How to proof the following function is always constant which satisfies $f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)\,dt} $? Suppose that $f(x)$ is a bounded continuous function on $\mathbb{R}$,and that there exists a positive number $a$ such that
$$f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)\,dt} $$
is constant. Can anybody show that  $f$ is necessarily constant ?

Starting with 
\begin{align}
f(x) + a \int_{x-1}^{x} f(t) \, dt = c
\end{align}
then, by differentiation,
\begin{align}
f'(x) + a \left( f(x) - f(x-1) \right) = 0.
\end{align}
Now, for this equation to be satisfied consider it in the form of 
\begin{align}
f'(x) = B = -a ( f(x) - f(x-1) ).
\end{align}
From the equation $f'(x) = B$ it is seen that $f(x) = Bx+c_{1}$. Now, 
\begin{align}
f(x) - f(x-1) &= - \frac{B}{a} \\
(Bx+c_{1}) - (Bx - B + c_{1}) &= - \frac{B}{a} \\
B &= - \frac{B}{a}.
\end{align}
If this is to be satisfied then $B = 0$ which then implies that $$f(x) = c_{1}.$$

The alternative is that given the equation
\begin{align}
f'(x) + a (f(x) - f(x-1) ) = 0
\end{align}
then let $f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdots$ to obtain
\begin{align}
0 &= [ a_{1} + 2 a_{2} x + \cdots ] + a [ a_{0} + a_{1} x + a_{2} x^{2} + \cdots ] - a[ a_{0} + a_{1} (x-1) + a_{2} (x-1)^{2} + \cdots] \\
&= [ a_{1} + 2 a_{2} x + \cdots] + a[ a_{1} + a_{2}(2x-1) +\cdots] \\
&= [ a_{1} - a(a_{2} - a_{3} + a_{4} - \cdots )] + [ 2 a_{2} + a(2 a_{2} - 3 a_{3} + \cdots) ] x + [ 3(1+a)a_{3} + \cdots ] x^{2} + \cdots .
\end{align}
All the coefficients of $x^{n}$, $0\leq n$, are zero which yields $a_{1} = a_{2} = \cdots = 0$. This then leaves $f(x) = a_{0}$ which is a constant. 
