Existence of a space $X$ such that $\sum_{k=1}^{\infty} |\langle x,e_k \rangle|^2 < \|x\|^2 $?

Question

Recall that

Theorem (Bessel inequality). Let $(e_k)$ be an orthonormal sequence in an inner product space $X$. Then for every $x \in X $, $$\sum_{k=1}^{\infty} |\langle x,e_k \rangle|^2 \le \|x\|^2 .$$

The proof results in $\le$ and not just $=$. Can $\le$ be 'reduced' to $=$? And if so, how? By example searching I couldn't find any space $X$ such that the relation is just $<$.

For some sapces that I knew, e.g. $X=\mathbb{R^n}$, equality holds. Is there any space such that $\sum_{k=1}^{\infty} |\langle x,e_k \rangle|^2 < \|x\|^2 ?$

Answer 1

In a Hilbert space, $\sum_0^\infty |\langle x, e_k\rangle|^2 = \|P(x)\|^2$ where $P$ is the orthogonal projection on the closed linear span of the $e_k$ and one has $$P(x) = \sum_0^\infty \langle x, e_k\rangle e_k$$ This closed linear span can be smaller than the whole space. Take any orthonormal sequence and remove some $e_k$'s and you have an example.

Answer 2

In general, equality holds (Parseval's theorem) when the $e_k$ form a complete orthonormal sequence. For instance the $e^{inx}$ for $n\in\Bbb Z$ is a complete orthonormal sequence in $L^2[0,1]$. If you take a complete orthonormal sequence, and discard an element, and call that $x$, then $\left<x,e_k\right>=0$ for all elements of the new sequence, but $\|x\|^2=1$.