Let $\{e_n\}_{n=1}^{\infty}$ be an orthonormal basis of the complex Hilbert space $\ell^2$. Fix complex numbers $\lambda_1,\lambda_2,\lambda_3,\dots$, let
$$ \mathscr{D}(T)=\{\sum_{n=1}^{\infty} x_n e_n\in \ell^2:\sum_{n=1}^{\infty}|\lambda_n x_n|^2< \infty\} $$
and define $T:\mathscr{D}(T) \rightarrow \ell^2$ as
$$ T\left( \sum_{n=1}^{\infty} x_n e_n \right) = \sum_{n=1}^{\infty} \lambda_n x_n e_n $$
for $\sum_{n=1}^{\infty} x_n e_n \in \mathscr{D}(T)$.
Determine the adjoint $T^*$ of $T$.
What I have done is the following.
Let $x,y\in\mathscr{D}(T)$. Then \begin{align*} |(Tx,y)| &=|(\sum_{n=1}^{\infty}\lambda_n x_n e_n,\sum_{n=1}^{\infty}y_n e_n)| \\ &=|(\lim_{m\rightarrow\infty}\sum_{n=1}^{m}\lambda_n x_n e_n,\lim_{m\rightarrow\infty} \sum_{n=1}^{m}y_n e_n)| \\ &=\lim_{m\rightarrow\infty} |(\sum_{n=1}^{m}\lambda_n x_n e_n,\sum_{n=1}^{m}y_n e_n)| \\ &=\lim_{m\rightarrow\infty} |\sum_{n=1}^{m}(\lambda_n x_n e_n,y_n e_n)| \\ &=\lim_{m\rightarrow\infty} |\sum_{n=1}^{m}( x_n e_n, \bar\lambda_n y_n e_n)| \\ &=\lim_{m\rightarrow\infty} |(\sum_{n=1}^{m} x_n e_n, \sum_{n=1}^{m} \bar\lambda_n y_n e_n)|\\ &=|(\sum_{n=1}^{\infty} x_n e_n, \sum_{n=1}^{\infty} \bar\lambda_n y_n e_n)|\\ &=|(x,\sum_{n=1}^{\infty} \bar\lambda_n y_n e_n)| \leq\|x\| \sqrt{\sum_{n=1}^{\infty}|\lambda_n y_n|^2} \end{align*} Where the third and forth equality follow from the continuity and linearity of the inner product respectively. Since $y\in\mathscr{D}(T)$ we have that $\sum_{n=1}^{\infty}|\lambda_n y_n|^2 < \infty$. It follows that the linear function $x\mapsto(Tx,y)$ is bounded and thus continuous. So we have $\mathscr{D}(T)\subset\mathscr{D}(T^*)$.
Now let $y\in\mathscr{D}(T^*)$ then $\Lambda x = (Tx,y)$ can be extended by the Hahn-Banach theorem 3.6 to a linear continuous function on $H$. By theorem 12.5 there exist an element $T^*y \in H$ such that $$ \Lambda x =(x,T^* y) $$ and $\|\Lambda\| = \|T^* y\|$. Since we also have that \begin{align*} \Lambda x &= (Tx,y) \\ &= (\sum_{n=1}^{\infty} \lambda_n x_n e_n, \sum_{n=1}^{\infty} y_n e_n) \\ &= \sum_{n=1}^{\infty} (\lambda_n x_n e_n,y_n e_n) \\ &= \sum_{n=1}^{\infty} (x_n e_n,\bar\lambda_n y_n e_n) \\ &= (x, \sum_{n=1}^{\infty} \bar\lambda_n y_n e_n). \\ \end{align*} It follows that $(x,T^* y) = (x, \sum_{n=1}^{\infty} \bar\lambda_n y_n e_n)$ and by theorem 12.7 that $T^* y=\sum_{n=1}^{\infty} \bar\lambda_n y_n e_n$. So $\sum_{n=1}^{\infty} |\lambda_n y_n|^2=\|\sum_{n=1}^{\infty} \bar\lambda_n y_n e_n\|^2 = \|(T^* y)\|^2 < \infty$ and $y \in \mathscr{D}(T)$. We conclude that $\mathscr{D}(T)=\mathscr{D}(T^*)$.
Furthermore since $\mathscr{D}(T)$ is dense $T^* y$ is uniquely defined.
My question now is if I'm done?
