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.

$\begingroup$ 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 $\mathbb{R}$. $\endgroup$– mathworker21Apr 23 at 3:39
1 Answer
Denote $F(x)=G(x)e^{x^2/2}$, this multiple is chosen in order to get an equation of the form $G(x+a)G(xa)=$ a given function. Indeed, you have $G(x)\in [0,e^{x^2/2}]$ and $$G(xa)e^{x^2/2+a^2/2}+e^{xa}G(x+a)e^{x^2/2+a^2/2}=e^{a^2/2}\\ G(x+a)G(xa)=e^{(x+a)^2/2}e^{x^2/2}.$$ One such function $G=:G_0$ is straightforward: $$G_0(t)=\sum_{k=0}^\infty (1)^ke^{(tka)^2/2},$$ then we are given that $GG_0$ is $2a$periodic. On the other hand, $G(x)$ goes to 0 when $x$ goes to $\infty$, and so does $G_0$. Thus $GG_0$ 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 $$\lim_{n\to +\infty}G_0(x+2na)=\sum_{k\in \mathbb{Z}}(1)^k e^{(xka)^2/2}$$ depends on $x$ (it is closely related to Jacobi thetafunction) and is not in general equal to 0, so alas, your $F$ does not exist.

$\begingroup$ Sorry, I made an erroneous edit to your answer, which is now "under review". I don't know how to revert it. $\endgroup$ Apr 23 at 5:41

1$\begingroup$ @HjalmarRosengren never mind, I rejected this edit $\endgroup$ Apr 23 at 5:59


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