Solve $f(x)=\int_{x-1}^{x+1} f(t) \mathrm{d}t$

Question

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?

Answer 1

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).