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

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 $\sum_{k=1}^{\infty} |\langle x,e_k \rangle|^2 < \|x\|^2 ?$

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.
• If I understood it correctly, the reason is that the number of $(e_k)$'s is infinite even after dropping some of them we still have summation from $1$ to $\infty$; which is not working for a finite dimensional case. That is, inequality holds only because of not considering some terms $\langle x, e_k\rangle e_k$, otherwise it's always equality. Am I right? – user231343 Aug 29 '17 at 6:59
• It is rather the other way round. In a separable Hilbert space any orthonormal family $e_k$ can be completed into a complete orthonormal family by adding another sequence of $e_k$'s. The family is complete if one cannot add an orthonormal vector. In this case the closed linear span is the whole space. Note that an inner product space is more general than a Hilbert space. – Gribouillis Aug 29 '17 at 7:11
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 $\|x\|^2=1$.