2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $C,c>0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.