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The following is Exercise 5 page 36 in Functional Analysis book of Conway:

Find the adjoint of a diagonal operator (Exercise 1.8).

The aforementioned exercise 1.8 read:

Exercise 1.8. Let $\{e_n\}_{n\in \mathbb{N}}$ be the usual basis for $l^2$ and let $\{a_n\}_{n\in \mathbb{N}}$ be a sequence of scalars. Show that there is a bounded operator $A$ on $l^2$ such that $Ae_n = a_n e_n$ for all $n \in \mathbb{N}$ if and only if $\{a_n\}_{n\in \mathbb{N}}$ is uniformly bounded, in which case $||A|| =\sup{\{|a_n|: n \ge 1}\}$. This type of operator is called a diagonal operator or is said to be diagonalizable.

Well Exercise 1.8 is easy to solve but I am struggling with Exercise 5 : $\langle Ah,g \rangle = \langle h,A^*g \rangle \implies \langle A \sum b_n e_n, \sum c_n e_n \rangle = \langle \sum b_n a_n e_n ,c_n A^*e_n \rangle$. Any calculation further than that looks not rigorous and correct.

Mittens
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    Use $h=e_j$ and $g=e_k$. – MaoWao Jun 30 '21 at 12:36
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    $$\begin{align} \langle Ah,g\rangle&=\langle \sum_na_nb_n\mathbf{e}_n,\sum_nc_n\mathbf{e}_n\rangle=\sum_na_nb_n\overline{c_n}\ &=\sum_nb_n\overline{\overline{a_n}c_n}=\langle \sum_nb_n\mathbf{e}_n,\sum_n\overline{a_n}c_n\mathbf{e}_n\rangle\ &=\langle h,A^g\rangle \end{align}$$ From this you can see that $A^\mathbf{e}_k=\overline{a_k}\mathbf{e}_k$ for all $k$. – Mittens Jun 30 '21 at 13:10

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I will assume the convention that inner product is linear on first coordinate and anti-linear on the second one. (To adapt the solution below to the other convention is trivial).

Let us prove that $A^*$ is the linear operator defined by $A^*e_n = \bar a_n e_n$.

\begin{align*} \langle Ah,g \rangle &= \langle A \sum b_n e_n, \sum c_m e_m \rangle = \\ &= \sum_{n,m} b_n \bar c_m \langle A e_n, e_m \rangle = \\ & = \sum_{n,m} b_n \bar c_m \langle a_n e_n, e_m \rangle = \\ & = \sum_{n,m} b_n \bar c_m a_n \langle e_n, e_m \rangle = \\ & = \sum_n b_n \bar c_n a_n = \\ & = \sum_{n,m} b_m \bar c_n a_n \langle e_m, e_n \rangle = \\ & = \sum_{n,m} b_m \bar c_n \langle e_m, \bar a_n e_n \rangle = \\ &= \langle \sum b_m e_m, \sum c_n \bar a_n e_n \rangle = \\ & = \langle h, A^*g \rangle \end{align*}

Ramiro
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