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I want to check that my approach for this question makes sense. Consider the bounded linear map $T:\ell_1 \to c_0$ where $c_0$ is the space of sequences with limit zero, defined as $$ T((c_n)_n) = (c_1 + c_2 +c_3 +\dots, c_2 + c_3 + \dots, c_3 + \dots, \dots) $$ for any $(c_n)_n \in c_0$. I want to find its adjoint $T^*$. Since $c_0^* = \ell_1$ and $\ell_1^* = \ell_\infty$, the adjoint is the map: $T^*:\ell_1 \to \ell_{\infty}$ with the following property:

Let $(x_n) \in \ell_1$ and $(c_n) \in c_0$ then

\begin{align*} [T^*((x_n)) ] (c_n) = (x_n) [T((c_n))] \end{align*} expanding the right hand side gives:

\begin{align*} &=(x_n) [(c_1 + c_2 +c_3 +\dots, c_2 + c_3 + \dots, c_3 + \dots, \dots)]\\ &= (x_n) \left ( \sum_{j=n}^{\infty}c_j \right )_{n=1}^{\infty}\\ &=\sum_{n=1}^{\infty} x_n \sum_{j=n}^{\infty}c_j\\ &= \sum_{n=1}^{\infty} (n c_n)x_n \end{align*}

So $T^*$ is given explicitly by the action of the sequence $(n c_n)_n \in \ell_{\infty}$ on elements of $\ell_1$.

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Everything is perfect until the last equality, where you suddenly decided that $\sum_{j=n}^\infty c_j=nc_n$.

When you have $$\sum_{n=1}^{\infty} x_n \sum_{j=n}^{\infty}c_j,$$ you have $n\geq1$, $j\geq n$. So to switch the sums now you have $j\geq1$, $1\leq n\leq j$. Thus $$ \sum_{n=1}^{\infty} x_n \sum_{j=n}^{\infty}c_j=\sum_{j=1}^\infty c_j\left(\sum_{n=1}^j x_n\right). $$ That shows that (using $k$ in place of $n$ and $n$ in place of $j$ above) $$ T^*x=\left(\sum_{k=1}^n x_k\right)_n $$

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