Let $(a_n)$ be a sequence from $\mathbb{C}$, and define an operator $T$ on $l_2 = l_2(\mathbb{N})$ by
$Tu = (a_nu_n), \ u = (u_n) \in dom(T)$
$dom(T) = \{u = (u_n) \in l_2 | (a_nu_n) ∈ l_2\}$.
(i) Show that $T$ is a closed operator.
(ii) Give conditions under which $T$ is bounded.
(iii) Show that
$\sigma_p(T) = \{a_n | n ∈ N\}, \ \sigma (T) = \overline{\sigma_p(T)}$.
(iv) Determine $R(z; T)$ for $z \in \rho(T)$.
ad (i)
I tried to prove with triangle inequlity, but I missed something.
Let $u_n \to u$ and $Tu_n \to v$ and I have to show that $Tu=v$.
So I have $||Tu-v||=||Tu-v+Tu_n-Tu_n|| \le ||Tu_n-v||+||Tu-Tu_n||$ but I don't know how to show that $||Tu-Tu_n||$ goes to $0$.
ad (ii)
I have proved that $T$ is boundef iff $a_n$ is bounded.
ad (iii)
$(T-\lambda I)v=0 \iff \lambda = a_n$
I don't know how to close $\sigma_p(T)$ and how to find $\sigma(T)$, because it seems to me that it will be a set of $\lambda$ where $\sum_{n=1}^{\infty} \frac{1}{a_n-\lambda} = \infty$
ad (iv)
From (iii) $R(z;T)=(\frac{1}{a_n-\lambda})$
I will be really grateful if somebody tell me how to end those proofs.