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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.

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  • $\begingroup$ First step would be to look at the definition, of course. $\endgroup$
    – Jakobian
    Jul 26 '19 at 19:50
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$T^{*}: \ell^{1} \to L^{\infty}[0,1]$ 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 $((a_n)) ((b_n))=\sum_n a_nb_n$].

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  • $\begingroup$ I do not understand why you can claim $T^{*}((a_n))=\sum_{n}a_{n}t^{n}$, all we have shown so far is that $T^{*}((a_n))(g)=((a_n)) Tg=\sum_n a_n \int_0^{1} t^{n} g(t)\, dt = \int_0^{1} \sum_n a_nt^{n} g(t)\, dt$. I mean what about the integral. Something like $g \mapsto \int_0^{1} \sum_n a_nt^{n} g(t)\, dt$ would be the adjoint wouldn't it? $\endgroup$
    – SABOY
    Jul 28 '19 at 20:15
  • $\begingroup$ You have to think of the integral as $\sum a_n t^{n}$ acting on $g$. That comes from the identification of $L^{\infty}$ with the dual of $L^{1}$. $\endgroup$ Jul 28 '19 at 23:10

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