Define A as: $ A\equiv \left \{ x\in l^{2}:\sum_{1}^{\infty}n^{2}{\left |x\left (n\right )\right |}^{2}<{\infty} \right \} $ Define a linear operator $T: A\rightarrow l^{2}$ (where $l^{2}$ is sequence space) by: $Tx(n)=nx(n)$ for $x\in A $ and $n=1,2,...$ I am trying to show that T satisfies the (inner product) condition: $(Tx,y)=(y,Tx) \ \forall x,y \in A$ yet T is unbounded.
To show it is unbounded, I used the unit basis vector $e_{n}$ to suggest: $\left \|Te_{n} \right \|=\left \|ne_{n} \right \|=n\left \|e_{n} \right \|$ therefore it is unbounded. Is this a valid approach? Secondly I am unsure how to show $(Tx,y)=(y,Tx)$ as I essentially have $\sum_{1}^{\infty}nx(n)\bar{y(n)}=\sum_{1}^{\infty}y(n)\bar{nx(n)}$ and I am unsure how to prove equality--any helpful hints?
I assume you wanted to write $(Tx, y) =(y, Tx)$, as the other one, $(Tx, y) =(y, Tx)$ would only mean that $(Tx,y)$ is real, which is obviously true for arbitrary $T$ if you work over $\Bbb R$, but is false over $\Bbb C$.
Your argument for unboundedness is correct.
To show the self-adjoint property: $$(Tx, y) =\sum_n\overline{n x_n}y_n=\sum_n n\bar x_ny_n=(x, Ty)\,. $$