Comparing two summable conditions Let $X$ be a Banach space and $(f_n)_{n\ge1}$ be a sequence in $X^*$. There are two summable conditions:
(1)$\sum_{n\ge1} f_n(x)$ is summable for each x in $X$;
(2)$\sum_{n\ge1} \phi(f_n)$ is summable for each $\phi$ in $X^{**}$;
Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).
That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!
 A: I thought of an example, hopefully is correct.
Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=\ell^1$ and $X^{**}=\ell^\infty$ (with the usual norms). The functions $f_n$ are represented in $\ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^\infty$. We have to find $a_k^n$ such that:
(1) $\sum_{k\geq1}|a_k^n|<\infty$;
(2) For every $(b_k)_{k=1}^\infty\in c_0$ it holds
$$
\sum_{n\geq 1}\sum_{k\geq1}a_k^nb_k<\infty
$$
(Remark that the dual pairing between $\ell^1$ and $c_0$ is exactly $\left\langle a,b\right\rangle=\sum_{k\geq1}a_kb_k$);
(3) There exists $(c_k)_{k=1}^\infty\in\ell^\infty$ such that 
$$
\sum_{n\geq 1}\sum_{k\geq1}a_k^nc_k=\infty
$$
(Again, the dual pairing between $\ell^\infty$ and $\ell^1$ is the same as above).
Now, write 
$$
a_k^n=\frac{1}{n^2+k^2}
$$
Point (1) is trivially true. 
For point (2), fix $(b_k)\in c_0$ and let $\bar{k}$ be such that $|b_k|\leq1/n^2$ for $k\geq\bar{k}$. Now, since $a_k^n\geq0$ and using the fact that
$$
\sum_{k\geq\bar{k}}\frac{1}{k^2+n^2}|b_k|\leq\sum_{k\geq\bar{k}}\frac{1}{k^2}\frac{1}{n^2}=O\left(\frac{1}{n^2}\right)
$$
one has
$$
\left|\sum_{n\geq 1}\sum_{k\geq1}a_k^nb_k\right|\leq\sum_{n\geq 1}\sum_{k\geq1}a_k^n|b_k|
\leq\sum_{n\geq1}\left(\sum_{k<\bar{k}}a_k^n|b_k|+O\left(\frac{1}{n^2}\right)\right)
$$
The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
\sum_{k<\bar{k}}\sum_{n\geq1}a_k^n|b_k|=\sum_{k<\bar{k}}|b_k|\left(\sum_{n\geq1}a_k^n\right)
$$
which is finite since $\sum_{n\geq1}a_k^n<\infty$.
Now, for point (3) we could simply choose the element $(1,1,\dots)\in\ell^\infty$, so that all we need to prove is that 
$$
\sum_{n\geq 1}\sum_{k\geq 1}\frac{1}{n^2+k^2}=\infty.
$$
I suspected it was true from a comparison with the $\mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $\sum_{n,k\geq 1}\frac{1}{n^2+k^2}$ divergent?.
I hope it works!
P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $\ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $\mathbb{N}\times\mathbb{N}$, in our case
$$
a(n,k)=\frac{1}{n^2+k^2}
$$
whose sections $a(\cdot,k_0):\mathbb{N}\to\mathbb{R}$ and $a(n_0,\cdot):\mathbb{N}\to\mathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $\mathbb{N}\times\mathbb{N}$.
A: I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.
I worked out another example. Take $X=\ell^1$, hence $X^*=\ell^\infty$ (with the usual norms). Let $f_n\in\ell^{\infty}$ be functionals defined as $$f_{2k-1}^i=0\qquad i\lt k
$$$$f_{2k-1}^i=1\qquad i\ge k$$$$f_{2k}^i=0\qquad i\le k$$$$f_{2k}^i=-1\qquad i\gt k$$
That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$
Let $(A_n)_{n=1}^{\infty}$ be another sequence in $\ell^\infty$ defined as$$A_n=\sum_{k=1}^{n}f_k$$
Then for every $x=(x_i)_{i=1}^\infty\in \ell^1$ and every $k \in \mathbb{N}^*$ we have$$A_{2k-1}(x)=\sum_{i=1}^{\infty}x_i$$$$A_{2k}(x)=\sum_{i=1}^{k}x_i$$$$\sum_{n=1}^{\infty}f_n(x)=\lim_{n\to\infty}A_n(x)=\sum_{i=1}^{\infty}x_i$$
by this we have checked that $\sum_{n=1}^{\infty}f_n(x)$ is summable for each $x \in \ell^1$.
Now pick a $\phi\in\ell^{\infty*}$ so that $\sum_{n=1}^{\infty}\phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $\ell^{\infty*}$ is exactly). To do this first define $\phi$ for $g=(g^i)_{i=1}^{\infty}\in c$(taken as a subspace of $\ell^{\infty}$) as$$\phi(g)=\lim_{i\to\infty}g^i$$
You can easily check that $\phi\in c^*$. By Hahn-Banach Theorem $\phi$ has a norm-preserving extension on $l^{\infty}$, which we still denote as $\phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$\phi(f_n)=\lim_{i\to\infty}f^i=(-1)^n$$
and $\sum_{n=1}^{\infty}(-1)^n$ is obviously asummable. This finishes the example.
