Let $f \in L^{1}[0,1]$ and now define $(f_{n}^{$})_{n\geq0}$ as the sequence of moments, i.e.
$f_{n}^{$}=\int_{0}^{1}f(t)t^{n}dt$. Show:
$1.$ $T:f \mapsto (f_{n}^{$})_{n\geq0}$ is a well-defined, bounded, linear operator from $L^{1}[0,1]$ to $c_{0}$.
$2.$ Find its adjoint $T^{x}: \ell^{1}\to L^{\infty}[0,1]$
on $1.$ linearity is clear from the linearity of the integral. Further, for any $f \in L^{1}[0,1]$ note that for $\vert f(t)t^{n}\vert 1_{[0,1]}(t)\leq\vert f(t)\vert1_{[0,1]}(t)$ and hence by dominated convergence:
$\lim\limits_{n \to \infty}\vert\int_{0}^{1}f(t)t^{n}dt\vert\leq \lim\limits_{n \to \infty}\int_{0}^{1}\vert f(t)t^{n}\vert dt=\int_{0}^{1}\lim\limits_{n \to \infty}\vert f(t)t^{n}\vert dt=0$ and hence $(f_{n}^{$})_{n\geq0}\in c_{0}.$
For boundedness (it is not explicitly stated which norm I should use, so I assume the supremum norm) $\vert\vert Tf\vert\vert_{\infty}=\vert\vert (f_{n}^{$})_{n\geq0}\vert\vert_{\infty}=\int_{0}^{1}\vert f(t)\vert dt\Rightarrow \vert\vert T\vert\vert_{*}\leq1$
$2.$ I do not know where to start to find the adjoint.
$T^{*}: \ell^{1} \to L^{\infty}[0,1]$ is defined by $T^{*}((a_n))(g)=((a_n)) Tg=\sum_n a_n \int_0^{1} t^{n} g(t)\, dt = \int_0^{1} \sum_n a_nt^{n} g(t)\, dt$. Hence $T^{*}((a_n))=\sum_n a_nt^{n}$. [When we regard $(a_n) \in \ell^{1}$ as a continuous linear functional on $c_0$ we have $((a_n)) ((b_n))=\sum_n a_nb_n$].