Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation

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


Proof

Linearity

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

\(\ds \map \Lambda {k \sequence {a_n} + \sequence {b_n} }\) \(=\) \(\ds \map \Lambda {\sequence {k a_n + b_n} }\) $P$-sequence space is a vector space
\(\ds \) \(=\) \(\ds \sequence {\lambda_n \paren{k a_n + b_n} }\)
\(\ds \) \(=\) \(\ds \sequence {\lambda_n k a_n + \lambda_n b_n }\)
\(\ds \) \(=\) \(\ds k \sequence {\lambda_n a_n} + \sequence {\lambda_n b_n}\) $P$-sequence space is a vector space
\(\ds \) \(=\) \(\ds k \map \Lambda {\sequence {a_n} } + \map \Lambda {\sequence {b_n} }\)

By linearity of transformations:

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


Continuity

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

\(\ds \norm {\Lambda \mathbf x }_2^2\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \size {\lambda_n a_n}^2\) Definition of P-Norm
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \size {\lambda_n}^2 \size {a_n}^2\) Complex Modulus of Product of Complex Numbers
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \size {a_n}^2\) Definition of Supremum of Sequence
\(\ds \) \(=\) \(\ds \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \sum_{n \mathop = 1}^\infty \size {a_n}^2\)
\(\ds \) \(=\) \(\ds \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \norm {\mathbf x}_2^2\) Definition of P-Norm

By Continuity of Linear Transformation between Normed Vector Spaces:

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

By definition:

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

$\blacksquare$


Sources