Question: Let $H$ be Hilbert, $(\phi_n)_{n=1}^\infty$ an orthonormal basis for $H$. Define $T(\phi_n) = \dfrac{\phi_{n}}{n}$.
How can I get a general formula for $Tx,\ x \in H$?
Solution: The obvious answer is $Tx = \displaystyle\sum_{n=1}^\infty \langle \phi_n , x \rangle \dfrac{\phi_n}{n}$, but I can't justify this.
\begin{align*}Tx &= T \left( \sum_{n=1}^\infty \langle \phi_n, x \rangle \phi_n \right) \\ &= T \left( \lim \limits_{k \to \infty} \sum_{n=1}^k \langle \phi_n, x \rangle \phi_n \right) \\ &\stackrel{*}{=} \lim \limits_{k \to \infty} T \left( \sum_{n=1}^k \langle \phi_n, x \rangle \phi_n \right) \\ &= \lim \limits_{k \to \infty} \sum_{n=1}^k \langle \phi_n, x \rangle T \phi_n \\ &= \sum_{n=1}^\infty \langle \phi_n , x \rangle \dfrac{\phi_n}{n} \end{align*} and step $*$ follows from boundedness of $T$ . . . but I can't show $T$ is a bounded operator since defining $Tx$ relies on boundedness. . .
Likely just a conceptual error on my part; please let me know if you have any input.