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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?).

Consider a 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?

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?).

Source Link
W2S
W2S
  • 103
  • 5

Rate of convergence

Consider a 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?