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?
 A: Hint 1: $1=\frac{1}{2}+\frac{1}{2}$
Hint 2: $a=1 \cdot a = (\frac{1}{2}+\frac{1}{2}) \cdot a$
A: assuming $m$ to be an integer.
from your observation
$$\alpha (x)-\alpha (x-1)= C$$
where $C$ is a constant.
differentiating both sides
 $$\alpha '(x)-\alpha '(x-1)= 0$$
from this we get the condition that either $\alpha '(x)$ is a periodic function with $1$ as its period or a constant. so $\alpha (x)$ is either a periodic or an equation representing a line. 
but if you consider the initial conditions. if $\alpha (0)=1$ then $\alpha (m)=0 $ is contradicting for a periodic function. therefore the resultant should represent an equation of line.
A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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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}
$$\color{#0000ff}{\large\alpha\pars{x} = 1 - {x \over m}}$$
A: 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 $a = \mu/\lambda$
