Example of a maximal orthonormal set in a non-Hilbert space

Question

We know that the following statements are equivalent in a Hilbert space for an orthonormal set $\left\lbrace e_{\alpha} | \alpha \in \Delta \right\rbrace$:

1. For each $x \in H$, we have

$$\| x \|^2 = \sum\limits_{\alpha \in \Delta} \left| \langle x, e_{\alpha} \rangle \right|^2$$

2. The set $\left\lbrace e_{\alpha} | \alpha \in \Delta \right\rbrace$ is a maximal orthonormal set in $H$.

I am trying to find an example where the equivalence fails if the underlying space is not complete. Clearly, (1) always implies (2) whether or not the space is complete. So, the equivalence can fail when we have a maximal orthonormal set but the Parseval's identity (1) is not satisfied.

I was thinking about $\mathscr{C} \left[ -1, 1 \right]$, the set of all real-valued continuous functions on the interval $\left[ -1, 1 \right]$, equipped with the inner product

$$\langle f, g \rangle = \int\limits_{-1}^{1} f \left( x \right) g \left( x \right) \mathrm{d}x.$$

This is not a Hilbert space. Now, the problem is to get a suitable maximal orthonormal set. I tried using the Gram-Schmidt orthonormalization process on the set $\left\lbrace 1, x, x^2, \cdots \right\rbrace$. But, the first few vectors obtained from this process are not in a "pattern" that can be generalized. Is there a way we can get a maximal orthonormal set in $\mathscr{C} \left[ -1, 1 \right]$ for which the Parseval's identity does not hold?

Answer 1

The two conditions are equivalent even if $H$ is an incomplete inner product space. To see this apply the equivalence to the completion of $H$ (and recall that $H$ is dense in its completion).