Supremum Operator Norm Need not be Attained

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $\sequence {\lambda_n}_{n \mathop \in \N}$ be a bounded sequence in $\Bbb K$ such that:

$\ds \forall n \in \N_{> 0} : \lambda_n = 1 - \frac 1 n$.

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 there is no $\mathbf x \in \ell^2$ such that:

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

and:

$\norm {\Lambda \mathbf x}_2 = \norm \Lambda$


Proof

Suppose $\ds \lambda_n = 1 - \frac 1 n$ with $n \in \N_{> 0}$.

By Supremum Operator Norm of Diagonal Operator over 2-Sequence Space:

\(\ds \norm \Lambda\) \(=\) \(\ds \sup_{n \mathop \in \N_{>0} } \set {1 - \frac 1 n}\)
\(\ds \) \(=\) \(\ds 1\)

Suppose:

$\mathbf x = \tuple {a_n}_{n \mathop \in \N_{> 0} } \in \ell^2$

such that $\norm {\mathbf x}_2 \le 1$ and $\norm {\Lambda \mathbf x}_2 = 1$.

Suppose $0 = a_2 = a_3 = \ldots$.

Then

\(\ds \Lambda \mathbf x\) \(=\) \(\ds \tuple {0 \cdot a_1, \frac 1 2 \cdot 0, \frac 2 3 \cdot 0 \ldots}\)
\(\ds \) \(=\) \(\ds \mathbf 0\)

Also $\norm {\Lambda \mathbf x}_2 = 0$.

This is in contradiction with the assumption $\norm {\Lambda \mathbf x}_2 = 1$.

Hence, at least one of the terms $a_2, a_3, \ldots$ must be nonzero.

However:

\(\ds 1\) \(\ge\) \(\ds \norm {\mathbf x}_2^2\)
\(\ds \) \(=\) \(\ds \size {a_1}^2 + \size {a_2}^2 + \size {a_3}^3 + \ldots\)
\(\ds \) \(\ge\) \(\ds \size {a_2}^2 + \size {a_3}^3 + \ldots\)
\(\ds \) \(>\) \(\ds \frac 1 4 \size {a_2}^2 + \frac 4 9 \size {a_3}^3 + \ldots\)
\(\ds \) \(=\) \(\ds \norm {\Lambda \mathbf x}_2^2\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(>\) \(\ds 1\)

This is a contradiction.

Hence, there exists no nonzero $\mathbf x \in \ell^2$ such that $\norm {\mathbf x}_2 \le 1$ and $\norm {\Lambda \mathbf x}_2 = \norm \Lambda$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Supremum_Operator_Norm_Need_not_be_Attained&oldid=617781"