# 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?

• $f(x) = A \cos(cx) + B \sin(cx)$ works where $c \approx 1.895494267$ is the positive solution of $c = 2 \sin c$. There should also be complex solutions $c$ of this equation, giving rise to further linearly independent solutions $f(x) = (A \cos sx + B \sin sx) \, e^{tx}$ where $c = s+it$. Jan 31 '14 at 18:37
• crossposted at math.stackexchange.com/questions/658615/… Jan 31 '14 at 18:51

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.
• Sure. When you differentiate, Fourier transform obtains one new root, $0$. Feb 1 '14 at 16:43