Equivalence of Definitions of Schauder Basis

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.

Let $\set {e_n : n \in \N}$ be a countable subset of $X$.

The following definitions of the concept of Schauder Basis are equivalent:

Definition 1

We say that $\set {e_n : n \in \N}$ is a Schauder basis for $X$ if and only if:

for each $x \in X$, there exists a unique sequence $\sequence {\map {\alpha_j} x}_{j \mathop \in \N}$ in $\Bbb F$ such that:

$\ds x = \sum_{j \mathop = 1}^\infty \map {\alpha_j} x e_j$

where convergence of the infinite series is understood in $\struct {X, \norm \cdot}$.

Definition 2

We say that $\set {e_n : n \in \N}$ is a Schauder basis for $X$ if and only if:

$(1): \quad$ for each $x \in X$, there exists a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ such that:

$\ds x = \sum_{j \mathop = 1}^\infty \alpha_j e_j$

$(2): \quad$ whenever $\sequence {\alpha_j}_{j \mathop \in \N}$ is a sequence in $\Bbb F$ such that:

$\ds \sum_{j \mathop = 1}^\infty \alpha_j e_j = 0$

we have $\alpha_j = 0$ for each $j \in \N$

where convergence of the infinite series is understood in $\struct {X, \norm \cdot}$.

Proof

Definition 1 implies Definition 2

Suppose that:

for each $x \in X$, there exists a unique sequence $\sequence {\map {\alpha_j} x}_{j \mathop \in \N}$ in $\Bbb F$ such that:

$\ds x = \sum_{j \mathop = 1}^\infty \map {\alpha_j} x e_j$

Clearly $\set {e_n : n \in \N}$ then satisfies $(1)$ of Definition 2.

Suppose the sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ is such that:

$\ds 0 = \sum_{j \mathop = 1}^\infty \alpha_j e_j$

Note that:

$\ds 0 = \sum_{j \mathop = 1}^\infty 0 e_j$

By hypothesis, this expansion is unique, so we have:

$\alpha_j = 0$

for each $j \in \N$, as required.

So $\set {e_n : n \in \N}$ also satisfies $(2)$ of Definition 2.

$\Box$

Definition 2 implies Definition 1

Suppose that:

$(1): \quad$ for each $x \in X$, there exists a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ such that:

$\ds x = \sum_{j \mathop = 1}^\infty \alpha_j e_j$

$(2): \quad$ whenever $\sequence {\alpha_j}_{j \mathop \in \N}$ is a sequence in $\Bbb F$ such that:

$\ds \sum_{j \mathop = 1}^\infty \alpha_j e_j = 0$

we have $\alpha_j = 0$ for each $j \in \N$

Let $x \in X$.

From $(1)$, there exists a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ such that:

$\ds x = \sum_{j \mathop = 1}^\infty \alpha_j e_j$

We show that this expansion is unique.

Let $\sequence {\beta_j}_{j \mathop \in \N}$ be another sequence in $\Bbb F$ such that:

$\ds x = \sum_{j \mathop = 1}^\infty \beta_j e_j$

Subtracting, we obtain:

$\ds 0 = \sum_{j \mathop = 1}^\infty \paren {\beta_j - \alpha_j} e_j$

Using $(2)$, we then have:

$\beta_j - \alpha_j = 0$ for each $j \in \N$

That is:

$\beta_j = \alpha_j$ for each $j \in \N$.

So the expansion is unique, and $\set {e_n : n \mathop \in \N}$ satisfies the requirement of Definition 1.

$\blacksquare$