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Let $X$ be an inner product space, let $x \in X$ be non-zero, and let $M$ be an uncountable orthonormal subset of $X$. Then what can we say about the cardinality of the following set? $$ \{ \ v \in M \colon \ \langle x, v \rangle \neq 0 \ \}.$$

I know that, for any orthonormal sequence $(e_k)$ in $X$, the Bessel inequality $$\sum_{k=1}^\infty \vert \langle x, e_k \rangle \vert^2 \leq \Vert x \Vert^2$$ holds and hence, for all $m \in \mathbb{N}$, the the set $$\{ \ k \in \mathbb{N} \ \colon \ \vert \langle x, e_k \rangle \vert > \frac{1}{m} \ \}$$ is at most finite.

But I'm struggling to proceed from the countable case to the uncountable one.

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If $M$ is an orthonormal set (the normalisation is not necessary, by the way, that the elements are mutually orthogonal suffices), the set

$$N(x) = \{ v\in M : \langle x,v\rangle \neq 0\}$$

is at most countable.

For a finite subset $F\subset M$, we define

$$x_F := \sum_{v\in F} \langle x,v\rangle\cdot v.$$

Then

\begin{align} 0 &\leqslant \lVert x-x_F\rVert^2\\ &= \langle x-x_F, x-x_F\rangle\\ &= \langle x,x\rangle - \langle x_F,x\rangle - \langle x,x_F\rangle + \langle x_F,x_F\rangle\\ &= \lVert x\rVert^2 - \sum_{v\in F} \bigl\langle \langle x,v\rangle\cdot v,x\bigr\rangle - \sum_{v\in F} \bigl\langle x, \langle x,v\rangle\cdot v\bigr\rangle + \sum_{v\in F} \bigl\langle \langle x,v\rangle\cdot v,x_F\bigr\rangle\\ &= \lVert x\rVert^2 - 2\sum_{v\in F} \lvert\langle x,v\rangle\rvert^2 + \sum_{v\in F} \langle x,v\rangle\langle v,x_F\rangle\\ &= \lVert x\rVert^2 - 2\sum_{v\in F} \lvert\langle x,v\rangle\rvert^2 + \sum_{v\in F} \lvert \langle x,v\rangle\rvert^2\\ &= \lVert x\rVert^2 - \sum_{v\in F} \lvert\langle x,v\rangle\rvert^2. \end{align}

Now, for an arbitrary $c > 0$, consider

$$M_c(x) = \{ v\in M : \lvert\langle x,v\rangle\rvert \geqslant c\}.$$

For any finite subset $F$ of $M_c(x)$, we have by the above

$$c^2\cdot \operatorname{card} F \leqslant \sum_{v\in F} \lvert\langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2,$$

and therefore $\operatorname{card} F \leqslant \frac{\lVert x\rVert^2}{c^2}$. It follows that $M_c(x)$ is a finite set with $\operatorname{card} M_c(x) \leqslant \frac{\lVert x\rVert^2}{c^2}$, since otherwise it would have a finite subset with more than $\frac{\lVert x\rVert^2}{c^2}$ elements. Since further

$$N(x) = \bigcup_{m=1}^\infty M_{1/m}(x)$$

exposes $N(x)$ as a union of countably many finite sets, it follows that $N(x)$ is at most countable.

We had shown and used Bessel's inequality

$$\sum_{v\in F} \lvert \langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2,\tag{1}$$

for finite subsets $F$ of the orthonormal set $M$ above. Bessel's inequality generalises to arbitrary orthonormal sets, we have

$$\sum_{v\in M} \lvert\langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2\tag{2}$$

for every orthonormal set $M$ in an inner product space $H$, and every $x\in H$, but one needs to define the sum of (possibly) uncountably many terms for the left hand side of $(2)$ to make sense.

Generally, that leads to the theory of summable families in abelian topological groups or topological vector spaces, but in the case here, where all terms are non-negative real numbers, a simpler definition suffices, and we can meaningfully assign a sum - either a non-negative real number or $+\infty$ - to every family of non-negative real numbers by defining

$$\sum_{\alpha\in A} a_\alpha := \sup \left\{ \sum_{\alpha\in F} a_\alpha : F\text{ is a finite subset of } A\right\}\tag{3}$$

if $A$ is a set and $a_\alpha$ is a non-negative real number for every $\alpha \in A$.

With this definition, $(2)$ is an immediate consequence of the fact that $(1)$ holds for every finite subset of $M$.

We note that a family $\{ a_\alpha : \alpha \in A\}$ of non-negative real numbers that has a finite sum $S$ can have at most countably many strictly positive members, for the number of members that are not smaller than $\frac{1}{n}$ is bounded by $n\cdot S$ and hence finite. A similar fact is true for summable families in metrisable abelian topological groups or metrisable topological vector spaces, there a summable family can contain at most countably many non-zero members. This does not necessarily hold for summable families in non-metrisable groups or vector spaces, there a summable family can have uncountably many non-zero members.

We further note that definition $(3)$ of the sum of a family of non-negative real numbers is compatible with the familiar definition as the limit of the partial sums for the case of a countable family that is indexed by natural numbers. If $(a_n)_{n\in \mathbb{N}}$ is a sequence of non-negative real numbers, then each partial sum $\sum_{n=0}^k a_n$ of the series $\sum_{n=0}^\infty a_n$ is the sum of a finite subfamily, hence less than or equal to the sum according to definition $(3)$. Conversely, each finite subset $F$ of $\mathbb{N}$ is contained in the initial segment $\{ n : n \leqslant \max F\}$ of $\mathbb{N}$, and therefore

$$\sum_{n\in F} a_n \leqslant \sum_{n=0}^{\max F} a_n \leqslant \lim_{k\to\infty} \sum_{n=0}^k a_n,$$

which shows that the supremum of the sums of finite subfamilies is less than or equal to the limit of the partial sums.

Incidentally, this proves that series with non-negative terms can be reordered arbitrarily without changing the sum, since definition $(3)$ makes no reference to any particular ordering of the family.

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  • $\begingroup$ But, @Daniel Fischer, what if we require the sum to be the supremum of sums of all the (convergent) infinite series of the Forier co-efficients of $x$ with respect to elements of $M$? $\endgroup$ – Saaqib Mahmood May 23 '15 at 3:58
  • $\begingroup$ @SaaqibMahmuud That will come out to the same sum, but what is the most important thing here doesn't need to consider the whole sum. For every finite subset of $M$, we have the inequality, and that gives a finite bound of the number of $v\in M$ with $\lvert\langle x,v\rangle\rvert \geqslant \frac{1}{m}$. Therefore, at most countably many $\langle x,v\rangle$ can be non-zero. $\endgroup$ – Daniel Fischer May 23 '15 at 9:31
  • $\begingroup$ I would appreciate if you can write out the argument in full detail. What you've written in your answer is clear as far as the fact any finitely many of the inner products do satisfy the Bessel inequaloity, but I'm unable to convince myself as to how the Bessel inequality holds when we sum over the entire uncountable set $M$. Your definition of this "uncountable" sum as a supremum also does not seem to me to generalise the "infinite series" definition of the sum of countably many numbers. $\endgroup$ – Saaqib Mahmood May 23 '15 at 13:43
  • $\begingroup$ @SaaqibMahmuud I think I've written it out in full detail. I have now started with the proof of countability using only finite sums, and treat Bessel's inequality and sums of possibly uncountable families of non-negative real numbers afterwards. I have so far not treated the more general theory of summable families, but if requested, I can include a short summary of that. $\endgroup$ – Daniel Fischer May 24 '15 at 0:51
  • $\begingroup$ please do. Please also include a good book as a reference for such questions. $\endgroup$ – Saaqib Mahmood May 25 '15 at 20:20

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