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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.

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What examples do you know of normed spaces which aren't Banach spaces? What examples do you know of sequences in them which don't converge? –  Qiaochu Yuan Oct 9 '12 at 6:50
    
@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. –  blindman Oct 9 '12 at 9:30

2 Answers 2

up vote 2 down vote accepted

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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Dear Sir. You should write $$\displaystyle\|x\|=\sup_{n\in\mathbb{N}}|x_n|$$. –  blindman Oct 9 '12 at 10:17
    
Dear Sir. Thank you for your nice solution. Please take your time to check my solution and give some your useful comments. –  blindman Oct 9 '12 at 10:25

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 $$ n_1<n_2<\ldots<n_k<\ldots $$ Let $y_k=x_{n_{k+1}}-x_{n_k}$. Then $$ \sum_{k=1}^{\infty}\|y_k\|=\sum_{k=1}^{\infty}\|x_{n_{k+1}}-x_{n_k}\|\leq\sum_{k=1}^{\infty}\frac{1}{2^k}<+\infty. $$ Hence $\sum_{k=1}^{\infty}\|y_k\|$ is convergent. We observe that $$ \sum_{k=1}^{m}\|y_k\|=x_{n_{m+1}}-x_{n_1} $$ We conclude that $\displaystyle\sum_{k=1}^{\infty}y_k$ is not convergent. If this sequence is convergent then $\{x_{n_m}\}$ is convergent. In addition $\{x_n\}$ is a Cauchy sequence. Then $\{x_n\}$ is convergent which is an absurd.

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 $$ x_n(t)=\begin{cases} 1& 0\leq t\leq \frac{1}{2}\\ 0& \frac{1}{2}+\frac{1}{2n}\leq t\leq 1\\ n+1-2nt& \frac{1}{2}\leq t \leq \frac{1}{2}+\frac{1}{2n}. \end{cases} $$

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