This is Example 1.16 taken from the book "The Mathematical Language of Quantum Theory" by Heinosaari and Ziman.
Let $l^2(\mathbb{N})$ be the Hilbert space of functions $f$ over the naturals $\mathbb{N}$, including zero. The purpose of the example is to prove that in an infinite Hilbert space $H$ (in this case $l^2(\mathbb{N}$), it is no longer true that a linear mapping $T:X\rightarrow H$ defined in a subspace $X\subset H$ has an extention to a bounded operator $\tilde{T}:H\rightarrow H$.
I will rephrase the example and stop and then ask at the point where I don't understand the reasoning.
Example 1.16. For each $f\in l^2(\mathbb{N})$, define a function $Nf:\mathbb{N}\rightarrow \mathbb{C}$ by the formula
$$ (Nf)(n) = nf(n). $$
It may happen that $Nf$ is not a vector in $l^2(\mathbb{N}).$ For instance, let $f(0)=0$, $f(n)=1/n$ otherwise. Then $f\in l^2(\mathbb{N})$ but $Nf\notin l^2(\mathbb{N})$. The set
$$ D(N) = \lbrace f\in l^2(\mathbb{N}):Nf\in l^2(\mathbb{N}) \rbrace $$
is a linear subspace of $l^2(\mathbb{N})$ and $N$ is a linear mapping from >$D(N)$ into $l^2(\mathbb{N})$.
- We have just said that $Nf\notin l^2(\mathbb{N})$, why $D(N)$ is not simply the empty set?
For each $k\in \mathbb{N}$, we have $N\delta_k = k\delta_k$ ($\delta_k$ is the canonical basis in $l^2(\mathbb{N})$), and hence $||N\delta_k||=k||\delta_k||$. This shows that there could not be a bounded operator $\tilde{N}$ on $l^2(\mathbb{N})$ which would be an extention of $N$.
- Why $||N\delta_k||=k||\delta_k||$ implies(?) the non-existence of a bounded operator $\tilde{N}$. As I see it, this last equation fits perfectly fine in the definition of bounded operator.
Question 1: the "for instance" gives a single example of a sequence $f \notin D(N)$, so that $D(N) \not= \ell^2(\mathbb N)$. That is pretty far from implying $D(N) = \emptyset$. For instance, as mentioned in the next paragraph each $\delta_k \in D(N)$.
Question 2: if there were to be an extension $\tilde N$ there would be a constant $C$ satisfying $\|\tilde Nf\| \le C \|f\|$ for all $f \in \ell^2(\mathbb N)$, and $\tilde N = N$ on $D(N)$. With $f = \delta_k$ this implies $$k \|\delta_k\| = \|N\delta_k\| = \|\tilde N \delta_k\| \le C \|\delta_k\|$$ forcing $C \ge k$, but $k$ is arbitrary.