I am struggling with the following problem. Let $f$ be a smooth real function with only a finite number of zero. Let $\sigma>0$ a real number. Is it possible to show that the function $$\displaystyle R_\sigma(x) \triangleq \frac{\int_\mathbb{R}sf(s)g_\sigma(x-s)ds}{\int_\mathbb{R}f(s)g_\sigma(x-s)ds}$$
(where $g_\sigma(x) = \frac{e^{-x^2/2\sigma^2}}{\sqrt{2\pi}}$) is bounded on $\mathbb{R}$ ?
Thank you very much!
It is not. Consider, for example, $f(s) = \sin(s) + 2$. Then $f(s)$ doesn't have any zeroes, denominator is continuous, periodic and positive - thus bounded and separated from $0$, while numerator, when $x = 2\pi k + \pi / 2$, is greater than
$$\int\limits_{2\pi k + \pi / 4}^{2\pi k + 3 \pi / 4} s f(s) g_\sigma(x - s)\, ds = \int\limits_{-\pi / 4}^{\pi / 4} (t + 2 \pi k + \pi/2)\cdot(\sin t + 2) \cdot g_\sigma(t)\, dt \geq 2\pi k \cdot \pi / 2 \cdot g_\sigma(\pi / 4)$$
which grows unboundly.