Supremum Operator Norm Need not be Attained

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 $\Lambda : \ell^2 \to \ell^2$ be a mapping 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 Explicit Form of Supremum Operator Norm of Elementwise Multiplication of 2-Sequence by Bounded Sequence:

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

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:

This is a contradiction.

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