If $E$ is an infinite vector space, how to interprete $\sum_{i}x_ie_i$?

Question

• Let $E$ a vector space of infinite dimension and let $(e_i)_{i\mathbb N}$ a basis. How is interpreted $$\sum_{i\in\mathbb N}x_ie_i \ \ ?$$

• Suppose now that $E$ is a Banach space with norm $\|\cdot \|$. I suppose that $$\sum_{i\in\mathbb N}x_ie_i=\lim_{n\to \infty }\sum_{i=1}^n x_ie_i,$$ in the $\|\cdot \|$ sense, i.e. $$\forall \varepsilon>0, \exists N: \forall n\in \mathbb N, n\geq N\implies \left\|\sum_{i\in\mathbb N}x_ie_i-\sum_{i=1}^nx_ie_i\right\|<\varepsilon.$$

But since $E$ is a vector space, $\sum_{i\in\mathbb N}x_ie_i\in E$ and thus exist (by definition of a vector space), but I guess it could happen that $$\lim_{n\to \infty }\sum_{i=1}^n x_ie_i$$ doesn't exist or is infinite, does it ? So how can we manage this case ?

Answer 1

Answer for the case of normed linear spaces: if you are defining 'basis' the way you do in finite dimensional spaces (the so called Hamel basis) then you can form inifinite sums (in the case of normed linear space) only when you know that the series converges. There are examples where every element has unique expansion as a convergent inifinite sum in terms of a sequence $\{e_n\}$. For infinite dimensional spaces there are various types of bases: orthogonal bases, Schauder bases etc. In the case where the vector space has no norm/metric/topology you can only form finite sums.

Answer 2

If $E=\text{Span}\{e_i\mid i\in\mathbb N\}$ has infinite dimension, then $$E=\bigoplus_{i\in\mathbb N} \text{Span}\{e_i\}:=\left\{\sum_{i=1}^n x_ie_i\mid x_i\neq 0 \text{ for a finite number of } x_i\right\}.$$

Answer 3

For the second case, if $E$ is a Hilbert space and the collection $\{e_i\}$ is orthonormal (I.e., $\left<e_i, e_j\right> = \delta_{ij}$), then $\sum x_i e_i$ converges if and only if $\sum |x_i|^2$ converges. By the triangle inequality, if the set $ \{\|e_i\|\}$ is bounded, then the series converges if $\sum |x_i|^2$ converges. So I guess the point is that if your coefficients are summable you can make the series converge by normalizing your basis elements.

Answer 4

Another nonequivalent definition is:

Let $(v_n)_{n\in\mathbb{N}}$ be a sequence of vectors in a normed space $X$. We say that $\sum_{n\in\mathbb{N}} v_n = v$ for some $v \in X$ if for every $\varepsilon > 0$ there exists a finite subset $F_0 \subseteq \mathbb{N}$ such that for every finite subset $F \subseteq \mathbb{N}$ with $F \supseteq F_0$ we have $$\left\|v - \sum_{n\in F}v_n\right\| < \varepsilon$$

It turns out that $\sum_{n\in\mathbb{N}} v_n = v$ is equivalent to the statement that $$\sum_{n=1}^\infty v_{\sigma(n)} = v$$ for all permutations $\sigma : \mathbb{N} \to \mathbb{N}$.

In this case we say that $\sum_{n\in\mathbb{N}} v_n$ converges unconditionally to $v$.