I'm solving some exercises for my Functional Analysis Exam. Here is one on which I am stuck:
Let $T: l_1 \to c_0$ be linear operator defined as $x_n \to \sum_{k\geqslant n} x_k$. Show that $T \in B(\ell_1,c_0)$ and compute $T^*$,the adjoint of $T$
Of course $T \in B(\ell_1,c_0)$ and $\| T \| \leqslant 1$. But I am unable to find the adjoint.
For any $\varphi \in c_0^{\ast}, \exists y\in \ell^1$ such that $$ \varphi((x_n)) = \sum_{n=1}^{\infty} x_ny_n $$ Then $$ T^{\ast}(\varphi)((x_n)) = \varphi(T(x_n)) = \sum_{n=1}^{\infty} \left(\sum_{k=n}^{\infty} x_k\right) y_n = \sum_{n=1}^{\infty} x_n \left(\sum_{k=1}^n y_k \right) =: \psi((x_n)) $$ Hence, $T^{\ast}(\varphi) = \psi$ where $\psi$ is as above.
If you identify $c_0^{\ast} \cong \ell^1$ and $\ell^1 \cong \ell^{\infty}$, then one can write $T^{\ast} : \ell^1 \to \ell^{\infty}$ by $$ T^{\ast}((y_n)) = (z_n) \text{ where } z_n = \sum_{k=1}^n y_k $$ Note that $$ |z_n| \leq \|y\|_1 \quad\forall n\in \mathbb{N} $$ and hence $T^{\ast} \in B(\ell^1, \ell^{\infty})$ with $\|T^{\ast}\| \leq 1$.
Is this the kind of thing you were looking for?