I am trying to verify a claim in the example below, taken from these notes: https://www.uio.no/studier/emner/matnat/math/MAT4400/v19/pensumliste/ela-190523.pdf.
Notation:
$\ell^p(\mathbb{N}) = \{ (x(n))_n \subset \mathbb{N} : \|(x(n))_n\|_p < \infty \}$.
$\ell^\infty(\mathbb{N}) = \{ (x(n))_n \subset \mathbb{N} : \|(x(n))_n\|_\infty < \infty \}$.
$\|(x(n))_n\|_p = \left( \sum_{n \in \mathbb{N}} \lvert x(n) \rvert^p \right)^{1/p}$.
$\|(x(n))_n\|_\infty = \sup_{n \in \mathbb{N}} \lvert x(n) \rvert$.
$\mathcal{B}(X) =$ the set of bounded linear operators from $X$ into $X$
$\overline{\mathcal{F}(X,X)} =$ the closure of the set of bounded finite rank operators from $X$ to $X$.
$\mathcal{K}(X,X) =$ the set of compact operators from $X$ to $X$.
$\|\cdot\|$ denotes the operator norm.
I am trying to show the first claim, i.e.
$$\|M_\lambda \| = \| \lambda \|_\infty.$$
Attempt:
$\|M_\lambda \| \leq \| \lambda \|_\infty$:
\begin{align*} \|M_\lambda(x) \|_p &= \|(\lambda(n) x(n))_n \|_p \\ &\leq \sup_{n \in \mathbb{N}} \lvert \lambda(n) \rvert \|(x(n))_n\|_p \\ &= \|\lambda\|_\infty \|(x(n))_n\|_p \end{align*}
Hence $\|M_\lambda\| \leq \|\lambda\|_\infty$.
Now to show equality:
Fix a $\lambda \in \ell^\infty(\mathbb{N})$.
Let $N$ denote the first index $n$ for which $\lvert \lambda(n) \rvert = \sup_{n \in \mathbb{N}} \lvert \lambda(n) \rvert$
Let $(x(n))_n \in \ell^p$ be defined by $x(n) = \begin{cases} 1 \ \text{if} \ n = N, \\ 0 \ \text{otherwise}. \end{cases}$
Then
\begin{align*} \|M_\lambda(x)\|_p &= \left( \sum_{n \in \mathbb{N}} \lvert \lambda(n)x(n) \rvert^p \right)^{1/p} \\ &= \left( \lvert \lambda(N) \cdot 1 \rvert^p\right)^{1/p} \\ &= \lvert \lambda(N) \rvert \\ &= \|\lambda\|_\infty. \end{align*}
Hence $\|M_\lambda\| = \|\lambda\|_\infty$.
Correct? My feeling is that $N$ being the first index for which something is equal to its supremum sounds kind of iffy...
The mistake is assuming there exists $n \in \mathbb{N}$ such that $|\lambda(n)|=\|\lambda\|_{\infty}.$ For example, let $\lambda(n)=1-\dfrac1n.$ Then $\lambda \in l^{\infty}(\mathbb{N})$ and $\|\lambda\|_{\infty}=1$ but $\not\exists\ n \in \mathbb{N}$ such that $\lambda(n)=1.$
Correct argument: Fix $\epsilon >0$ and $M=\|\lambda\|_{\infty}.$ Then there exists $N \in \mathbb{N}$ such that $|\lambda(N)|\geq M-\epsilon.$ Let $(x(n))_n \in \ell^p$ be defined by $$x(n) = \begin{cases} 1 \ \text{if} \ n = N, \\ 0 \ \text{otherwise}. \end{cases}$$
Then \begin{align*} \|M_\lambda(x)\|_p &= \left( \sum_{n \in \mathbb{N}} \lvert \lambda(n)x(n) \rvert^p \right)^{1/p} \\ &= \left( \lvert \lambda(N) \cdot 1 \rvert^p\right)^{1/p} \\ &= \lvert \lambda(N) \rvert \\ &\geq M-\epsilon. \end{align*}
Thus $\|M_{\lambda}\|\geq M-\epsilon$ for every $\epsilon >0.$ Hence $\|M_{\lambda}\|\geq M.$