Difference and functional equations Lemma

A solution of (A-r)f=0 is a geometric sequence fixed by its value at 0

Statement

Let $A$ be the shift operator on functions $f:\mathbb{Z}\to\mathbb{C}$ , so that $(Af)(n)=f(n+1)$ . Let $r$ be a nonzero number, and let $f$ satisfy the operator equation $(A-r)f=0$ , that is, $f(n+1)=r\,f(n)$ for all integers $n$ . Put $c=f(0)$ . Then $f(n)=c\,r^n$ for every integer $n$ .

Remarks

The values of $f$ at nonnegative $n$ follow from the rule $f(n+1)=rf(n)$ by induction upward from $0$ . The values at negative $n$ follow by induction downward from $0$ , using $f(n)=r^{-1}f(n+1)$ . This step is where $r\neq 0$ is needed. If $r=0$ , the claim can fail: a function with $f(0)=c$ and $f(n)=0$ for $n\ge 1$ is a solution, but its values at negative $n$ are not fixed by $c$ .

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Shift operator concept
A solution of (A-r)f=0 is a geometric sequence fixed by its value at 0 statement

Tag	`GGSHNC4CGZ`
URI	`m2://GGSHNC4CGZ`
Kind	statement, lemma
Writing	W0 stub
MSC	39A06
Status	This copy of the site has no build data. Run `m2 status GGSHNC4CGZ`.
Uses	Shift operator
Changed	4 November 2026

A solution of (A-r)f=0 is a geometric sequence fixed by its value at 0

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