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Thanks to the Riemann theorem we know that absolute convergence and unconditional convergence are the same for $\mathbb{R}$. In all the Frechet spaces absolute convergence implies unconditional convergence. There are counterexamples for the converse implication in some spaces - for example in $\ell_p$ ($p>1$) the series $\sum_{n=1}^\infty \frac{1}{n}e_n$ is unconditionally convergent but not absolutely convergent (where $e_n$ is $1$ for $n$ and $0$ otherwise). However it doesn't work for $\ell_1$. Here is my question:

Does unconditional convergence imply absolute convergence in $\ell_1$?

I can't come up with any argument or a counterexample. I'm particularly interested in the real case if it matters.

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up vote 6 down vote accepted
+500

Here is an explicit example of an unconditionally convergent series in $\ell_1$ that is not absolutely convergent.

Let $$ A_0=\left[\matrix {1}\right] $$ and define $x_1=(1,0,0,\ldots)$.

Set $$\textstyle A_1=\left[\matrix {{1\over2}&\phantom{-}{1\over2}\cr{1\over2} &-{1\over2}\cr}\right]. $$

Define $x_2$ and $x_3$ to be the vectors in $\ell_1$ corresponding to the rows of $A_1$ padded with a zero in the first coordinate and padded with zeros on the right.

The rows of $A_1$ are essentially the first two Rademacher functions seen as elements in $\ell_1$. By Khintchine's inequality, we have $$ \biggl\Vert\,\sum_{i=2}^3 c_i x_i\,\biggr\Vert_{\ell_1} \le K \Bigl(\,\sum_{i=2}^3 c_i^2\,\Bigr)^{1/2}, $$ for any choice of scalars $c_i$, where $K$ is the constant in Khintchine's inequality.

Let $$\textstyle A_2=\left[\matrix { {1\over8} &\phantom{-}{1\over8} &\phantom{-}{1\over8} &\phantom{-}{1\over8} &\phantom{-} {1\over8} &\phantom{-} {1\over8} &\phantom{-}{1\over8} &\phantom{-} {1\over8} \cr {1\over8} &\phantom{-} {1\over8} &\phantom{-}{1\over8} &\phantom{-}{1\over8} &-{1\over8} & -{1\over8} &-{1\over8} &-{1\over8} \cr {1\over8} &\phantom{-} {1\over8} &-{1\over8} &-{1\over8} &\phantom{-}{1\over8} &\phantom{-} {1\over8} &- {1\over8} & -{1\over8} \cr {1\over8} &- {1\over8} &\phantom{-}{1\over8} &-{1\over8} &\phantom{-}{1\over8} &-{1\over8} &\phantom{-}{1\over8} &-{1\over8} \cr }\right]. $$

Define $x_4$, $x_5$, $x_6$, and $x_7$ to be the vectors in $\ell_1$ corresponding to the rows of $A_2$ padded with zeroes in the first three coordinates and padded with zeros on the right.

The rows of $A_2$ are essentially the first four Rademacher functions seen as elements in $\ell_1$. By Khintchine's inequality $$ \biggl\Vert\,\sum_{i=4}^7 c_i x_i\,\biggr\Vert_{\ell_1} \le K \Bigl(\,\sum_{i=4}^7 c_i^2\,\Bigr)^{1/2}, $$ for any choice of scalars $c_i$.

Next we construct $A_3$ to be the $8\times128$ "Rademacher matrix". We define $x_8$, $x_9$, $\ldots\,$, $x_{15}$ to be the vectors in $\ell_1$ corresponding to the rows of $A_3$ padded appropriately (disjointly supported from the previous $x_i$). We then have $$ \biggl\Vert\,\sum_{i=8}^{15} c_i x_i\,\biggr\Vert_{\ell_1} \le K \Bigl(\,\sum_{i=8}^{15} c_i^2\,\Bigr)^{1/2}, $$ for any choice of scalars $c_i$.

$$\vdots$$

Consider the sum $\sum\limits_{i=1}^\infty {1\over i} x_i$.

Let $n$ be an integer. Choose the largest integer $m$ with $2^m\le n$ and set $$ c_i=\cases{\textstyle{1\over i},&$i\ge n$\cr 0,\strut& otherwise.} $$ We then have, using the "disjointness" of the $A_i$, for any choice of signs $\{\epsilon_i\}$: $$\eqalign{ \biggl\Vert\sum_{i=n}^\infty \, {\textstyle{1\over i}}\epsilon_i x_i\biggr\Vert_{\ell_1}\, &=\sum_{i=m}^\infty \,\, \biggl\Vert \,\sum_{x_j\in A_i} \epsilon_j{c_j} x_j\,\biggr \Vert_{\ell_1} \cr &\le K\sum_{i=m}^\infty \Bigr( \,\sum_{j=2^i}^{2^{i+1}-1} {\textstyle{(\epsilon_j c_j)^2}}\,\Bigl)^{1/2}\cr &\le K\sum_{i=m}^\infty \Bigr( \,\sum_{j=2^i}^{2^{i+1}-1} {\textstyle{1\over j^2}}\,\Bigl)^{1/2}\cr &\le K\sum_{i=m}^\infty \bigl(\textstyle{1\over2^i}\bigr)^{1/2}\cr &= {K (1/\sqrt2)^{m-1}\over (\sqrt2-1)}\cr &\buildrel{n\rightarrow\infty}\over\longrightarrow0. } $$ It follows that the sum $\sum\limits_{i=1}^\infty {1\over i} x_i$ is unconditionally convergent in $\ell_1$.

But, noting that the $\ell_1$-norm of any $x_i$ is 1, we have $$ \sum\limits_{i=1}^\infty \bigl\Vert {\textstyle{1\over i}}x_i\bigr\Vert_{\ell_1} =\sum\limits_{i=1}^\infty{\textstyle{1\over i}}=\infty; $$ and thus $\sum\limits_{i=1}^\infty {1\over i} x_i$ is is not absolutely convergent in $\ell_1$.

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This is very nice! –  t.b. Mar 15 '12 at 1:48
    
Great, just what I needed! Thank you. –  savick01 Mar 15 '12 at 18:02
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No it does not. More generally, there are unconditionally convergent series which are not absolutely convergent in any infinite-dimensional Banach space. A proof of this fact can be found in this 1950 article by A. Dvoretzky and C. A. Rogers.

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Thank you Alex! +1. And thanks to Community [;)], since I have no access to jstor. Alex, can you propose a direct counterexample? I haven't read through the whole article carefully yet, but it seems complex and I hope that for $\ell_1$ there might be a self-explanatory counterexample. –  savick01 Mar 14 '12 at 22:14
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