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 $X$.

Definition 1 implies Definition 2
Suppose that:


 * for each $x \in X$, there exists a unique sequence $\sequence {\map {\alpha_j} x}_{j \in \mathop \N}$ 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.

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.