. Taking suprema proves both contractive inequalities.

Given $x$ and $\epsilon >0$, step 3.1 gives finite-support $u$ with [given, L0, L3, step 3.1, step 4.1] $\mathrm{\parallel }x-u{\mathrm{\parallel }}_J<\epsilon$. For $N$ beyond its support, step 4.1 gives $\mathrm{\parallel }x-{\mathrm{\Pi }}_{N}x{\mathrm{\parallel }}_J=\mathrm{\parallel }(I-{\mathrm{\Pi }}_N)(x-u){\mathrm{\parallel }}_J\le \mathrm{\parallel }x-u{\mathrm{\parallel }}_J<\epsilon$. Coordinatewise uniqueness in the labeling of [L0] is immediate, so [L3] makes $(e_n)$ a Schauder basis.

Finite-support sequences with rational coordinates form a countable set. [given, step 3.1, step 5.1] They are dense by step 3.1 and finite-dimensional rational approximation, so $J$ is separable.