# A functional equation

I am working on some physics problem and got stuck with the following equation: Let $$a$$ be a very small positive number. Is there a bounded function $$F$$, $$0 \leq F \leq 1$$, such that for all $$x \in \mathbb{R}$$, $$F(x - a) e^{x a} + (1 - F(x+a)) e^{-x a} = e^{a^2/2}.$$ I had never seen anything like that before. Any references are most welcome.

• You don't want measurability or anything, so the question comes down to whether one can specify values of $F$ on $[0,2a)$ (all between $0$ and $1$) so that $0 \le F \le 1$ always holds when using the functional equation to extend $F$ to all of $\mathbb{R}$. Apr 23 at 3:39

Denote $$F(x)=G(x)e^{x^2/2}$$, this multiple is chosen in order to get an equation of the form $$G(x+a)-G(x-a)=$$ a given function. Indeed, you have $$G(x)\in [0,e^{-x^2/2}]$$ and $$G(x-a)e^{x^2/2+a^2/2}+e^{-xa}-G(x+a)e^{x^2/2+a^2/2}=e^{a^2/2}\\ G(x+a)-G(x-a)=e^{-(x+a)^2/2}-e^{-x^2/2}.$$ One such function $$G=:G_0$$ is straightforward: $$G_0(t)=\sum_{k=0}^\infty (-1)^ke^{-(t-ka)^2/2},$$ then we are given that $$G-G_0$$ is $$2a$$-periodic. On the other hand, $$G(x)$$ goes to 0 when $$x$$ goes to $$-\infty$$, and so does $$G_0$$. Thus $$G-G_0$$ is a periodic function which goes to 0 when $$x$$ goes to $$-\infty$$. The only such function is identical 0. Therefore $$G\equiv G_0$$, and the question reduces to whether $$G_0(x)\in [0,e^{-x^2/2}]$$ for all $$x$$. But $$\lim_{n\to +\infty}G_0(x+2na)=\sum_{k\in \mathbb{Z}}(-1)^k e^{-(x-ka)^2/2}$$ depends on $$x$$ (it is closely related to Jacobi theta-function) and is not in general equal to 0, so alas, your $$F$$ does not exist.