Let $g_1, g_2 \in \mathbb{R}^{[0,5]}$ be defined by $g_1(x) = x$ and $g_2(x) = 1-x$ for each $x \in [0,5]$.
Find a second-order homogeneous linear difference equation on $[0,3]$ such that {$g_1, g_2 $} forms a fundamental set of solutions to that equation on$[0,3]$.
My work on this question:
I started with the general proof of this questions. Let me state the question and the proof I conducted to the general format. I believe it is easier to replicate through the general format.
Let $g_1, ..., g_n \in \mathbb{R}^{[a,b+n]}$ with the property that $\forall x \in [a,b] : C[g_1(x), ..., g_n(x) ] \neq 0 $, where a, b $\in \mathbb{R}$ with b - a $\ge$ n. construct n th order homogeneous linear difference equation on [a,b] such that {$g_1, ..., g_n $} is a fundamental set of solutions of this equation on [a,b], if possible.
$\mathbf{PROOF}$
Assume that $g_1, ..., g_n \in \mathbb{R}^{[a,b+n]}$ are such that $\forall x \in [a,b] $ : $C[g_1(x), ..., g_n(x) ] \neq 0 $. Now assume that y(x+n)+...+$p_1(x)$y(x+1) +$p_0(x)$y(x) = 0 such that $g_1, ..., g_n$ are all solutions of this equation on [a,b] in which case $g_1, ..., g_n$ will be a fundamental system of solutions. But then, we have, for each x $\in$ [a,b], $p_0(x)g_1(x)+...+p_{n-1}(x)g_1(x+n-1)= - g_1(x+n)$, $p_0(x)g_n(x)+...+p_{n-1}(x)g_n(x+n-1)= - g_n(x+n)$. But now, again for each $x \in [a,b]$, the determinant of the coefficient matrix of the above system is nothing but $C[g_1(x), ..., g_n(x) ] \neq 0 $. So, one can solve the above sytem for $p_0(x)+...+p_{n-1}(x)$ in terms of $g_1, ..., g_n $, obtaining the desired n th order linear difference equation.
