# 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:

$$T^{**}a= [T_{mn}][a_n] = \left[ \sum_{n=1}^\infty T_{mn}a_n\right]$$

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Now answered at MO: mathoverflow.net/questions/83977/… – B. Salkas Dec 21 '11 at 0:19
B. Salkas: could you post this as an answer and accept it so that the question can be considered closed, please? – t.b. Dec 21 '11 at 2:18
@t.b. I think there may be a system-imposed time limit on how soon one can post/accept one's own answer to a question. – Willie Wong Dec 21 '11 at 8:18

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.
See also my answer here for Schur's criteria when operators given by matrices map $c \to c$ and $\ell^{\infty} \to c$. – t.b. Dec 23 '11 at 19:25
Thanks! ${}{}{}$ – t.b. Dec 24 '11 at 22:30