# Rate of convergence

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?).


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 $$|z|\to\infty$$. $$\Box$$

• Thank you very much for your very interesting answer. Sorry, maybe my statement was not clear enough, but I wasn't trying to get a bound that works simultaneously for all $f\in L^1(\mathbb{R})$. Instead, for each fixed $f\in L^1$, I was wondering if it's possible to get a bound. I think it would be slightly different, right? In other words, $\delta$ might depends on some norm of $f$ (for example depending on $\Vert f\Vert_{L^1}$, I think this would rule out your counterexample, right?). Does that have any sense or maybe I am getting confused?
– W2S
Nov 16, 2020 at 17:29
• For me the most important part is that, since $f$ is fixed, as $\vert y\vert\gg1$ grows, you are really somehow approaching the mean of $f$. My problem with your counterexample is that, since for each $y\in\mathbb{R}$ you are re-scaling $f$, then of course you will never approach the mean of $f$. I hope I am explaining myself sufficiently clear.
– W2S
Nov 16, 2020 at 17:36
• @W2S : You said "without any additional hypothesis on $f$" and the examples of rates you gave contained no information about $f$. So, effectively, the desire expressed in your question was for a bound valid for all $f$ in $L^1$ with $\int f=0$. This question has been fully answered. If you wanted to actually ask some other question, you should assume the responsibility for what you actually posted, rather than change your post to invalidate a valid answer. Instead, you may want to post your modified question separately. Nov 16, 2020 at 18:52
• @W2S : BTW, no bound small for large $|y|$ in terms of $\|f\|_1$ (or, say, $\|f\|_p$ for any other $p>0$) is possible either. However, that would be an answer to another post. Nov 16, 2020 at 18:55