We have known that "A normed space $X$ is a Banach space if and only if each absolutely convergent series in X converges". We would like to find an explicitly incomplete normed space and an explicitly series in that space such that the given series is absolutely convergent but not convergent.
Series in incomplete normed space
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0@Qiaochu Yuan: I have known that $X=C^{L}_{[0,1]}$, space of continuous functions on $[0,1]$ with $\|x\|_X=\int_0^1|x(t)|dt$ is a normed space which is not Banach space. I can't construct a series in this space which is absolutely convergent but not convergent. – 2012-10-09
2 Answers
This seems to me to be a relatively simple example:
$X=$ set of all real sequences with finite support (i.e., there are only finitely many non-zero elements)
$\|x\|=\sup\limits_{n\in\mathbb N} |x_n|$
Consider the sequence $a_n=(0,\dots,0,\frac1{n^2},0,0,\dots)$ and the series $\sum a_n$ in $X$.
This series is absolutely convergent, since $\sum \|a_n\|= \sum\frac1{n^2}$.
It cannot be convergent in $X$. Take any sequence $x$ with finite support. This means that there is $n_0$ such that $x_n=0$ for each $n\ge n_0$. If $s_n=\sum\limits_{k=1}^n a_k$ denotes the $n$-th partial sum, we have $\|s_n-x\| \ge \frac1{n_0^2}$ for each $n\ge n_0$. So w have $\|s_n-x\|\not\to0$ and $s_n\not\to x$.
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0Dear Sir. Thank you for your nice solution. Please take your time to check my solution and give some your useful comments. – 2012-10-09
Following the hint of Qiaochu Yuan i found the solution for my question. The construction of a series in an incomplete normed space that is absolutely convergent but not convergent follows from the proof of the theorem "A normed space X is a Banach space if and only if each absolutely convergent series in X converges". Let $X$ be an incomplete normed space and $\{x_n\}\subset X$ is a Cauchy sequence that is not convergent. For every $k\geq 1$ there exists $n_k\geq k$ such that $ \|x_p-x_q\|<\frac{1}{2^k} \quad \forall p,q\geq n_k. $ Without of loss generality we can assume that $ n_1
Note. We can choose $X=C^{L}_{[0,1]}$, the space of continuous functions on $[0,1]$ with the norm given by $ \|x\|_{X}=\int_0^1|x(t)|dt. $ Let $\{x_n(t)\}_{n\in \mathbb{N}}\subset X$ be given by $ x_n(t)=\begin{cases} 1& 0\leq t\leq \frac{1}{2}\\ 0& \frac{1}{2}+\frac{1}{2n}\leq t\leq 1\\ n+1-2nt& \frac{1}{2}\leq t \leq \frac{1}{2}+\frac{1}{2n}. \end{cases} $