1
$\begingroup$

I have what is in essence a basic analysis question.

To make working out a certain example a bit easier I found that I need to find existence of a function $f\in C^\infty(\mathbb{R})$ with the following properties:

  1. $f$ is increasing
  2. $f(x)=0$ for all $x\leq 0$
  3. $f(x)=1$ for all $x\geq 5$
  4. $\frac{f(x)}{x}<1$ for all $x>0$
  5. For any $y>-1$ the function $F(x,y)=\frac{f(x)}{f(x+f(x)y)}$ which is immediately well-defined and smooth for $x>0$ can be extended to a smooth function on $\{(x,y)| y>-1\}$.

To find a function $f$ that satisfies the first 4 conditions is relatively simple by tweaking one that looks like $e^{-\frac{1}{x^2}}$. So the main question is whether $f$ exists such that 5. is satisfied. After some discussion it seems like it should be possible even setting $F(x,y)=1$ for $x\leq 0$, but I have not been able to convince myself fully yet. The idea is to rewrite the denominator as $f(x(1+\frac{f(x)}{x}y))$ and use the fact that $\frac{f(x)}{x}$ vanishes to order $n$ at $x=0$ to deduce $n$ times differentiability of $F(x,y)$. Then, since $f^{(n)}(0)=0$ for arbitrary $n$ we find that smoothness.

Any insights would be appreciated of course!

$\endgroup$
3
  • 1
    $\begingroup$ The formula $\frac{f(x+f(x)y)}{f(x)}=1+y\int_0^1 f'(x+tyf(x))\,dt$ may be quite useful ;-) $\endgroup$
    – fedja
    Apr 12, 2018 at 23:52
  • $\begingroup$ Indeed it does seem useful! In fact doesn't this mean that 1-4 imply 5? $\endgroup$ Apr 16, 2018 at 13:16
  • 1
    $\begingroup$ Under the assumption that $f'$ is confined to some interval $[0,a]$ with $a<1$ it certainly does. $\endgroup$
    – fedja
    Apr 16, 2018 at 14:01

1 Answer 1

1
$\begingroup$

$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$

It is straightforward, but a bit tedious, to construct a function $f\in C^\infty(\R)$ satisfying conditions 1--4 and such that $f(x)=e^{-1/x}$ for $x$ in a right open neighborhood $N$ of $0$ (see details at the end of this answer). Then \begin{equation} F(x,y)=e^{1/z-1/x}, \end{equation} where $z:=x+ye^{-1/x}$; everywhere here, $y>-1$ and $x$ is a small enough positive real number (depending on $y$) such that $x$ and $z$ are both in $N$. Note that $z'_x=1+ye^{-1/x}/x^2$ and $z'_y=e^{-1/x}$, whence we have the crucial observation: \begin{multline*} \Big(\frac1z-\frac1x\Big)'_x =\frac1{x^2}-\frac1{z^2}\Big(1+e^{-1/x}\frac y{x^2}\Big) =\frac{(z-x)(z+x)}{x^2 z^2}-e^{-1/x}\frac y{x^2 z^2} \\ =y e^{-1/x}\Big(\frac1{xz^2}+\frac1{zx^2}-\frac1{x^2z^2}\Big). \end{multline*}

So, we see that \begin{align*} F'_x(x,y)&=F(x,y)p_1(1/x,1/z,y,e^{-1/x})e^{-1/x},\\ F'_y(x,y)&=F(x,y)q_1(1/x,1/z,y,e^{-1/x})e^{-1/x} \end{align*} for some polynomials $p_1$ and $q_1$. So, by induction, all partial derivatives of $F(x,y)$ of orders $\ge1$ are of the form $F(x,y)p(1/x,1/z,y,e^{-1/x})e^{-1/x}$ for some polynomials $p$.

Also, $|z-x|\ll e^{-1/x}$, $z\sim x$, $F(x,y)\to1$, which implies that all partial derivatives of $F(x,y)$ of orders $\ge1$ go to $0$; the convergence here is for $x\downarrow0$ uniformly over all $y$ in any compact subset of $(-1,\infty)$. This implies that $F\in C^\infty(\R\times(-1,\infty))$, with $F(x,y):=1$ for $x\le0$.


Added: details concerning the first sentence of this answer. For all real $x$, let \begin{equation} f(x) :=\int_\R g(x + \ep u(x)t)Cu(t)\,dt, \end{equation} where \begin{equation} g(x) := e^{-1/x} \ii{0 <x\le2}+[e^{-1/2} +(x-2)(1-e^{-1/2})]\,\ii{2 < x \le 3}+\ii{x > 3} \end{equation} \begin{equation} u(x) := \exp\Big\{-\frac1{(x - 1)(4 - x)}\Big\}\,\ii{1 < x < 4}, \end{equation} $C := 1/\int_\R u(x)\,dx$, $\ii{\cdot}$ is the indicator, and $\ep$ is a positive real number small enough so that $1+4\ep u'> 0$ (whence $x+\ep u(x)t$ is increasing in $x \in\R$ for each $t$ in the interval $(1,4)$).

Then $f$ is in $C^\infty(\R)$, satisfies conditions 1--4, and $f(x)=e^{-1/x}$ for $x\in(0,1)$. In particular, $f$ is increasing -- because $g$ is so and $x+\ep u(x)t$ is increasing in $x \in\R$ for each $t\in(1,4)$. Also, $f(x)=e^{-1/x}<x$ for $x\in(0,1)$, $g\le1$ and hence $f\le1$ on $\R$, and so, $f(x)<x$ for $x>1$. That $f$ is $C^\infty$ on the interval $(1,4)$ follows because \begin{equation} f(x) =\int_\R g(y)Cu\Big(\frac{y-x}{\ep u(x)}\Big)\,\frac{dy}{\ep u(x)} \end{equation} for $x\in(1,4)$.

$\endgroup$
4
  • $\begingroup$ I have added details on how to construct a function $f\in C^\infty(\mathbb R)$ satisfying conditions 1--4 and such that $f(x)=e^{-1/x}$ for $x$ in a right open neighborhood of $0$. $\endgroup$ Apr 15, 2018 at 3:27
  • $\begingroup$ Hey Iosif, I appreciate the great answer! There seems to be a small gap since z may not be in N for large y, but one can always find a small enough neighborhood of (0,y) for which z is in N, so the argument still holds. I just wanted to give you time to edit it, before I mark as an answer, but will do so anyway if you don't manage to make it back here. $\endgroup$ Apr 16, 2018 at 13:14
  • $\begingroup$ Hey again, after closer inspection I don't quite get the expression for $F'_x$. Why does it have an overall factor of $e^{-\frac{1}{x}}$? $\endgroup$ Apr 16, 2018 at 15:19
  • $\begingroup$ @NiekdeKleijn : Thank you for comments. I have now made sure that $z$ is in $N$ and also added details on the (indeed crucial) fact that the expression for $F'_x$ indeed contains the factor $e^{-1/x}$. $\endgroup$ Apr 16, 2018 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.