Let $A$ be a linear second-order difference operator acting on the space of complex sequences as
$$(Af)\_{n}=f\_{n-1}+a\_{n}f\_{n}+f\_{n+1}, \quad n\in\mathbb{Z},$$
where $a\_{n}\in\mathbb{C}$. Further, let for fixed $k\in\mathbb{N}\_{0}$, $f^{(0)},f^{(1)},\dots,f^{(k)}$ are nonzero sequences satisfying
$$ Af^{(0)}=0 \quad \mbox{ and }\quad Af^{(j)}=j\cdot f^{(j-1)}, \quad \mbox{for } j=1,2,\dots,k.$$
Suppose that the same holds true for another tuple of nonzero sequences $g^{(0)},g^{(1)},\dots,g^{(k)}$.


Finally, for $j=0,1,\dots,k$, denote by $C^{(j)}$ the sequence whose elements are given by
$$C\_{n}^{(j)}:=\sum\_{i=0}^{j}\binom{j}{i}C\_{n}(f^{(i)},g^{(j-i)})$$
where $C\_{n}(f,g)=f\_{n}g\_{n+1}-f\_{n+1}g\_{n}$ is the Casoratian.


The problem is to find a proof of the following statement (which I strongly believe is true).


**Claim:** If $C^{(0)}=C^{(1)}=\dots=C^{(k)}=0$, then $g^{(k)}$ is a linear combination of the elements $f^{(0)},f^{(1)},\dots,f^{(k)}$.



A problem from linear algebra and difference equations
I'm going to use the notation defined above here. When a letter is used alone, it denotes the entire sequence. 


Claim: If $C^{[j]}(f, g) = 0$ for all $j \leq k$, then $g^{[k]}$ is a linear combination of the $f^{[i]}, i \leq k$.


We work by induction. This is obvious for $k = 0$. Assume the statement is true up to $k - 1$. Let $f, g$ be solutions to the relevant equations such that $C^{[j]} = 0$ for $j \leq k$. Then $g^{[0]} = a f^{[0]}$ for some $a$. Let $g' = g - af$; note that as the summed Casoratians are antisymmetric and bilinear, $C^{[j]}(f, g') = C^{[j]}(f, g)$. Then $g'^{[0]} = 0$. Let $g''^{[j]} = g'^{[j + 1]}$; then $g''$ satisfies all relevant relations. Then $0 = C^{[j]}(f, g') = C^{[j - 1]}(f, g'')$, so $C^{[j']}(f, g'') = 0$ for $j' \leq k - 1$. By the induction hypothesis, $g''^{[k - 1]}$ is a linear combination of the $f^{[i]}$ for $i \leq k - 1$, so $g'^{[k]}$ is, so $g^{[k]} = g'^{[k]} + a f^{[k]}$ is a linear combination of the $f^{[i]}, i \leq k$. 


