Find a cut-off function ${\psi _\varepsilon }$ such that $x{\psi _\varepsilon }'\left( x \right)$ is bounded uniformly

Question

In my recent works in PDEs, I'm interested in finding a family of cut-off functions satisfying following properties:

For each $\varepsilon >0$, find a function ${\psi _\varepsilon } \in {C^\infty }\left( \mathbb{R} \right)$ which is a non-decreasing function on $\mathbb{R}$ such that:

1. ${\psi _\varepsilon }\left( x \right) = \left\{ {\begin{array}{*{20}{l}} {0 \mbox{ if } x \le \varepsilon ,}\\ {1\mbox{ if } x \ge 2\varepsilon ,} \end{array}} \right.$ and
2. The function $x \mapsto x{\psi _\varepsilon }'\left( x \right)$ is bounded uniformly with respect to $\varepsilon$ as $\varepsilon \to 0$.

The main problem here is ${\psi _\varepsilon }'\left( x \right) \to \infty $ for some $x \in \left( {\varepsilon ,2\varepsilon } \right)$ as $\varepsilon \to 0$. I also start with this function to define explicitly ${\psi _\varepsilon }$ in the interval $\left( {\varepsilon ,2\varepsilon } \right)$ but my attempts to adjust the referenced function failed.

Can you find an example of these cut-off functions?

Thanks in advanced.

Answer 1

Let $\psi\in C^{\infty}(\mathbb{R})$ be any non-decreasing smooth function satisfying $\psi(x) = 0$ for $x\le 1$ and $\psi(x) = 1$ for $x\ge 2$, and set $\psi_{\epsilon}(x) = \psi(x/\epsilon)$, so that $\psi_{\epsilon}(x) = 0$ for $x\le\epsilon$ and $\psi_{\epsilon}(x) = 1$ for $x\ge 2\epsilon$. Then $\psi_{\epsilon}'(x) = \frac{1}{\epsilon}\psi'(\frac{x}{\epsilon})$. Notice that $\psi'$ is zero outside $[1,2]$, and hence $$\max\limits_{x\in\mathbb{R}}{|x\psi_{\epsilon}'(x)|} = \max\limits_{x\in\mathbb{R}}{\left|\frac{x}{\epsilon}\psi'\left(\frac{x}{\epsilon}\right)\right|} = \max\limits_{\frac{x}{\epsilon}\in[1,2]}{\left|\frac{x}{\epsilon}\psi'\left(\frac{x}{\epsilon}\right)\right|} = \max\limits_{x\in[1,2]}{|x\psi'(x)|}, $$ i.e. $x\mapsto x\psi_{\epsilon}'(x)$ is uniformly bounded in $\epsilon$.

Answer 2

You can start with the piecewise affine Lipschitz function $$ h_\varepsilon(x) := \begin{cases} 0, & \text{if}\ |x| \leq 5\varepsilon/4,\\ 1, & \text{if}\ |x| \geq 7\varepsilon/4 \end{cases} $$ (and affine for the remaining values of $x$). For this function you have $|x h_\varepsilon'(x)| \leq 4$.

Let $\varphi_\rho$ be the standard family of mollifiers.

For every $\rho \in (0, \varepsilon/4)$ the mollified function $$ \psi_\varepsilon := \varphi_\rho \ast h_\varepsilon $$ is a $C^\infty$ function satisfying your requirements.

Namely, since $\psi_\varepsilon' = \varphi_\rho \ast h_\varepsilon'$, by the definition of convolution and the fact that $\varphi_\rho\geq 0$ you get $$ \left| \psi_\varepsilon'(x) \right| = \left| \int \varphi_\rho(x-y) h_\varepsilon'(y) dy \right| \leq \int \varphi_\rho(x-y) | h_\varepsilon'(y) | dy \leq 2 \int \varphi_\rho(x-y) dy = 4. $$

Answer 3

First let $\Psi(x) = e^{-1/(1-x^2)}$ for $|x|<1$ and $=0$ otherwise (as here). This satisfies $\Psi \in C_c^\infty(\mathbb R)$ and $0 \leq \Psi(x) \leq e^{-1}$.

Then let $A = \int_{-\infty}^{\infty} \Psi(x) \, dx$ so that $\rho(x) = A^{-1} \int_{-\infty}^{x} \Psi(t) \, dt$ satisfies $\rho \in C^\infty(\mathbb R)$ with $\rho(x)=0$ for $x<-1$ and $\rho(x)=1$ for $x>1$, and $|\rho'(x)| \leq (Ae)^{-1}$.

Finally set $\psi_{\epsilon}(x) = \rho(\frac{2x}{\epsilon}-3).$ Then $\psi_\epsilon \in C^\infty(\mathbb R)$ with $\psi_\epsilon(x)=0$ for $x<\epsilon,$ $\psi_\epsilon(x)=1$ for $x>1,$ and $|\psi_\epsilon'(x)| = \frac{2}{\epsilon} |\rho'(\frac{2x}{\epsilon}-3)| \leq 2(Ae\epsilon)^{-1}$ so $|x \psi_\epsilon'(x)| \leq 2(Ae\epsilon)^{-1} \cdot 2\epsilon = 4(Ae)^{-1}.$

Thus, this $\psi_\epsilon$ satisfies your wishes.

Answer 4

Take the construction proposed here and use $\psi_\epsilon(x)=f(\frac x\epsilon)$. Then, $\psi^\prime_\epsilon $ is supported on $(\epsilon,2\epsilon)$, and bounded by $(1+s)\epsilon^{-1}$ for any $s>0$ you wish a priori. As a consequence, $0\leq x\psi^\prime_\epsilon < 2\epsilon(1+s)\epsilon^{-1} = 2(1+s)$, uniformly.