Find the adjoint of an operator $T\colon \ell^1\to\ell_0^\infty$

Question

Let $\ell_0^\infty=\{x=(x_k)_{k\in\Bbb{N}};x_k\to0\}$. I already proved that the operator $T\colon\ell^1\to\ell_0^\infty$ given by $$T(x_1,x_2,x_3,...):=\left(\sum_{j=1}^\infty x_j,\sum_{j=2}^\infty x_j,\sum_{j=3}^\infty x_j,...\right)$$ is well defined and belongs to $B(\ell^1;\ell_0^\infty)$. Of course $(Tx)_k=\sum_{j=k}^\infty x_j$ is the k-th coordinate of the vector $Tx$.

Now I need to find the k-th coordinate of the adjoint of $T$, in the sense that, if $T^\prime$ denotes the adjoint of $T$, it has the form $T^\prime\colon(\ell_0^\infty)^\prime\to(\ell^1)^\prime$. But we already konw that we can make the identification $$(\ell^1)^\prime=\ell^\infty\qquad\text{and}\qquad(\ell_0^\infty)^\prime=\ell^1,$$ hence our adjoint operator is $T^\prime\colon\ell^1\to\ell^\infty$.

The thing is, I never face this kind of problem in my life before, and in the references that my teacher is usying don't have anything like it.

Thanks for any atention or help! :)

Answer 1

$T'((a_n)) (e_i)=(a_n) (Te_i)=(a_n) (1,1,...,1,0,0...)=a_1+a_2+...+a_i$. Here there are $i$ one's in $(1,1,...,1,0,0...)$ and $e_i$ is the sequnce with $1$ in the $i-$ th place and $0$ elsewhere.

I am writing $(a_n)(b_n) $ for the value of $(a_n) \in \ell^{1}=(\ell_0^{\infty})'$ at $(b_n) \in \ell_0^{\infty}$ so $(a_n)(b_n) =\sum_n a_nb_n$.

Answer 2

$T'$ sends an $f\in (\ell_0^\infty)'$ to a $T'f\in (\ell^1)'$ given by $T'f(x)=f(Tx).$ To each such $f$ there is a unique $y\in \ell^1$ such that $f(z)=\sum y_kz_k.$ On the other hand, to $T'f$ there is a unique $w\in \ell^\infty$ such that $T'f(x)=\sum w_kx_k.$

To determine $w$ all we need to do is set $x=j^{th}$ canonical basis vector for $\ell^1$ and put the information in the previous paragraph together. We get $w_j=(f(Tx))_j=(f\overset{k\ \text{times}}{\overbrace{(1,1,\cdots, 1}},0,0,0,\cdots ,))_j=\sum_{k=1}^jy_k$.

Finally, using the identifications $(\ell^1)^\prime=\ell^\infty\ \text{and}\ (\ell_0^\infty)^\prime=\ell^1,$ we have a formula for the adjoint, namely $y_j\mapsto \sum_{k=1}^jy_k.$