Scale a cutoff $\psi \in C^\infty(0, \infty)$ so that $\int^\infty_0 (\psi'_n)^2 r dr \to 0$ as $n \to \infty$.

Question

Let $\psi \in C^\infty((0,\infty); [0,1])$ be supported away from zero and identically one near $[1, \infty]$.

Is it possible to make some nonlinear scaling $\psi_n = \psi \circ f_n(r)$ of $\psi$ to achieve $$ \int^\infty_0 (\partial_r \psi_n)^2 r dr \to 0 \qquad \text{as $n \to \infty$?} $$ Here, I want the $f_n$ to be chosen such that $\psi_n \in C^\infty((0,\infty); [0,1])$ and is supported away from zero and identically one near $[1, \infty]$.

Note that by, change of variables, using something like $\psi_n = \psi(2^n r)$ merely gives $\int^\infty_0 (\psi'_n)^2 r dr = \int^\infty_0 (\psi')^2 r dr$, so a cleverer choice of $f_n$ is required.

For some context, the reason why I want to make this construction is because I am trying to understand some properties of the Friedrichs extension of the operator $-\partial^2_r - (2r)^{-2}: L^2(0,\infty) \to L^2(0,\infty)$.

Answer 1

From your question, I gather that here are the constraints :

1. $\psi(x)=0$ on $(-\infty,\epsilon]$,
2. $\psi(x)=1$ on $[1-\epsilon, +\infty)$,
3. $\psi(\mathbb R)=[0,1].$
4. $\psi \in C^\infty$.

I will add a common feature of cutoff functions :

5. $\psi$ is stricly increasing on $[\epsilon,1-\epsilon].$

Under these hypothesis, the smallest you can make $$ \int_0^\infty (\psi^\prime_n)^2 r \textrm{d}r \textrm{ is }\frac{1}{\ln(\frac{1-\epsilon}{\epsilon})}. $$ When $\epsilon$ tends to zero, it tends to zero like $\ln(\epsilon)^{-1}$.

Let me explain. Because of 1,2,3 and 5, $\psi$ limited to $[\epsilon,1-\epsilon]$ is a bijection onto $[0,1]$. Let $\varphi$ be its inverse: $\psi\circ\varphi$ is the identity on $[0,1]$, it is simply the function $\psi\circ\varphi(x)=x$ for all $x\in[0,1]$.

Let $g_n$ be any other cutoff function, satisfying 1,2,3,4 (and 5 or not as desired). Then, let $$f_n = \varphi \circ g_n$$ We have $$ \psi\circ f_n =g_n, $$ so it satisfies all the requirements, and the original $\psi$ has de facto disappeared.

So your question becomes.

what is the infimum of $$ I(u) = \int_{\mathbb R} r(u^\prime(r))^2 \textrm{d}r = \int_\epsilon^{1-\epsilon} r(u^\prime(r))^2 \textrm{d}r, $$ among all smooth $u$ satisfying 1,2,3,4 (and 5 or not as desired)?

Let us forget 5 for the moment. It will turn out that it is satisfied by the minimiser corresponding to $1,2,3$ and a weaker constraint than $4$. When looking for a minimizer, if it exists (it certainly does in $H^1_r(\epsilon, 1-\epsilon)=\{ u : u \in L^2(\epsilon,1-\epsilon), u^\prime\in L^2(\epsilon,1-\epsilon) \}$, you find the so-called Euler-Lagrange equation satisfied by the minimiser $u_\star$, that is \begin{equation}\label{1} (ru_\star^\prime)^\prime =0,\quad u_\star(\epsilon)=0, \textrm{ and } u_\star(1-\epsilon)=1.\qquad(1) \end{equation} To remind you how it goes, you notice that $I(u)$ is bounded below by $0$, and so you can look for an infimum. By various arguments (convexity of the integrand w.r.t $u$ for example) you can show that a minimiser exists and is unique. Then, you write what it means that $$ I(v)\geq I(u_\star). $$ for any $v$ satisfying 1,2,3 and being in $H^1(\epsilon,1-\epsilon)$. Take any $h$ which satisfies $h(\epsilon)=h(1-\epsilon)=0$, and write $v=u_\star + \theta h$, $\theta\neq0$. This means $$ \theta^2 I(h) + 2\theta\int_\epsilon^{1-\epsilon} u_\star^\prime(r) h^\prime(r) r \textrm{d} r \geq 0. $$ This quadratic function in $\theta$ can only be of constant sign if $$ \int_\epsilon^{1-\epsilon} u_\star^\prime(r) h^\prime(r) r \textrm{d} r =0, $$ for any $h$ as prescribed, which implies (1).

Now, we can solve $(1)$ explicitly :

$$ u_\star=\ln\left(\frac x{\epsilon}\right)\left(\ln\left(\frac {1-\epsilon}{\epsilon}\right)\right)^{-1}, $$ and $$ I(u_\star) = \frac{1}{\ln(\frac{1-\epsilon}{\epsilon})}>0. $$ The candidate, $u_\star$ extended by $0$ on the left and $1$ on the right, is not in $C^\infty(\mathbb R)$: that can be remedied by a mollification, up to an arbitrarily small error.

Answer 2

Building from an observation in another answer, it suffices to use $$ \psi_n(r) = \psi \big(\ln\big( \tfrac{r}{\epsilon} \big) \big[\ln(\tfrac{1-\epsilon}{\epsilon})\big]^{-1}\big), \qquad \epsilon = 2^{-n}. $$ Then each $\psi_n \in C^\infty((0, \infty);[0,1])$ is still zero near $r = 0$, and identically one near $[1, \infty]$. And by change of variables, $$ \int (\psi'_n)^2 r dr = \big[\ln(\tfrac{1-\epsilon}{\epsilon})\big]^{-1}\int(\psi')^2 dr \to 0, \qquad \text{as $n \to \infty$.} $$