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I was reading a paper by Carlitz and he defined the operator $E$ by $Ef(s)=f(s-1)$ which is a shift operator mostly encountered in Calculus, and later he managed to check the commutativity of operators of the form $E + \alpha (s)$ so he wrote : $$[E + \alpha (s)][E + \beta(s)] = E^2 + [\alpha(s)+\beta(s-1)]E + \alpha (s) \beta (s)$$

The question is: How $E \beta (s) = \beta (s-1) E$, why not $E\beta(s)=\beta(s-1)$ only?

Rodrigo de Azevedo
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Abdelhay Benmoussa
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3 Answers3

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Because they are operators, so $\beta$ is the operator of multiplication by $\beta$. You have $$ [(E\beta)f](s)=[E(\beta f)](s)=\beta(s-1)\,f (s -1)=\beta(s -1)\,(Ef)(s) $$

Martin Argerami
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Simply, as an operator identity, you have the distributive law by application on a general last element c of the representation space V of functions with a scalar product structure

$\forall \ b, \ c, \ b c \in V:$ $$\begin{align} & E\left(\ \left( E \ + b(s) \mathbb 1 \right) \ \left( c(s) \right) \ \right) \\& = E (E c(s)) \ + E \left(b(s) \ c(s)\right)\ \\ &= \ E^2 c(s) \ + \ E (b(s)) \ (E c(s)) \\&= c(s-2) + b(s-1)\ c(s-1) = \left(E^2 + (E b(s)) E\right) \ \left(c(s)\right) \end{align} $$

The fact that V is a vector space and the product is a tensor product is generally suppressed in the literature by automatic simplification in an algebra of complex valued functions.

Roland F
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The "coefficients" $\,\alpha,\beta\,$ in these linear operators are linear scale-by $\:\!f\:\!$ operations associated to a function $\:\!f,\,$ i.e. $\,{\bf f}(g) := f g\,$ (we write the operation in bold to distinguish it from the function). Below are the commutation rules for a shift operator $\bf E$ (and a derivative $\bf D$ for comparison).

$\!\begin{align}{\bf Lemma} \qquad \color{#0a0}{{\bf D}(fg) = f {\bf D}g + ({\bf D}f)g}\ \Rightarrow\ &\color{#c00}{\bf D\,f = f\,D\!+\!f'},\, \ f' := {\bf D}f\\[.4em] \color{#0a0}{{\bf E}(fg) = {\bf E}f\,{\bf E}g}\ \Rightarrow\ &\color{#c00}{\bf E\,f \:\!= f^{\dagger}E},\qquad\ f^{\dagger} := {\bf E}f\\ \end{align}$

by $\ (\color{#c00}{{\bf Df}})g = {\bf D}({\bf f}(g)) = \color{#0a0}{{\bf D}(fg) = f\, {\bf D}g + f' g} = ({\color{#c00}{\bf f D + f'}})g$

$\ \ \ \ \ \ ({\color{#c00}{\bf Ef}})g = {\bf E}({\bf f}(g)) = \color{#0a0}{{\bf E}(fg) = {\bf E}f\:{\bf E}\:\!g} = f^{\dagger}{\bf E}g = (\color{#c00}{{\bf f^{\dagger}}{\bf E}})g$

Bill Dubuque
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