Supremum Operator Norm of Diagonal Operator over 2-Sequence Space
Theorem
Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ 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 $\Lambda : \ell^2 \to \ell^2$ 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 $\ds \norm \Lambda = \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Proof
From Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation:
- $\ds \norm {\Lambda \mathbf x }_2^2 \le \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \norm {\mathbf x}_2^2$
Therefore:
- $\ds \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} \norm {\mathbf x}_2$
Suppose $\norm {\mathbf x}_2 \le 1$.
Then:
- $\ds \norm {\Lambda \mathbf x}_2 \le \sup \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n}$
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:
- $\ds \norm {\Lambda} \le \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Let $\mathbf e_n \in \ell^2$ such that:
- $\mathbf e_n = \tuple {\underbrace{0, \ldots, 0}_{n - 1}, 1, 0, \ldots}$
Then:
\(\ds \forall n \in \N_{> 0} : \ \ \) | \(\ds \norm \Lambda\) | \(=\) | \(\ds \norm \Lambda \cdot 1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm \Lambda \norm {\mathbf e_n}_2\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \norm {\Lambda \mathbf e_n}_2\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\tuple {\underbrace{0, \ldots, 0}_{n - 1}, \lambda_n \cdot 1, 0 \ldots} }_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\lambda_n}\) |
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 $\ds \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Hence:
- $\ds \norm \Lambda \ge \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Combining the results yields:
- $\ds \norm \Lambda = \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$