Why is $l^2$ space needed in definition of weighted shift?

Question

Today I've come across a definition of weighted shift on Wolfram MathWorld.

They define the weighted shift as the operator working with elements from the standard orthonormal basis the space $X$. And $X = l^2$.

My question is - Since the weighted shift is simply this operation:

$$ T e_n = \alpha_n e_{n+1}, $$

$T$ being the continuous linear operator and $e$ being an element from the space, where $T$ is defined... Isn't it enough to define this simple operation?

What's the exact point of defining elements for the operator like the elements of the space orthonormal basis and the space being especially $l^2$?

Answer 1

The choice to use $l^2$is mostly so that they can make use of the inner product notation - such an operator makes plenty of sense on $l^1,l^∞$ and so on.