Question

How can operators appear without operand?

Answer 1

It's not "common factor". It's definition of subtraction on operators. In E-1, the 1 is an (identity) operator that takes any function f just to f itself, and the minus is a binary operation _on two operators_ that produce a new operator. Say T-U, it takes f to a new function (Tf)-(Uf), where the minus here is the subtraction of two functions (the point-wise subtraction).

Answer 2

Yes, and it is saying that the Δ operator is equivalent to the operator (E - 1).

Applying the operator (E - 1) to f just means Ef - 1f = Ef - 1. I think you could interpret that almost as a definition.

That is, the operator (A + B) is defined to be the operator that produces Af + Bf when applied to f.

Then you can do the rearrangement E = 1 + Δ  and that's valid operator arithmetic because of ordinary arithmetic on the underlying expressions Δf, Ef and 1f.

Answer 3

Δ and E are just functions whose inputs and outputs are other functions. Δ = E - 1 is just an equality of functions, just like sin^(2) + cos^(2) = 1 which you are probably more familiar with.

>Autor basically takes f(x) common and after that cancels f(x). How is that possible?

if f and g are functions, the equation f = g is the same as saying: for all input values x, f(x) = g(x). the same is true here, except the inputs to Δ and E are functions, not numbers. Δ = E - 1 means: for all input functions f, Δ(f) = (E-1)(f), which in turn means: for all functions f and inputs x, Δ(f)(x) = (E-1)(f)(x).