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?

Another way could be re-writing the original problem
$$
\cosh(\frac{d}{d\alpha})\alpha(x) = \cosh(\frac{d}{dx})\alpha(x)
$$
Since $\cosh(x) = \frac{1}{2}(e^{x} + e^{-x})$ and $e^{\frac{d}{dx}}\alpha(x) = \alpha(x+1)$, $$\frac{1}{2}(e^{\partial_\alpha} + e^{-\partial_\alpha})\alpha(x) = \frac{1}{2}\left(\alpha(x) + 1 + \alpha(x) - 1\right) = \alpha(x) = \frac{1}{2}(e^{\partial_x} + e^{-\partial_x})\alpha(x) = \frac{1}{2}\left(\alpha(x+1) + \alpha(x-1)\right)$$  You can then cancel all like terms on the right in the first expression to arrive at
$\frac{d}{d\alpha}\alpha = \frac{d}{dx}\alpha = 1$, so $\alpha(x) = x + C$. 
Another thought is that the original equation implies that
$$
\alpha(x) = \frac{1}{2}(\alpha(x+\lambda) + \alpha(x-\lambda))
$$
is true for all integers $\lambda$. Also $\cosh(\mu\frac{d}{d\alpha})\alpha(x) = \alpha(x)$ for any $\mu\in\mathbb{R}$ This means that
$$
\cosh(\mu\frac{d}{d\alpha})\alpha(x) = \cosh(\lambda\frac{d}{dx})\alpha(x)
$$
works just as well to describe the problem and so the general solution to 
$$
\mu = \lambda\frac{d\alpha}{dx}
$$
is $\alpha(x) = ax + b$ for some constants $b$ and $a = \mu/\lambda$
