Simple functional equation: $\frac 1 2 [\alpha(x - 1) + \alpha(x + 1)]$, $\alpha(0) = 1$, $\alpha(m) = 0$ I have a simple functional equation:
$$
\alpha(x)
=
\frac 1 2 [\alpha(x - 1) + \alpha(x + 1)]\,,
\qquad \alpha(0) = 1\,,\quad\alpha(m) = 0
$$
I know it has a linear solution $\alpha(x) = ax + b$ but I don't have any idea how to prove that this is the only solution. Is there any way I can derive this solution from the initial equation?

In your last comment to date, you rightly observe that for $b(x)=a(x)-a(x-1)$ you get the equation
$$b(x+1)=b(x)$$
Its solution are all functions with period 1.
Now 
$$a(x+n)=a(x+n-1)+b(x)=a(x+n-2)+2\,b(x)=...=a(x)+nb(x)$$
using the periodicity of $b$. This leads to the idea to consider 
$$c(x)=a(x)-x\,b(x).$$ 
It satisfies the discrete dynamic
\begin{align}
c(x+1)&=a(x+1)-(x+1)\,b(x+1)=a(x)+b(x+1)-(x+1)\,b(x+1)\\
&=a(x)-x\,b(x)=c(x),
\end{align}
so it is again a periodic function with period $1$. The general solution has thus the form
$$a(x)=c(x)+x\,b(x)$$
with $b$ and $c$ any 1-periodic functions. This specializes to the linear solution in the case that $b$ and $c$ are constant functions.
