# Boundendess of $f(x-y)/f(x)$ for $f(x)$ vanishing at $\infty$

Assume that $f(x)$ is a strictly positive, continuous and bounded function on $\mathbb R^d$ vanishing at infinity, and such that $f(x)\to 0$ as $|x|\to\infty$. Can we always tell that $f(x+y)/f(x)$ is bounded in $x$ when $|y|\leq 1$? If not, then what are the additional assumptions sufficient to tell that?

## 1 Answer

The answer in general is no. Consider $f(x)=e^{-x^2}$ on $\mathbb{R}$. Then $$\frac{f(x-1)}{f(x)}=e^{2x-1}$$ is unbounded as $x\to\infty$. This happens because $f(x)$ decays at a very fast rate as $x\to\infty$.

A more complicated example is the following. Let $f\colon\mathbb{R}\to\mathbb{R}$ be such that $\lim_{|x|\to\infty}f(x)=0$ and $f(2\,k)=3^{-k}$, $f(2\,k+1)=2^{-k}$ for $k\in\mathbb{N}$. Then $$\frac{f(2\,k+1)}{f(2\,k)}=\Bigl(\frac32\Bigr)^k.$$

A sufficient condition would be $f$ decreasing and $$|f(x)|\ge C\,e^{-c|x|}$$ for some constants $C,c>0$.

• Thank You for very nice examples. Could I ask for some tip on how to prove the sufficiency? – lucas Nov 26 '16 at 9:12