Supremum Operator Norm of Diagonal Operator over 2-Sequence Space

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

Let $\sequence {\lambda_n}_{n \mathop \in \N}$ 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 $\Lambda : \ell^2 \to \ell^2$ be a diagonal operator such that:


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

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Then $\ds \Lambda \in \map {CL} {\ell^2}$ and $\ds \norm \Lambda = \sup_{n \mathop \in \N} \size {\lambda_n}$

Proof
From Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation:


 * $\ds \norm {\Lambda \mathbf x }_2^2 \le \paren {\sup_{n \mathop \in \N} \size{\lambda_n} }^2 \norm {\mathbf x}_2^2$

Therefore:


 * $\ds \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N} \size{\lambda_n} \norm {\mathbf x}_2$

Suppose $\norm {\mathbf x}_2 \le 1$.

Then:


 * $\ds \norm {\Lambda \mathbf x}_2 \le \sup \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N} \size{\lambda_n}$

By definition of supremum operator norm:


 * $\norm \Lambda = \sup \set {\norm {\Lambda \mathbf x}_2 : \forall \mathbf x \in \ell^2 : \norm {\mathbf x}_2 \le 1 }$

Hence:


 * $\ds \norm {\Lambda} \le \sup_{n \mathop \in \N} \size {\lambda_n}$

Let $\mathbf e_n \in \ell^2$ such that:


 * $\mathbf e_n = \tuple {\underbrace{0, \ldots, 0}_{n - 1}, 1, 0, \ldots}$

Then:

Take the supremum of both sides over $n \in N$.

On the left-hand dize nothing happens because it is an $n$-independent real number.

On the left-hand dize we have $\ds \sup_{n \mathop \in \N} \size {\lambda_n}$

Hence:


 * $\ds \norm \Lambda \ge \sup_{n \mathop \in \N} \size {\lambda_n}$

Combining the results yields:


 * $\ds \norm \Lambda = \sup_{n \mathop \in \N} \size {\lambda_n}$