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

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Let $alpha, beta >0.$




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





real-analysis analysis functions examples-counterexamples




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closed as off-topic by user223391, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel Jul 21 at 14:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.












  • I think this is a "bump function" and I've seen them somewhere.
    – coffeemath
    Jul 21 at 1:04










  • You have to impose $alpha < beta - alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap.
    – Winther
    Jul 21 at 1:27










  • @Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect?
    – Math Learner
    Jul 21 at 1:45






  • 1




    math.stackexchange.com/questions/198748/…
    – Cave Johnson
    Jul 21 at 1:52






  • 1




    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 at 1:53














up vote
1
down vote

favorite












Let $alpha, beta >0.$




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





real-analysis analysis functions examples-counterexamples




share|cite|improve this question













closed as off-topic by user223391, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel Jul 21 at 14:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.












  • I think this is a "bump function" and I've seen them somewhere.
    – coffeemath
    Jul 21 at 1:04










  • You have to impose $alpha < beta - alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap.
    – Winther
    Jul 21 at 1:27










  • @Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect?
    – Math Learner
    Jul 21 at 1:45






  • 1




    math.stackexchange.com/questions/198748/…
    – Cave Johnson
    Jul 21 at 1:52






  • 1




    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 at 1:53












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $alpha, beta >0.$




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





real-analysis analysis functions examples-counterexamples




share|cite|improve this question













Let $alpha, beta >0.$




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






real-analysis analysis functions examples-counterexamples





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 21 at 1:19
























asked Jul 21 at 0:53









Math Learner

1619




1619




closed as off-topic by user223391, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel Jul 21 at 14:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by user223391, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel Jul 21 at 14:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Xander Henderson, John Ma, Taroccoesbrocco, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.











  • I think this is a "bump function" and I've seen them somewhere.
    – coffeemath
    Jul 21 at 1:04










  • You have to impose $alpha < beta - alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap.
    – Winther
    Jul 21 at 1:27










  • @Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect?
    – Math Learner
    Jul 21 at 1:45






  • 1




    math.stackexchange.com/questions/198748/…
    – Cave Johnson
    Jul 21 at 1:52






  • 1




    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 at 1:53
















  • I think this is a "bump function" and I've seen them somewhere.
    – coffeemath
    Jul 21 at 1:04










  • You have to impose $alpha < beta - alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap.
    – Winther
    Jul 21 at 1:27










  • @Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect?
    – Math Learner
    Jul 21 at 1:45






  • 1




    math.stackexchange.com/questions/198748/…
    – Cave Johnson
    Jul 21 at 1:52






  • 1




    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 at 1:53















I think this is a "bump function" and I've seen them somewhere.
– coffeemath
Jul 21 at 1:04




I think this is a "bump function" and I've seen them somewhere.
– coffeemath
Jul 21 at 1:04












You have to impose $alpha < beta - alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap.
– Winther
Jul 21 at 1:27




You have to impose $alpha < beta - alpha$ as otherwise the $f(x) = 0$ and $f(x)=1$ regions overlap.
– Winther
Jul 21 at 1:27












@Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect?
– Math Learner
Jul 21 at 1:45




@Winther: Thanks. Can you explain a bit more? Even after assuming this, what one can expect?
– Math Learner
Jul 21 at 1:45




1




1




math.stackexchange.com/questions/198748/…
– Cave Johnson
Jul 21 at 1:52




math.stackexchange.com/questions/198748/…
– Cave Johnson
Jul 21 at 1:52




1




1




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 at 1:53




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 at 1:53










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










With $u<v:$ Let $f(x)=0$ for $xleq u.$ Let $f(x)=1$ for $xgeq v.$



Let $ nin 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 $f(x)=frac 1Kint_u^x g(y)dy.$



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



Note: $Kne 0$ because either (i) $g(y)>0$ for all $yin (u,v)$ or (ii) $g(y)<0$ for all $yin (u,v).$






share|cite|improve this answer





















  • 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 at 15:52






  • 1




    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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
    – DanielWainfleet
    Jul 22 at 18:03


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










With $u<v:$ Let $f(x)=0$ for $xleq u.$ Let $f(x)=1$ for $xgeq v.$



Let $ nin 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 $f(x)=frac 1Kint_u^x g(y)dy.$



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



Note: $Kne 0$ because either (i) $g(y)>0$ for all $yin (u,v)$ or (ii) $g(y)<0$ for all $yin (u,v).$






share|cite|improve this answer





















  • 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 at 15:52






  • 1




    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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
    – DanielWainfleet
    Jul 22 at 18:03















up vote
1
down vote



accepted










With $u<v:$ Let $f(x)=0$ for $xleq u.$ Let $f(x)=1$ for $xgeq v.$



Let $ nin 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 $f(x)=frac 1Kint_u^x g(y)dy.$



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



Note: $Kne 0$ because either (i) $g(y)>0$ for all $yin (u,v)$ or (ii) $g(y)<0$ for all $yin (u,v).$






share|cite|improve this answer





















  • 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 at 15:52






  • 1




    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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
    – DanielWainfleet
    Jul 22 at 18:03













up vote
1
down vote



accepted







up vote
1
down vote



accepted






With $u<v:$ Let $f(x)=0$ for $xleq u.$ Let $f(x)=1$ for $xgeq v.$



Let $ nin 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 $f(x)=frac 1Kint_u^x g(y)dy.$



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



Note: $Kne 0$ because either (i) $g(y)>0$ for all $yin (u,v)$ or (ii) $g(y)<0$ for all $yin (u,v).$






share|cite|improve this answer













With $u<v:$ Let $f(x)=0$ for $xleq u.$ Let $f(x)=1$ for $xgeq v.$



Let $ nin 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 $f(x)=frac 1Kint_u^x g(y)dy.$



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



Note: $Kne 0$ because either (i) $g(y)>0$ for all $yin (u,v)$ or (ii) $g(y)<0$ for all $yin (u,v).$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 21 at 1:57









DanielWainfleet

31.6k31643




31.6k31643











  • 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 at 15:52






  • 1




    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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
    – DanielWainfleet
    Jul 22 at 18:03

















  • 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 at 15:52






  • 1




    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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
    – DanielWainfleet
    Jul 22 at 18:03
















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 at 15:52




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 at 15:52




1




1




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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
– DanielWainfleet
Jul 22 at 18:03





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 $r xin (u,v),$ then the $m$th derivative $f^(m)(x)$ is the non-zero constant $m!A$ when $xin (u,v)$.... but $f^(m)(x)=0$ when $x<u.$
– DanielWainfleet
Jul 22 at 18:03



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