Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation

Theorem
Let $\Bbb K = \set {\R, \C}$.

Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.

Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.

Suppose $\Lambda : \ell^2 \to \ell^2$ is a diagonal operator such that:


 * $\Lambda \tuple {a_1, a_2, a_3, \ldots} = \tuple {\lambda_1 a_1, \lambda_2 a_2, \lambda_3 a_3, \ldots}$

Then $\Lambda \in \map {CL} {\ell^2}$.

Linearity
Let $\sequence {a_n}_{n \mathop \in \N_{> 0} }, \sequence {b_n}_{n \mathop \in \N_{> 0} } \in \ell^2$.

By linearity of transformations:


 * $\Lambda \in \map \LL {\ell^2}$

Continuity
For $\tuple {a_n}_{n \mathop \in \N_{> 0} }$:

By Continuity of Linear Transformation between Normed Vector Spaces:


 * $\Lambda \in \map C {\ell^2}$.

By definition:


 * $\Lambda \in \map {CL} {\ell^2}$