I am trying to solve this:
Let $$\dfrac{u(x+1)+u(x-1)}{2} = f(x) \tag 1$$
and $$\dfrac{u(x+4)+u(x-4)}{2} = g(x) \tag 2$$
Express $u$ in terms of $f$ and $g$.
My question is if the following steps are justified (I think they're not):
Make the substitutions $x\to x+4$ and $x \to x+1$ in $(1)$ and $(2)$ respectively
Let $$\dfrac{u(x+5)+u(x+3)}{2} = f(x+4) \tag 3$$
$$\dfrac{u(x+5)+u(x-3)}{2} = g(x+1) \tag 4$$
Subtract $(4)$ from $(3)$:
$$\dfrac{u(x+3)-u(x-3)}{2} = f(x+4)-g(x+1) \tag 5$$
I continue and get the solution doing a few more of these substitutions, but I am just not quite sure if the substitutions are valid.
I don't think that I can just combine $3$ and $4$, this is just abuse of notation, right? Because actually I am setting $x=a+4$ and $x=b+1$, so I can't pretend like $a$ and $b$ are independent.
Hint: $$u(x+4) + u(x-4) = (u(x+4) + u(x+2)) - (u(x+2) + u(x)) - (u(x) + u(x-2)) + (u(x-2) + u(x-4)) + 2 u(x)$$