Consider a function **fixed** function $f\in L^1(\mathbb{R})$ such that $$
\int\_{\mathbb{R}}f(x)dx=0
$$
Now define the following function: $$
F(y)=\int\_{\mathbb{R}} f(x)\mathrm{sech}\Big(\frac{x}{\exp(y)}\Big)dx.
$$
Then, by definition of $F$ is clear that $$
\lim\_{y\to\infty}F(y)=0.
$$
I am wondering if it is possible to prove any rate of convergence for $F(y)$, like for example $$
\hbox{for some }\, \alpha\in(0,1), \quad F(y)=O\Big(y^{-\alpha}\Big) \qquad \hbox{for } \vert y\vert\gg 1,
$$
without any additional hypothesis on $f$. Is that possible? For me it sounds like it should be the case, since intuitively the $\mathrm{sech}$ is behaving as $1$ on growing bounded sets as $y$ grows. Then, since $f$ is fixed, then, for $y\gg1$ sufficiently large, the integral $F(y)$ should satisfies $$
F(y)\approx \int\_{\mathbb{R}} f(x)=0.
$$
However, I would like to understand the rate of convergence of this property as I change the scaling. In other words, if I define now $$
F\_2(y):=\int\_{\mathbb{R}}f(x)\mathrm{sech}\Big(\dfrac{x}{g(y)}\Big)dx,
$$
for some $g(y)\in C^\infty(\mathbb{R})$ growing sufficiently fast. Then, I am wondering how could I write the rate of convergence of $F(y)\to 0$ in terms of the growth of $g(y)\to+\infty$. For example, something like $$
F\_2(y)=O\big(\log(g(y)\big) \quad \hbox{for } \vert y\vert\gg 1?
$$
I've been thinking about it but I am quite lost, does anyone has any comments that might help?


**PS:** Here the rate of convergence might depends on some norm of $f$ as well (for example on its $L^1$ norm?).



Rate of convergence
$\newcommand\de\delta\newcommand\R{\mathbb R}$The answer is no: without any additional hypothesis on $f$, no such rate of convergence exists.


Indeed, suppose the contrary: that
$$|F(f)(y)|\le\de(y)\tag{1}$$
for some some function $\de\colon\R\to\R$ such that $\de(z)\to0$ as $|z|\to\infty$, all $f\in L^1(\R)$ with $\int\_{\R}f(x)\,dx=0$, and all real $y$, where
$$F(f)(y):=\int\_{\R}\frac{f(x)\,dx}{\cosh(xe^{-y})}.$$  

Take any $g\in L^1(\R)$ with $\int\_{\R}g(x)\,dx=0$ such that
$$a:=F(g)(0)=\int\_{\R}\frac{g(x)\,dx}{\cosh(x)}>0;$$
for instance, one may take $g=1\_{[0,1]}-1\_{[-2,-1]}$.
For each real $y$, let $g\_y(x):=g(xe^{-y})$ for all real $x$. Then for each real $y$ we have $g\_y\in L^1(\R)$ and $\int\_{\R}g\_y(x)\,dx=0$, whereas
$$F(g\_y)(y)=\int\_{\R}\frac{g(xe^{-y})\,dx}{\cosh(xe^{-y})}=e^y\int\_{\R}\frac{g(z)\,dz}{\cosh(z)}=ae^y.$$
So, by (1), $\de(y)\ge ae^y$, which contradicts the condition that $\de(z)\to0$ as $|z|\to\infty$. $\Box$


