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 ?

Below, we will see that the function $f$ is actually the restriction of an entire function to $\Bbb{R}$, i.e. the sume of a convergent power-series with infinite radius of convergence. Once this is shown, the (sadly rather downvoted) answer of @Leucippus actually becomes a valid argument.
For this, let $K := \Vert f \Vert_\sup$. By assumption, $K < \infty$. We will show by induction on $n \in \Bbb{N}_0$ that $\Vert f^{(n)} \Vert_\sup \leq K \cdot |2a|^n$ for all $n \in \Bbb{N}_0$. For $n=0$ this is trivial.
As also noted in the other posts, continuity of $f$ implies that $x \mapsto \int_{x-1}^x f(t) dt$ is continuously differentiable, which implies that $f$ is continuously differentiable with $f'(x) = -a \cdot (f(x) - f(x-1))$.
By induction, $$f^{(n)}(x) = -a \cdot [f^{(n-1)}(x) - f^{(n-1)}(x-1)].$$
Indeed, for $n=1$, this is what we just noted. In the induction step, we get
$$
f^{(n+1)}(x) = \frac{d}{dx} (-a) \cdot [f^{(n-1)}(x) - f^{(n-1)}(x-1)] = (-a) \cdot [f^{(n)}(x) - f^{(n)}(x-1)].
$$
By induction hypothesis, this yields
$$
|f^{(n+1)}(x)| \leq |a| \cdot [|f^{(n)}(x)| + |f^{(n)}(x-1)|] \leq 2|a| \cdot \Vert f^{(n)} \Vert_\sup \leq |2a|^{n+1}.
$$
But this implies
$$
\sum_{n=0}^\infty \left|\frac{f^{(n)}(0) \cdot x^n}{n!} \right| \leq \sum_{n=0}^\infty \frac{(2|ax|)^n}{n!} = \exp(2|ax|) < \infty,
$$
so that the power series $$g(x) := \sum_{n=0}^\infty \frac{f^{(n)}(0) \cdot x^n}{n!}$$ around $0$ of $f$ converges absolutely on all of $\Bbb{R}$.
It remains to show $f = g$. To this end, note that the Lagrange form of the remainder for Taylor's formula (see http://en.wikipedia.org/wiki/Taylor's_theorem#Explicit_formulae_for_the_remainder) yields
$$
|f(x) - g(x)| \xleftarrow[k \to \infty]{} \left| f(x) - \sum_{n=0}^k \frac{f^{(n)}(0) x^n}{n!} \right| = \left| \frac{f^{(k+1)}(\xi_L)}{(k+1)!} \cdot |x|^{k+1}\right| \leq \frac{(2|xa|)^{k+1}}{(k+1)!} \xrightarrow[k\to\infty]{} 0,
$$
because even $\sum_k \frac{(2|xa|)^{k+1}}{(k+1)!} = \exp(2|xa|) < \infty$. Hence, $f = g$ is the restriction of an entire function to $\Bbb{R}$.
