Order of Operators - Question in Proof of Abel's Transformation - Summation by Parts

Question

Given a sequence $\left(a_n \right)$, the operators $\Delta, E$ are defined by $$\Delta a_n=a_{n+1}-a_n,\\Ea_n=a_{n+1}.$$

Let $\left(u_n \right)$ and $\left(v_n \right)$ be two sequences. Let $\Delta_1, E_1$ denote the restriction of $\Delta, E$ to $u_n$, and let $\Delta_2, E_2$ denote the restriction of $\Delta, E$ to $v_n$.

In the proof of Abel's transformation (page 21 of Finite Difference Equations, 1st Ed. by H. Levy and F. Lessman), they write

$$\Delta \left(u_n v_n\right)=u_{n+1}\Delta_{2}v_n+v_n\Delta_{1}u_n=\left(E_{1}\Delta_{2}+\Delta_{1}\right)u_n v_n.$$

Then the authors simplify this as

$$\Delta \left(u_n v_n\right)=E_1 \color{green}{\Delta_2} \left(1+\color{blue}{E_{1}^{-1}\Delta_{1}\Delta_{2}^{-1}}\right)u_nv_n. \tag{$1$}$$

• Does the order of the operators in the expression $\color{blue}{E_{1}^{-1}\Delta_{1}\Delta_{2}^{-1}}$ matter? [I think yes, in general, but perhaps not in this case because the $\color{green}{\Delta_2}$ outside of the parentheses will destroy any constant that appears upon replacing $E_{1}^{-1}\Delta_{1}\Delta_{2}^{-1}$ with any operator formed by reordering the three operators.]

From $(1)$, we have

$$\Delta ^{-1} \left(u_n v_n\right)=\frac{E_{1}^{-1}\Delta_{2}^{-1}}{1+E_{1}^{-1}\Delta_1 \Delta_{2}^{-1}}u_n v_n, \tag{$2$}$$

and then they operate on both sides of $(2)$ with $1+E_{1}^{-1}\Delta_1 \Delta_{2}^{-1}$ to obtain

$$\Delta^{-1} \left(u_{n}v_{n} \right) = \left(E_{1}^{-1}\Delta_{2}^{-1}-\color{red}{\Delta^{-1}\Delta_{1}E_{1}^{-1}\Delta_{2}^{-1}}\right)u_n v_n. \tag{$3$}$$

• In $(3)$, how do they get the $\color{red}{\Delta^{-1}\Delta_{1}E_{1}^{-1}\Delta_{2}^{-1}}$ term? [I get $E_{1}^{-1}\Delta_{1}\Delta_{2}^{-1}\Delta^{-1}.$]

Answer 1

First question: Does the order of the operators in the expression $\color{blue}{E_{1}^{-1}\Delta_{1}\Delta_{2}^{-1}}$ matter? The answer is no, it does not matter.

• The operators $\Delta,\Delta_1,\Delta_2,E,E_1,E_2$ are linear. We have for instance \begin{align*} \Delta_1\left(cu_n+du_m\right)&= c\left(u_{n+1}-u_n\right)+d\left(u_{m+1}-u_m\right)\\ &=c\Delta_1 \left(u_n\right)+d\Delta_1 \left(u_m\right) \end{align*} where c,d are constant values and linearity can be similarly shown for the other operators.

• The operators $E_1^{-1}$ and $\Delta_1$ commute. We obtain \begin{align*} \color{blue}{\left(E_1^{-1}\Delta_1\right)}\left(u_nv_n\right)&=v_n\left(E_1^{-1}\Delta_1\right)\left(u_n\right)\tag{1.1}\\ &=v_nE_1^{-1}\left(u_{n+1}-u_n\right)\\ &=v_n\left(u_n-u_{n-1}\right)\tag{1.2}\\ &=v_n\Delta_1\left(u_{n-1}\right)\\ &=v_n\Delta_1E_1^{-1}u_n\\ &\,\,\color{blue}{=\left(\Delta_1E_1^{-1}\right)}\left(u_nv_n\right)\tag{1.1} \end{align*} and commutativity follows. In (1.1) we use the linearity of the operators $\Delta_1$ and $E_1$, since due to the restriction of these operators the other sequence elements $v_n$ are considered to be constants.

Now we want to calculate $E_1^{-1}\Delta_1\Delta_2^{-1}$ and $\Delta_2^{-1}E_1^{-1}\Delta_1$. We recall the following identities \begin{align*} \Delta_2^{-1}\left(v_n\right)&=\sum_{r=1}^{n-1}v_r+c&\text{c const.}\tag{2.1}\\ \Delta_2\Delta_2^{-1}\left(v_n\right)&=v_n&\text{per def.}\\ \Delta_2^{-1}\Delta_2\left(v_n\right)&=v_n+d&\text{d const.}\\ \end{align*}

We obtain \begin{align*} &\color{blue}{\left(E_1^{-1}\Delta_1\Delta_2^{-1}\right)\left(u_nv_n\right)}\\ &\qquad=\left(E_1^{-1}\Delta_1\right)\Delta_2^{-1}\left(u_nv_n\right)\\ &\qquad=\left(E_1^{-1}\Delta_1\right) u_n\left(\Delta_2^{-1}\left(v_n\right)\right)\tag{$\to$ linearity}\\ &\qquad=\left(E_1^{-1}\Delta_1\right) \left(u_n\left(\sum_{r=1}^{n-1}v_r+c\right)\right)\tag{$\to$ (2.1)}\\ &\qquad=\left(\sum_{r=1}^{n-1}v_r+c\right)\left(E_1^{-1}\Delta_1\right) \left(u_n\right)\tag{$\to$ linearity}\\ &\qquad\,\,\color{blue}{=\left(\sum_{r=1}^{n-1}v_r+c\right)\left(u_n-u_{n-1}\right)}\tag{$\to$ (1.2)}\\ \\ &\color{blue}{\left(\Delta_2^{-1}E_1^{-1}\Delta_1\right)\left(u_nv_n\right)}\\ &\qquad=\Delta_2^{-1}\left(\left(E_1^{-1}\Delta_1\right)\left(u_nv_n\right)\right)\\ &\qquad=\Delta_2^{-1} v_n\left(E_1^{-1}\Delta_1\right)\left(u_n\right)\tag{$\to$ linearity}\\ &\qquad=\left(\Delta_2^{-1}\right)\left( v_n\left(u_n-u_{n-1}\right)\right)\tag{$\to$ (1.2)}\\ &\qquad=\left(u_n-u_{n-1}\right)\left(\Delta_2^{-1}\right)\left( v_n\right)\tag{$\to$ linearity}\\ &\qquad\,\,\color{blue}{=\left(u_n-u_{n-1}\right)\left(\sum_{r=1}^{n-1}v_r+c\right)}\tag{$\to$ (2.1)}\\ \end{align*} and the claim follows.

Second question: In $(3)$, how do they get the $\color{red}{\Delta^{-1}\Delta_{1}E_{1}^{-1}\Delta_{2}^{-1}}$ term? [I get $E_{1}^{-1}\Delta_{1}\Delta_{2}^{-1}\Delta^{-1}$]. Here are some aspects. The operator identity stated in (2) \begin{align*} \Delta ^{-1}=\frac{E_{1}^{-1}\Delta_{2}^{-1}}{1+E_{1}^{-1}\Delta_1 \Delta_{2}^{-1}} \end{align*} indicates, that both, multiplication from the left-hand side as well as multiplication from the right-hand side with \begin{align*} \frac{1}{1+E_{1}^{-1}\Delta_1 \Delta_{2}^{-1}}=\left(1+E_{1}^{-1}\Delta_1 \Delta_{2}^{-1}\right)^{-1}\tag{3.1} \end{align*} is admissible. In (3) we have multiplication from the right-hand side. We obtain from (3) \begin{align*} \color{blue}{\Delta^{-1}}&\color{blue}{=E_1^{-1}\Delta_2^{-1}-\Delta^{-1}\Delta_1E_1^{-1}\Delta_2^{-1}}\\ \Delta^{-1}+\Delta^{-1}\Delta_1E_1^{-1}\Delta_2^{-1}&=E_1^{-1}\Delta_2^{-1}\\ \Delta^{-1}\left(1+\Delta_1E_1^{-1}\Delta_2^{-1}\right)&=E_1^{-1}\Delta_2^{-1}\\ \Delta\Delta^{-1}\left(1+\Delta_1E_1^{-1}\Delta_2^{-1}\right)&=\Delta E_1^{-1}\Delta_2^{-1}\\ \color{blue}{1+\Delta_1E_1^{-1}\Delta_2^{-1}}&\color{blue}{=\Delta E_1^{-1}\Delta_2^{-1}}\tag{4.1}\\ \end{align*}

We check the validity of (4.1) by evaluating left-hand side and right-hand side at $u_nv_n$.

We obtain from the left-hand side of (4.1) \begin{align*} &\color{blue}{\left(1+\Delta_1E_1^{-1}\Delta_2^{-1}\right)(u_nv_n)}\\ &\qquad=u_nv_n+\left(\Delta_1E_1^{-1}\right)(u_n)\left(\Delta_2^{-1}\left(v_n\right)\right)\tag{$\to$ linearity}\\ &\qquad =u_nv_n+\left(\Delta_1E_1^{-1}\right)(u_n)\left(\sum_{r=1}^{n-1}v_r+c\right)\tag{$\to$ (2.1)}\\ &\qquad =u_nv_n+\left(\sum_{r=1}^{n-1}v_r+c\right)\left(\Delta_1E_1^{-1}\right)(u_n)\tag{$\to$ linearity}\\ &\qquad\,\,\color{blue}{=u_nv_n+\left(\sum_{r=1}^{n-1}v_r+c\right)\left(u_n-u_{n-1}\right)}\tag{$\to$ (1.2)}\\ \end{align*} and from the right-hand side \begin{align*} &\color{blue}{\left(\Delta E_1^{-1}\Delta_2^{-1}\right)\left(u_nv_n\right)}\\ &\qquad=\Delta E_1^{-1}\left(u_n\right)\Delta_2^{-1}\left(v_n\right)\tag{$\to$ linearity}\\ &\qquad=\Delta E_1^{-1}\left(u_n\right)\left(\sum_{r=1}^{n-1}v_r+c\right)\tag{$\to$ (2.1)}\\ &\qquad=\Delta \left(\sum_{r=1}^{n-1}v_r+c\right)E_1^{-1}\left(u_n\right)\tag{$\to$ linearity}\\ &\qquad=\Delta \left(\sum_{r=1}^{n-1}v_r+c\right)u_{n-1}\\ &\qquad=\left(\sum_{r=1}^{n}v_r+c\right)u_{n}-\left(\sum_{r=1}^{n-1}v_r+c\right)u_{n-1}\tag{$\to \Delta$}\\ &\qquad\color{blue}{=u_nv_n+\left(\sum_{r=1}^{n-1}v_r+c\right)\left(u_n-u_{n-1}\right)} \end{align*} showing the validity of (4.1).

Note: Not part of the answer, but a nice symmetry which follows from OPs stated product formula at the beginning: \begin{align*} \Delta \left(u_n v_n\right)=u_{n+1}\Delta_{2}v_n+v_n\Delta_{1}u_n=\left(E_{1}\Delta_{2}+\Delta_{1}\right)u_n v_n \end{align*} Since $\Delta\left(u_n v_n\right)=\Delta\left(v_n u_n\right)$ and since $E_1=\Delta_1+1$ and $E_2=\Delta_2+1$ we have \begin{align*} \color{blue}{\Delta}&=E_{1}\Delta_{2}+\Delta_{1}\\ &=E_2\Delta_1+\Delta_2\\ &\,\,\color{blue}{=\Delta_1+\Delta_2+\Delta_1\Delta_2} \end{align*}