13.9 Schauder Bases
Definition 13.9.1 (Schauder Basis).label Let $E$ be a separable Banach space over $K \in \RC$ and $\seq{x_n}\subset E$, then $\seq{x_n}$ is a Schauder basis of $E$ if for each $x \in E$, there exists a unique $\seq{\lambda_n(x)}\in K^{\natp}$ such that
The sequence of mappings $\seq{\lambda_n}\subset K^{E}$ is the coefficient forms of $\seq{x_n}$, and the Schauder basis $\seq{x_n}$ is normalised if $\norm{x_n}_{E} = 1$ for all $n \in \natp$.
Proposition 13.9.2.label Let $E$ be a separable Banach space over $K \in \RC$, $\seq{x_n}\subset E$ be a normalised Schauder basis, and $\seq{\lambda_n}\subset K^{E}$ be its coefficient forms, then:
- (1)
$\seq{\lambda_n}\subset E^{*}$ is equicontinuous.
- (2)
For each $N \in \natp$ and $x \in E$, let $P_{N}x = \sum_{n = 1}^{N} \dpn{x, \lambda_n}{E}x_{n}$, then $P_{N} \to \text{Id}$ uniformly on compact sets as $N \to \infty$.
Proof, [III.9.6, SW99]. By uniqueness of the basis decomposition, $\seq{\lambda_n}\subset \hom(E; K)$. For each $x \in E$, let
then $\norm{x}_{E}\le \norm{x}_{E'}$, and $E$ is complete with respect to $\norm{\cdot}_{E'}$. By the Open Mapping Theorem, $\norm{\cdot}_{E}$ is equivalent to $\norm{\cdot}_{E'}$, and there exists $C \ge 0$ such that $\norm{x}_{E'}\le C\norm{x}_{E}$ for all $x \in E$.
(1): For each $N \in \natp$ and $x \in E$, since $\seq{x_n}$ is normalised,
Hence $\seq{\lambda_n}\subset E^{*}$ with $\sup_{n \in \natp}\norm{\lambda_n}_{E^*}\le 2C$.
(2): Since $\sup_{N \in \natp}\norm{P_N}_{L(E; E)}\le C$, $\seq{P_N}$ is equicontinuous. By the Arzelà-Ascoli Theorem, $P_{N} \to \text{Id}$ uniformly on compact sets as $N \to \infty$.$\square$
Corollary 13.9.3.label Let $E$ be a separable Banach space over $K \in \RC$ with a Schauder basis, then $E$ enjoys the approximation property.
Proof. By Proposition 13.9.2, there exists $\seq{P_N}\subset L(E; E)$ such that $P_{N} \to \text{Id}$ uniformly on compact sets as $N \to \infty$.$\square$
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