In most books/websites, Proposition 2 (see below) is either stated for a Hilbert space or proved via Riesz-Fischer. Does the follow approach (which seems to work in an inner product space) fall down somewhere?
I would be very grateful for any comments.
Let $X$ be a complex inner product space.
Suppose $S=(e_n : n \in \mathbb{N})$ is an orthonormal sequence in $X.$
Proposition 1. $S$ is a total sequence (i.e. the linear span of $S$ is dense in $X$) if and only if $$x=\sum_{k=1}^\infty(x,e_k)e_k \; \; \; \; \forall x \in X.$$
Proof. (if) Clear.
(only if) Let $\varepsilon>0$ be given. There exist $c_1,\ldots,c_N \in \mathbb{C}$ such that $$\left\|\,x-\sum_{i=1}^Nc_ie_i\,\right\|<\varepsilon.$$ However, by the "minimum property of Fourier coefficients" we have $$\left\|\,x-\sum_{i=1}^N(x,e_i)e_i\,\right\|\leq\left\|\,x-\sum_{i=1}^Nc_ie_i\,\right\|$$ and so we are done. Q.E.D.
Proposition 2. $S$ is total sequence if and only if $$\|x\|=\sum_{k=1}^\infty|(x,e_k)|^2 \; \; \; \; \forall x \in X.$$
Proof. Observe that, for any $N \in \mathbb{N}$, we have $$\left\|\,x-\sum_{i=1}^N(x,e_i)e_i\,\right\|^2=\|x\|^2-\sum_{i=1}^N|(x,e_i)|^2.$$ The result now follows by combining this observation with Proposition 1. Q.E.D.