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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 $f^{(0)},f^{(1)},\dots,f^{(k)}$.

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  • $\begingroup$ Do you have the converse here - that any linear combination satisfying $A g^{(j)} = j g^{j - 1}$ has 0 Casoratian? If so, induction should work pretty directly. $\endgroup$
    – user44191
    Commented Oct 14, 2016 at 20:54
  • $\begingroup$ Unfortunately, nor the opposite direction has been established. Naturally, the proof by mathematical induction in $k$ was my first attempt. While the case $k=0$ is easy, I did not find the way how to treat the induction step. $\endgroup$
    – Twi
    Commented Oct 15, 2016 at 19:59
  • $\begingroup$ It may be easier to work with $f^{[j]} = \frac{f^{(j)}}{j!}$, so that $A f^{[j]} = f^{[j - 1]}$. $\endgroup$
    – user44191
    Commented Oct 21, 2016 at 6:55
  • $\begingroup$ In which case the summed Casoratian is $j! \sum_{i = 0}^j C_n(f^{[i]}, g^{[j - i]})$. $\endgroup$
    – user44191
    Commented Oct 21, 2016 at 6:58

1 Answer 1

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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 $f^{[i]}, i \leq k$.

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  • $\begingroup$ The converse is also true, through similar reasoning. $\endgroup$
    – user44191
    Commented Oct 21, 2016 at 7:34

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