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I am studying a proof in a differential geometry book where the author use a a smooth function satisfying some properties. Here is what it is claimed. $\forall 0 < \epsilon < \pi$ there exists a $C^{\infty}$-function $f : [0,1] \rightarrow \mathbb{R}$ satisfying the following properties

  1. $f(x)=0$, near $0$

  2. $f(x)= \pi - \epsilon$, near $1$

  3. $0 \leq f(t) < \pi - \epsilon, \forall t$

  4. $0 \leq f'(t) < \pi $

I was not able to show that such a function exists (althought it seems obvious). Here is what I have done.

I first define the bell shaped function $f$ as follows

$ f(x) = \begin{cases} c \cdot exp(\frac{-1}{1-(2x-1)^2}) & x \in (0,1) \\ 0 & else \\ \end{cases} $

Where $c = \frac{1}{\int_{- \infty}^{\infty} f(x) dx}$

It can be shown that $f$ is $C^{\infty}$. Now I set $h(x)= \int_{- \infty}^{x} f(t) dt$. Then I have that $h(x)=0$ if $x<0$ and that $h(x)=1$ if $x>1$. Also we have $0 < h(x) <1$.

My idea was now to find two parameters $a$ and $b$, possibly depending on $\epsilon$ and to do the same construction, but on an open interval $I \subset (0,1)$. Do I need to find out an explicit expression for $I$ ? Then I did the following construction

$ f(x) = \begin{cases} c \cdot exp(\frac{-1}{1-((2+\eta)x-(1+\kappa))^2}) & x \in I \\ 0 & else \\ \end{cases} $

The constant $c$ being defined as the value of the integral, just like before. Then I tried to set $h(x)= (\pi - \epsilon) \int_{- \infty}^{x} f(t) dt$

By taking the restriction of $h$ on $[0,1]$ I have constructed a function which satisfies properties 1 to 3 but unfortunately not the last one...

Did I make a mistake in my calculation ? Is there an easier way to show the existence of such functions ?

Many thanks for your help

Alain
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1 Answers1

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Take $\delta\in(0,\pi/2)$ and let $g\colon\Bbb R\to\Bbb R$ be $$ g(x)=\begin{cases} 0 & x\le\delta,\\ \dfrac{\pi-\epsilon}{\pi-2\,\delta}\,(x-\delta) & \delta<x<\pi-\delta\\ \pi-\epsilon & x\ge\pi-\delta \end{cases} $$ $g$ satisfies the requirements except that it is not smooth; it is only continuous. Take $\phi$ a $C^\infty$ function with compact support on $[-1,1]$ such that $\phi(x)\ge0$ and $\int_{-1}^1\phi(x)\,dx=1$ and let $\phi_t(x)=\frac{1}{t}\,\phi(\frac{x}{t})$, $t>0$. Then you can take $f=g\ast \phi_t$ for small enough $t$.

Julián Aguirre
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  • Thank you for your answer. I see how the function g looks like, I am still unsure why you use this $\delta \in (0, \pi /2)$. Is it to make sure that $g'(x) < \pi$ ? – Alain Oct 24 '18 at 19:04
  • Also if I may ask you for clarifications, are these $\phi _t (x)$ mollifiers ? I can probably create an example of such things using the exponential function used in my question.

    And I am quite unsure how to check that $f$ satisfies the properties. I checked that if I would derive $f$ I would get $g' \ast \phi_t$, right ? And how can I show that $f$ vanishes near $0$ ?

     – Alain Oct 24 '18 at 19:10
  • Yes, $\phi_t$ are mollifiers. The support of $\phi_t$ is $[-t,t]$. Taking for instance $t=\delta/2$, $g\ast\phi_t(x)=0$ if $x<\delta/2$. – Julián Aguirre Oct 24 '18 at 19:53