Solve $f(x)=\frac{1}{2}\int_{x-1}^{x+1}{f(t)}dt$ Let $f(x)\in\mathbb{C}(\mathbb{R})$,$\inf\lim\limits_{|x|\to\infty}{f(x)}=0$(that is $\inf\lim\limits_{x\to+\infty}{f(x)}=0$ and $\inf\lim\limits_{x\to-\infty}{f(x)}=0$), such that 
$$f(x)=\frac{1}{2}\int_{x-1}^{x+1}{f(t)}dt$$
Prove that:
$$f(x)\equiv 0,\  x\in\mathbb{R}$$
My attempt
I try to solve this equation, but I have trouble in it. Obviously, $f(x)= ax + b$ satisfies the equation. However, I don't know how to deal with the general case.
I want to get some help. Thanks.
 A: Let $x \in \mathbb{R}$. Then $$\left|\frac{1}{2}\int_{x-1}^{x+1}f(x)dx\right| \le \frac{1}{2}2\sup_{t \in [x-1,x+1]}|f(t)|$$
with equality iff $f$ is constant throughout $[x-1,x+1]$. 
Suppose $f$ is not identically zero. Then $\sup_{x \in \mathbb{R}}|f(x)| > 0$, and this maximum is attained (since $f(x) \to 0$), at some $\xi$ say. WLOG $f(\xi) > 0$. Then we have:
$$\sup_{t \in [\xi-1,\xi+1]}|f(t)| = f(\xi) \ge \frac{1}{2}\int_{\xi-1}^{\xi+1}f(t)dt$$
so $f$ is constant on $[\xi-1,\xi+1]$. By induction, $f$ is constant on $[\xi-n,\xi+n]$ for all $n$, and choosing $n$ large enough shows that this constant must be $0$. 
EDIT: This isn't quite complete - I've realised the OP only says $f(x) \to 0$ as $x \to \infty$ and not necessarily as $x \to -\infty$. Not obvious to me how to patch it yet.
Patch: By the argument above, we certainly have a $\xi$ which maximises $f$ on $[0,\infty)$. It follows that $f$ is constant and equal to zero on $[0,\infty)$. But then we can go backwards and get that $f$ is constant on $[-1,1]$, and $[-2,0]$, and $[-3,-1]$, and so on... So $f$ must be identically zero.
Edit 2: As @clark points out, this argument is still not complete. 
A: Answer to the edited question. As for the edited question which assumes $f(x) \to 0$ as $|x| \to \infty$, the equation is essentially a version of Laplace equation (and is indeed the Laplace equation corresponding to some discrete-time diffusion process, see below), and so, we can borrow various fancy ideas in the theory of harmonic functions. (I strongly recommend to read @clark's now-deleted answer, if you have enough reputation to see deleted answers.)
Just for fun, let me present a solution using a bit of probability theory. Let $(X_n)_{n=1}^{\infty}$ be a sequence of i.i.d. random variables with $X_n \sim \mathrm{Uniform}([-1,1])$ and $S_n = X_1 + \cdots + X_n$ be its partial sum. Then the assumption translates to $ f(x) = \mathbf{E}[f(x+X_1)] $, hence
$$ f(x) = \mathbf{E}[f(x + S_n)] $$
for any $n$. Since $f(x) \to 0$ as $|x| \to \infty$, $f$ is bounded and let $M \geq 0$ be a bound of $f$. Moreover, for any fixed $t > 0$, we know from Central Limit Theorem that $\mathbf{P}(|S_n| \leq t) \to 0$ as $n\to\infty$. So it follows that
\begin{align*}
|f(x)|
&= \mathbf{E}[|f(x+S_n)|\mathbf{1}_{\{|S_n| \leq t\}}] + \mathbf{E}[|f(x+S_n)|\mathbf{1}_{\{|S_n| > t\}}] \\
&\leq M \mathbf{P}(|S_n| \leq t) + \left( \sup_{|x-x'| > t} |f(x')| \right).
\end{align*}
Taking limsup as $n\to\infty$ followed by $t\to\infty$, we obtain $|f(x)| \leq 0$ and therefore $f \equiv 0$.

Answer to the original question. Only under the condition $\lim_{x\to+\infty} f(x) = 0$, there are indeed non-trivial solutions originating from delay-differential equation.
Let $\alpha$ be a non-zero solution of $\alpha = \sinh \alpha$ in $\mathbb{C}$ such that $\operatorname{Re}(\alpha) < 0$. (It is easy to check that there is no pure-imaginary solution except for zero. So if $\alpha \neq 0$ solves the equation, then so does $-\alpha$ and one of $\alpha$ or $-\alpha$ has negative real-part. The existence of such solution is not hard to prove. One such zero is numerically given as $-2.76868 + 7.49768 i$.) Then set
$$ f(x) = \operatorname{Re}(e^{\alpha x}). $$
Since $|f(x)| \leq |e^{\alpha x}| = e^{\operatorname{Re}(\alpha)x} \to 0$ as $x \to +\infty$, we know that $f(x) \to 0$ as $x \to +\infty$. Moreover,
$$ \frac{1}{2}\int_{x-1}^{x+1}f(x) \, \mathrm{d}x
= \frac{1}{2} \operatorname{Re}\left[ \int_{x-1}^{x+1} e^{\alpha t} \, \mathrm{d}t \right]
= \operatorname{Re}\left[ \frac{\sinh(\alpha)}{\alpha} e^{\alpha x} \right]
= f(x). $$
Therefore $f$ satisfies all the assumptions in the original question but is not identically zero.
