Example of an absolutely summable series that is not summable

Question

When I encountered Banach spaces I was presented with some proofs that link completeness, vector spaces, series and sequences (of partial sums). In particular I was presented with the following theorem:

A normed vector space X is complete if and only if every absolutely summable series in X is summable. That is, X is complete if and only if $\sum_{n=1}^{\infty}x_n$ converges in X whenever $\sum_{n=1}^{\infty}||x_n|| < \infty$.

The problem I am facing is that I can only imagine that every absolutely summable series in X is always summable, as that is what I was taught. (This is what I thought initially; I know now it is not true.)

I then did some searching online, as well as here, and came to the conclusion that this theorem is actually the defining factor in that statement; Completeness links absolutely summable series to summable series, and conversely, when absolutely summable series are summable they define a complete space. This must mean that all 'examples' I can think of (or in other words, the scope of my imagination) are examples in Banach spaces.

This however raises the following question:

Can anyone provide me with some examples of absolutely summable series in a normed vector space that are not summable, and explain why?

To clarify further, my goal is to understand what is the defining difference between being absolutely summable and summable. I suppose that the norm (or metric) used is key, as this defines when $\sum_{n=1}^{\infty}||x_n|| < \infty$. I just don't see it yet. If there is any way to link the explanation to Cauchy sequences, as most proofs use that approach, that would be great. Thanks!

Answer 1

Take $\mathbb Q$, which is a vector space over $\mathbb Q$, and let $0.a_1a_2\ldots$ be the decimal expansion of $\sqrt2-1$. Then $\sum_{n\geq1}a_n10^{-n}$ does not converge in $\mathbb Q$, but we have that $\sum_{n\geq1}\|a_n10^{-n}\|=\sum_{n\geq1}a_n10^{-n}$ converges (to $\sqrt2-1$, of course).

For a less trivial example, take $C_c(\mathbb R)\subset L^2(\mathbb R)$, approach $\exp(-x^2)\in L^2$ by a sequence $(u_n)$ of continuous functions with compact supports, rewrite this as a series $\sum_{n\geq1}f_n$ with $f_n=u_n-u_{n-1}$ (agreeing on $u_0=0$). Then the series is absolutely summable, but not summable, because the limit $\exp(-x^2)$ is not in $C_c(\mathbb R)$.

The point is that an absolutely convergent series need not converge since in a non-complete space a Cauchy sequence need not converge. Each non-complete space can be used to construct such an example. Note that an absolutely summable series defines a Cauchy sequence (of its partial sums). Hence it should converge in the completion $\overline X$ of your original space $X$. The only way that the sequence does not converge in $X$, is when $X\neq \overline X$, so when $X$ is not complete.

Answer 2

Quote by Václav Mordvinov: Each non-complete space can be used to construct such an example.

Then let's do this. Choose a Cauchy sequence $(f_n)$ in the non-complete normed space $X$, which doesn't converge. Then you find a subsequence $(f_{n_k})$ such that $\|f_{n_{k+1}} - f_{n_k}\|\le 2^{-k}$. That subsequence still doesn't converge since a Cauchy sequence with a convergent subsequence converges. Now, set $g_k := f_{n_{k+1}} - f_{n_k}$. Obviously, $\sum_k\|g_k\| < \infty$. However, $\sum_{k=1}^K g_k = \sum_{k=1}^K (f_{n_{k+1}} - f_{n_k}) = f_{n_{K+1}} - f_{n_1}$, which does not converge.

Answer 3

I will simply considere the following situation : take $B$ a Banach space and $V$ a linear subspace which is not complete. The last point is equivalent to say that $V$ is not closed. Then there exists a sequence of $V$ who has a limit in $B$ and the limit is not in $V$. Denote this sequence $(x_n)$ and $x$ the limit, fix $(y_n)$ the sequence of $V$ such that $$ \sum_{k=0}^n y_k = x_n. $$ You then have that the series $\sum y_n$ does not converges (otherwise $x$ would be in $V$) but since $$y_{n+1} = \sum_{k=0}^{n+1}y_k - \sum_{k=0}^n y_k = x_{n+1} - x_n $$ you have $$ \sum_{n \geq 0} ||y_n|| = ||x_0|| + \sum_{n=1}^\infty ||x_{n} - x_{n-1}|| \leq ||x_0|| + 2 \sum_{n \geq 0} ||x_n-x||. $$ This last series may not be convergent, but you can extract from $(x_n)$ a sequence that goes to $x$ at speed $1/n^2$. Up to replacing $(x_n)$ by one of its subsequence, you have a series of $V$ which does not converges but is absolutely summable.