Is this operator bounded and closed?

Let's define the subset of $\ell^2(\mathbb C)$ $$\mathcal D(A) = \left\{ {z \in {\ell ^2}\left( C \right),\sum\limits_{k = 1}^\infty {k^2{{\left| {{z_k}} \right|}^2} < \infty } } \right\},$$ and define the linear operator on $\mathcal D(A)$ by $$A\left( z \right) = \left( {{z_1},2{z_2},3{z_3}, \cdots } \right)$$ for $z\in \mathcal D(A)$. I want to know if this operator is bounded and closed.

What I know: If this operator is bounded, then it must hold that $${\left\| {A\left( z \right)} \right\|_2} \le c{\left\| z \right\|_2}$$ for all $z \in \mathcal D(A)$. Also for closedness, an operator is closed if the closure of the graph is equal to the graph. If I denote the graph by $\Gamma(A)$, this implies $\Gamma(A)=\bar \Gamma \left( A \right)$. The graph is the tuple $\left( {z,A\left( z \right)} \right)$ for all $z \in \mathcal D(A)$.

How could I apply these results to my problem to deduce the above? any tips are welcomed.

Solution 1:

consider $\left\| {A\left( z \right)} \right\|_2^2 = \left\| {\left( {{z_1},2{z_2},3{z_3}, \ldots } \right)} \right\|_2^2 = \sum\limits_{k = 1}^\infty {k{{\left| {{z_k}} \right|}^2}} $. However, $z \in \mathcal D(A)$, and therefore $\sum\limits_{k = 1}^\infty {{k^2}{{\left| {{z_k}} \right|}^2}} < \infty $ and hence $\left\| {A\left( z \right)} \right\|_2^2 < \sum\limits_{k = 1}^\infty {{k^2}{{\left| {{z_k}} \right|}^2}} < \infty $, therefore notice that the mapping does not explode. Lets try to bound it now w.r.t $\left\| z \right\|_2^2$. Lets consider $\sum\limits_{k = 1}^\infty {{k^2}{{\left| {{z_k}} \right|}^2}} = \alpha < \infty $. Remember that if a series converges, then the sequence of terms converges to 0, i.e., $\mathop {\lim }\limits_{k \to \infty } {k^2}{\left| {{z_k}} \right|^2} = 0$. Similarly, $\mathop {\lim }\limits_{k \to \infty } {k}{\left| {{z_k}} \right|^2} = 0$. Therefore, we can select large enough $N$ such that $$\sum\limits_{k = 1}^\infty {k{{\left| {{z_k}} \right|}^2}} = \sum\limits_{k = 1}^N {k{{\left| {{z_k}} \right|}^2}} + \sum\limits_{k = N + 1}^\infty {k{{\left| {{z_k}} \right|}^2}} \le \sum\limits_{k = 1}^N {k{{\left| {{z_k}} \right|}^2}} + \varepsilon \le \\2\sum\limits_{k = 1}^N {k{{\left| {{z_k}} \right|}^2}}. $$

The last inequality holds since $\varepsilon \ll \sum\limits_{k = 1}^N {k{{\left| {{z_k}} \right|}^2}} $ by assumption. Now, notice that $$\left\| {A\left( z \right)} \right\|_2^2 = \sum\limits_{k = 1}^\infty {k{{\left| {{z_k}} \right|}^2}} \le 2\sum\limits_{k = 1}^N {k{{\left| {{z_k}} \right|}^2}} \le 2\sum\limits_{k = 1}^N {N{{\left| {{z_k}} \right|}^2}} \le 2N\sum\limits_{k = 1}^N {{{\left| {{z_k}} \right|}^2}} \le 2N\left\| z \right\|_2^2$$

Therefore we proved that ${\left\| {A\left( z \right)} \right\|_2} \le \sqrt {2N} {\left\| z \right\|_2}$, so we conclude the operator is bounded.