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?

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Let's consider
$\ds{\alpha\pars{x + 1} - 2\alpha\pars{x}
+ \pars{1 - \epsilon^{2}}\alpha\pars{x - 1} = 0}$ where $\epsilon > 0$.
Later on, we'll recover the original equation in the limit $\ds{\epsilon \to 0^{+}}$. Solutions are $\ds{\propto \beta^{x}}$ such that
$\ds{\beta^{2} - 2\beta + \pars{1 - \epsilon^{2}} = 0}$ wich leads to
$$
\beta_{\pm}
= {-\pars{-2} \pm \root{\pars{-2}^{2} - 4\pars{1 - \epsilon^{2}}} \over 2}
=1 \pm \epsilon
$$
The general solution is
$\ds{\alpha\pars{x} = A\pars{1 + \epsilon}^{x} + B\pars{1 - \epsilon}^{x}}$. Let's impose the boundary conditions:
$$\left\{%
\begin{array}{rcrcl}
A & + & B & = & 1
\\
\pars{1 + \epsilon}^{m}A & + & \pars{1 - \epsilon}^{m}B & = & 0
\end{array}\right.
$$
Then
$$
A = {\pars{1 - \epsilon}^{m} \over \pars{1 - \epsilon}^{m} - \pars{1 + \epsilon}^{m}}
\,,\qquad
B = -\,{\pars{1 + \epsilon}^{m} \over \pars{1 - \epsilon}^{m} - \pars{1 + \epsilon}^{m}}
$$
and
$$
\alpha\pars{x}
=
{\pars{1 - \epsilon}^{m}\pars{1 + \epsilon}^{x}
-
\pars{1 + \epsilon}^{m}\pars{1 - \epsilon}^{x}
\over \pars{1 - \epsilon}^{m} - \pars{1 + \epsilon}^{m}}
$$
With the limit $\epsilon \to 0^{+}$:
\begin{align}
&\alpha\pars{x}
=\\[3mm]&\lim_{\epsilon \to 0^{+}}
{-m\pars{1 - \epsilon}^{m - 1}\pars{1 + \epsilon}^{x} + x\pars{1 - \epsilon}^{m}\pars{1 + \epsilon}^{x - 1}
-
m\pars{1 + \epsilon}^{m - 1}\pars{1 - \epsilon}^{x}
+
x\pars{1 + \epsilon}^{m}\pars{1 - \epsilon}^{x - 1}
\over
-m\pars{1 - \epsilon}^{m - 1} - m\pars{1 + \epsilon}^{m - 1}}
\\[3mm]&=
{-2m + 2x \over -2m}
\end{align}
$$\color{#0000ff}{\large\alpha\pars{x} = 1 - {x \over m}}$$
