The problem I'm trying to solve reads: Let $\{w_k\}_{k \in \mathbb{N}} \in \ell^{\infty}$, and define $T: \ell^2 \to \ell^2$, $T\left(\{x_n\}_{n\in\mathbb{N}}\right) = \{w_nx_n\}_{n \in \mathbb{N}}$. Compute $||T||$, $\sigma(T)$, and $T^*$.
So far I've come up with $||T|| = ||\{w_k\}||_{\infty}$ (using the definition of an operator norm) and $\sigma(T) = \{w_n: n \in \mathbb{N}\}$ (by computing the inverse of $\lambda I - T$).
I think I'm stuck on the third part, though. I started with $\langle Tx,y \rangle = \langle x,T^*y \rangle$ and got as far as $\sum_{n=1}^{\infty} x_n(w_n\overline{y_n} - \overline{T^*y_n}) = 0$. But I don't know what to do next.
Am I on the right track? And if not, where did I go wrong? I'm basically teaching myself this material because I've been too sick to go to class, so I'm not sure if I understand it or not.
I've worked on the third part of the problem some more; here is my proposed solution:
Let $x, y \in \ell^2$. \begin{align*} \langle Tx, y \rangle &= \langle x, T^*y \rangle\\ \langle wx,y \rangle &= \langle x, T^*y \rangle\\ \sum_{n=1}^{\infty} w_nx_n\overline{y_n} &= \sum_{n=1}^{\infty} x_n\overline{T^*y_n}\\ \sum_{n=1}^{\infty} (w_nx_n\overline{y_n} - x_n\overline{T^*y_n}) &= 0\\ \sum_{n=1}^{\infty} x_n(w_n\overline{y_n} - \overline{T^*y_n}) &= 0\\ \end{align*} Note that this equation holds for all $x,y \in \ell^2$; in particular, it holds for $x = \{e_i\}$ for each $i \in \mathbb{N}$ and $y$ such that only the $i$th term is nonzero. It follows that for each $i$, we have \begin{align*} \sum_{n=1}^{\infty} (w_n\overline{y_n} - \overline{T^*y_n}) &= w_i\overline{y_i} - \overline{(T^*y)_i} =0\\ w_i \overline{y_i} &= \overline{(T^*y)_i}\\ w\overline{y} &= \overline{T^*y}\\ \overline{w}y &= T^*y \end{align*}