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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:

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

B. Salkas
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  • Now answered at MO: http://mathoverflow.net/questions/83977/second-conjugate-operators-to-operators-on-c-0 – B. Salkas Dec 21 '11 at 00:19
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    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 02:18
  • @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 08:18

1 Answers1

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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.

Matthew Daws
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