How to choose smooth function $f:\mathbb R \to [0,1]$?

Question

Let $\alpha, \beta >0.$

Can we expect to choose differentiable $f:\mathbb R \to [0,1]$ such that $f(x+\alpha)=0$ if $x\leq 0,$ and $f(x-\alpha)=1$ if $x\geq \beta$? If so, can we make $f$ smooth?

I think this is a "bump function" and I've seen them somewhere. — coffeemath, Jul 21 '18 at 01:04
You have to impose $\alpha < \beta - \alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap. — Winther, Jul 21 '18 at 01:27
@Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect? — Math Learner, Jul 21 '18 at 01:45
https://math.stackexchange.com/questions/198748/c-infty-version-of-urysohn-lemma-in-bbb-rn — Cave Johnson, Jul 21 '18 at 01:52
For example $f(\alpha) = 0$ by your first definition and $f(\beta-\alpha) = 1$ by your second definition. If for example $\alpha = \beta - \alpha$ then you have two values for $f$ here. — Winther, Jul 21 '18 at 01:53

score 1 · Accepted Answer · answered Jul 21 '18 at 01:57

1

With $u<v:$ Let $f(x)=0$ for $x\leq u.$ Let $f(x)=1$ for $x\geq v.$

Let $ n\in \Bbb Z^+.$

Let $g(y)=(y-u)^{n+1} (y-v)^{n+1}.$ Let $K=\int_u^v g(y)dy.$

For $u<x<v$ let $f(x)=\frac {1}{K}\int_u^x g(y)dy.$

The $n$th derivative of $f$ exists and is continuous at all points.

Note: $K\ne 0$ because either (i) $g(y)>0$ for all $y\in (u,v)$ or (ii) $g(y)<0$ for all $y\in (u,v).$

answered Jul 21 '18 at 01:57

DanielWainfleet

59,529

Thanks. I like your trick. Please can you explain bit more what goes wrong if we want $f$ is smooth (that is $f$ is infinitely many times differentiable)? – Math Learner Jul 21 '18 at 15:52
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In my answer, replace $g(y)$ with $e^{ h(y) }$ where $h(y)= \frac {-1} {(y-u)^2(y-v)^2}$ to make $f$ infinitely smooth.... For infinite smoothness neither $f$ nor $g$ can be a polynomial, because if $f(x)=Ax^m+Bx^{m-1}+...$ is a non-constant polynomial of degree $m>0 $ when $r x\in (u,v),$ then the $m$th derivative $f^{(m)}(x)$ is the non-zero constant $m!A$ when $x\in (u,v)$.... but $f^{(m)}(x)=0$ when $x<u.$ – DanielWainfleet Jul 22 '18 at 18:03

How to choose smooth function $f:\mathbb R \to [0,1]$?

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