Second conjugate operators and their representations We can think about a bounded operator $T\colon c_0\to c_0$ as a double-infinite matrix $[T_{mn}]_{m,n\geq 1}$ which acts on a sequence $a=[a_1, a_2, a_3, \ldots ]\in c_0$ in the same way as usual (finite) matrices act on vectors ($n$-tuples of scalars), i.e.
$$ Ta=  [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right] $$
Suppose $a=[a_1, a_2, a_3, \ldots ]\in \ell^\infty = (c_0)^{**}$. Does the following formula still hold:
$$ T^{**}a=  [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right] $$
 A: Over at MO, Bill Johnson have an answer which works for any Banach space with a Shrinking Basis.  As no-one has given an answer here yet, I thought I'd show a more "simple minded" calculation which works in the special case of $c_0$.
Notice that the sum
$$ \sum_n T_{mn} a_n $$
only converges, for all $(a_n)\in c_0$, if $\sum_n |T_{mn}|<\infty$.  But this is not enough to ensure that $T$ is a bounded operator into $c_0$; for that, you need precisely that
$$ \sup_m \sum_n |T_{mn}|<\infty. $$
(Edit: Actually, this only ensures that $T$ maps into $\ell^\infty$.  For $T$ to map to $c_0$, we need that the rows of the matrix of $T$, treated as a sequence in $\ell^1$, is weak$^*$-null, equivalently, that each column of the matrix tends to $0$.)
This quantity is $\|T\|$.  So we have characterised which matrices can occur.
Now let $(f_m)\in\ell^1$, so
$$ \langle (f_m) , T(a_n) \rangle = \sum_m \sum_n f_m T_{mn} a_n. $$
The double sum is absolutely convergent, so we can re-arrange, and conclude that
$$ T^*(f_m) = \Big[ \sum_m T_{mn} f_m \Big]. $$
That is, $T^*$ is matrix multiplication by the tranpose of $T$ (as we might hope!)
So finally for $(x_n)\in\ell^\infty$,
$$ \langle T^{**}(x_n) , (f_m) \rangle
= \sum_n x_n \sum_m T_{mn} f_m. $$
Again, the double sum converges absolutely as
$$ \sum_{n,m} |T_{mn}| |f_m| |x_n|
= \sum_m |f_m| \sum_n |T_{mn}| |x_n|
\leq \|f\|_1 \sup_m \sum_n |T_{mn}| |x_n| $$
$$\leq \|f\|_1 \sup_m \|x\|_\infty \sum_n |T_{mn}|
= \|f\|_1 \|x\|_\infty \|T\|. $$
So we can re-arrange and conclude that, yes,
$$ T^{**}(x_n) = \Big[ \sum_m T_{mn} x_n \big]. $$
In particular, for fixed $m$, the sequence $(T_{mn})$ is in $\ell^1$, and so
the sum genuinely converges.
