I am wondering if the two definitions of Schauder bases are equivalent. First definition:
Definition 1. Given a Hilbert space $\mathcal{H}$, a Schauder basis is a countable set $S = \{ v_{n} \}$ such that every $v\in \mathcal{H}$ is uniquely written in the form $v = \sum \alpha_{n}v_{n}$ for some coefficients $\alpha_{n}\in F$.
Second definition:
Definition 2. Given a Hilbert space $\mathcal{H}$, a Schauder basis is a countable set $S = \{ v_{n} \}$ such that (1) every $v\in\mathcal{H}$ is of the form $v = \sum \alpha_{n}v_{n}$, and (2) if $\sum \alpha_{n}v_{n} = 0$ then $\alpha_{n} = 0$ for all $n$.
All references I've seen use the first definition, and the second one is my own.
I'm curious to know if the following proof is valid or if it has a mistake.
Proof. Showing #1 implies #2 is very simple. If a basis $S\subseteq \mathcal{H}$ satisfies #1, then the zero vector is uniquely written as an (infinite) linear combination of elements of $S$. But the linear combination where all coefficients are zero gives us the zero vector, so by uniqueness all instances where $\sum \alpha_{n}v_{n} = 0$ must have $\alpha_{n} = 0$ for all $n$. Hence $S$ satisfies #2.
Conversely, suppose $S\subseteq\mathcal{H}$ satisfies #2. By definition, it must then satisfy the spanning property, and it is only left to show that if
$$ v = \sum \alpha_{n}v_{n} = \sum \beta_{n} v_{n} $$
for any $v\in\mathcal{H}$ then $\alpha_{n} = \beta_{n}$ for all $n$.For this, we unpack the definition of the infinite sums. Take any $\epsilon > 0$.
By definition there exists $N' \in\mathbb{N}$ such that
\begin{align} & N > N' \implies \left\langle \sum_{n=1}^{N} \alpha_{n}v_{n} – v,\; \sum_{n=1}^{N} \alpha_{n}v_{n} – v \right\rangle < \epsilon. && (1) \end{align}
Likewise, there exists $N'' \in\mathbb{N}$ such that
\begin{align} & N > N'' \implies \left\langle \sum_{n=1}^{N} \beta_{n}v_{n} – v,\; \sum_{n=1}^{N} \beta_{n}v_{n} – v \right\rangle < \epsilon. && (2) \end{align}Let $M = \max(N', N'')$. In general, we have
\begin{align*}
&\left\langle \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n}, \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n} \right\rangle
\\[1.6ex]
&= \left\langle \left(\sum_{n=1}^{N} \alpha_{n}v_{n} – v\right) – \left( \sum_{n=1}^{N} \beta_{n}v_{n} – v\right), \left(\sum_{n=1}^{N} \alpha_{n}v_{n} – v\right) – \left( \sum_{n=1}^{N} \beta_{n}v_{n} – v\right) \right\rangle.
\end{align*}
By writing $A = \sum_{n=1}^{N} \alpha_{n}v_{n} – v$ and $B = \sum_{n=1}^{N} \beta_{n}v_{n} – v$, the above simplifies to
\begin{align*}
\left\langle \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n}, \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n} \right\rangle = \langle A, A \rangle – \langle A, B \rangle – \langle B, A \rangle + \langle B, B \rangle.
\end{align*}By positive-definiteness of the inner-product, the LHS is nonnegative and therefore the RHS is nonnegative, which means we can write
\begin{align*}
\left\langle \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n}, \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n} \right\rangle &= |\langle A, A \rangle – \langle A, B \rangle – \langle B, A \rangle + \langle B, B \rangle | \\
&\le \langle A, A \rangle + |\langle A, B\rangle | + |\langle B, A\rangle | + \langle B, B \rangle.
\end{align*}Now if $N > M$, then $\langle A, A\rangle, \langle B, B \rangle\le \epsilon$ due to $(1)$ and $(2)$. For the other two terms, the Cauchy-Schwartz inequality gives $|\langle A, B\rangle |^{2} = |\langle B, A\rangle |^{2} \le \langle A, A\rangle \langle B, B\rangle \le \epsilon\cdot\epsilon$.
Therefore, for $N > M$ we have
\begin{align*}
\left\langle \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n}, \sum_{n=1}^{N} (\alpha_{n}-\beta_{n})v_{n} \right\rangle \le 4\epsilon.
\end{align*}
By definition of infinite summation, we have
$$ \sum_{n=1}^{\infty} (\alpha_{n} – \beta_{n})v_{n} = 0. $$
Since the set $S = \{ v_{n} \}$ satisfies #2, it follows that $\alpha_{n} = \beta_{n}$, which proves that it satisfies #1 as well.
$$\tag*{$\blacksquare$}$$
Best Answer
More simply,
(i). If $S$ satisfies Def'n 1 and $\sum a_nv_n=0=\sum 0v_n$ then every $a_n=0$ by uniqueness, so $S$ satisfies Def'n 2.
(ii). If S satisfies Def'n 2 and $v=\sum a_nv_n=\sum a'_nv_n$ then $$0=\lim_{M\to \infty}\|v-\sum_{n=1}^Ma_nv_n\|=\lim_{M\to \infty}\|v-\sum_{n=1}^Ma'_nv_n\|.$$ So $0=\lim_{M\to \infty}\|\sum_{n=1}^M(a_n-a'_n)v_n\|.$ That is, $\sum (a_n-a'_n)v_n=0,$ so every $a_n-a'_n=0.$ So $S$ satisfies Def'n 1.