Does anyone know if there is a measurable function $f: {\mathbb R} \to {\mathbb R}$ such that
$$ |x+y|-|x-y| \geq f(x)f(y) $$
for any $x,y \in {\mathbb R}$.
Motivation : if $f$ exists, one can immediately deduce an easy answer to that other question.
Let $f$ be a function (no further assumptions, not even measurabilty) $\mathbb R\to \mathbb R$ with $$\tag1f(x)f(y)\le |x+y|-|x-y|\quad\text{for al }x,y\in\mathbb R.$$ Assume there is $y>0$ with $f(y)>0$. Then for $x\ge y$ we have $f(x)\le \frac{2y}{f(y)}$ and for $0\le x<y$, $f(x)\le \frac{2x}{f(y)}\le \frac{2y}{f(y)}$. Hence $f$ is bounded from above over the nonnegative reals. Since the functions $x\mapsto -f(x)$ and $x\mapsto f(-x)$ also satisfy $(1)$, we conclude that $f$ is bounded, say $|f(x)|\le M$ for all $x$. But if we plug $x=M^2+1>0$ and $y=-x$ into $(1)$, we get $$-M^2\le f(M^2+1)f(-M^2-1)\le-2M^2-2,$$ and hence $M^2\le-2$, a contradiction.
We conclude that no function satisfying $(1)$ exists.
