Solve $f(x)=\int_{x-1}^{x+1} f(t) \mathrm{d}t$ Solve $f(x)=\int_{x-1}^{x+1} f(t) \mathrm{d}t$. 
When $f$ is a function, it looks like the only solution is $f(x)=0$. But what if we allow distributions, such as the Dirac delta?
 A: By differentiating we obtain 
$$f'(x)=f(x+1)-f(x-1)$$
This type of equations was addressed in the MO question
On equation f(z+1)-f(z)=f'(z)
Let $\lambda$ be any (complex) root of the equation
$$\lambda=e^\lambda-e^{-\lambda},$$
which is equivalent to $z=2\sin z$, as Noam wrote.
to this $\lambda$ a solution $f(x)=e^{\lambda x}$ is associated.
So you have infinitely many exponential solutions. Any linear combination is
also a solution. Then, depending on your assumptions of $f$ you can consider various limits of those linear combinations in the appropriate topology for your
functions/distributions class.
Edit. For a complete theory of this kind of equations see "Fonctions moyenne-periodiques", a theory created by Delsart and Schwartz in 1940-s.
This is generalized in the modern theory of "equations of convolution-type", see,
for example Hormander, Linear Partial differential operators. In this theory one considers equations $u\star f=0$, where $u$ is a distribution. In your case,
$u=\delta-\chi,$ where $\chi$ is the characteristic function of $[-1,1]$. The method of solution is an appropriate version of Fourier--Laplace transform, depending on your class of functions/distributions. Ordinary Fourier transform of $u$ is $U(\lambda)=1-2\sin\lambda/\lambda$, whose roots give you the exponential solutions.
The question whether we obtain all solutions in this way is addressed by the Malgrange approximation theorem (Theorem 16.4.1 in Hormander).
