I am currently reading Hilbert Spaces and confused about a thing. Say, $C=\{e_\alpha : \alpha\in\mathcal{A}\}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $\sum_{\alpha\in\mathcal{A}}e_\alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?

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    $\begingroup$ Even when $\mathcal A$ is countably infinite the sum $\sum e_n$ does not exist either in the norm or weakly. $\endgroup$ Commented Jan 4, 2019 at 10:20
  • $\begingroup$ Oh, I see if it was a well defined vector, then the norm would be in $\mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy $\endgroup$
    – Partha
    Commented Jan 4, 2019 at 10:23
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    $\begingroup$ If $\sum e_n=x$, say in weak topology, then we have $\sum \langle y,e_n\rangle =\langle y,x\rangle$ for all $y$. You get a contradiction by taking $y=\sum \frac 1 j e_j$. $\endgroup$ Commented Jan 4, 2019 at 10:27

3 Answers 3


In an infinite dimensional Hilbert space the term $$ \sum_{a\in \mathcal A} e_a $$ is not well defined. Even for countable $\mathcal A$ this sum does not converge.

For defining uncountable sums it is usually required that at most countable many summands are nonzero and that the countable sum over the nonzero entries converges absolutely.


The sum $\sum_{\alpha\in A} v_\alpha$ does make sense. It is defined to converge to $L \in H$ if $$\forall \varepsilon > 0 \,\exists F_0 \subseteq A \text{ finite such that }\forall F \subseteq A \text{ finite}, F \supseteq F_0 \text{ we have} \left\|\sum_{\alpha\in F}v_\alpha - L\right\| < \varepsilon$$

However, the sum $\sum_{\alpha\in A} e_\alpha$ of an orthonormal set only converges if $A$ is finite.

Indeed, assume that $\sum_{\alpha\in A} e_\alpha = L$. For $\varepsilon = \frac12$ there exists $F_0 \subseteq A$ finite such that for all $F \subseteq A$ finite, $F \supseteq F_0$ implies $\left\|\sum_{\alpha\in F} e_\alpha -L\right\| < \frac12$.

For any $F_0 \subseteq F \subseteq A$ finite we have $$\sqrt{|F \setminus F_0|}= \left\|\sum_{\alpha\in F\setminus F_0}e_\alpha\right\| = \left\|\sum_{\alpha\in F}e_\alpha - \sum_{\alpha\in F_0}e_\alpha\right\| \le \left\|\sum_{\alpha\in F}e_\alpha - L\right\| +\left\|L- \sum_{\alpha\in F_0}e_\alpha\right\| < 1$$

so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.


The background of your question is how do you define $\displaystyle \sum_{i \in I} c_i$ in a Banach space for any set $I$?

The usual way is to say that the sum $\displaystyle \sum_{i \in I} c_\alpha$ of a set $\mathcal C = \{c_i \ ; \ i \in I\}$ of vectors exists when following Cauchy criteria is met:

$$(\forall \epsilon > 0) \, (\exists J_0 \in \mathcal F(I)) \, (\forall K \in \mathcal F(I \setminus J_0)) \, \left\Vert \displaystyle \sum_{k \in K} c_k \right\Vert< \epsilon $$

Where $\mathcal F(A)$ is defined as the sets of finite subsets of $A$.

This has interesting consequences.

  1. Such a sum $\displaystyle \sum_{i \in I} c_\alpha$ does make sense only if the set of non zero elements of $\mathcal C$ is at most countable.
  2. For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $\epsilon = 1/2$ in the definition above; you'll get a contradiction.

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