How to construct a bump function where one plateau is not at zero?











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Let $f : mathbb{R} to mathbb{R}$ be a linear function with $f(x_0)=0$, $f(epsilon)=1$, where $varepsilon >0$ fulfills $x_0-varepsilon > varepsilon>0$. I am trying to construct a smooth function $psi: mathbb{R} to [0,1]$, which fullfills $psi(x)=begin{cases} 1, text{ for } x le varepsilon - r_0 \ f(x), text{ for } varepsilon +r_0 le x le x_0-varepsilon -r_1 \ 0, text{ for } x ge x_0 -varepsilon + r_1 end{cases}$,



where $r_0, r_1 >0$ are to be chosen in such a way that the open neighbourhoods $(varepsilon - r_0, varepsilon + r_0)$ and $(x_0 - varepsilon - r_1, x_0-varepsilon +r_1)$ are non-empty and disjoint.



I am now looking for a way to construct two bump functions, which will make my function smooth (by which I mean $C^{infty}$) in those intervals. I know that I can construct a bump function by using the function
$ f(x)=begin{cases}e^{-frac{1}{x}}&text{for $x ge 0$}\ 0&text{for $x le 0$.}end{cases}$
and then taking $g(x)=frac{f(x)}{f(x) + f(1-x)}=begin{cases}1&text{for $x ge 1$} \ 0 &text{for $x le 0$ } end{cases}$.



I know how to change the shape by shifting etc. In particular I am familiar with the construction in Loring W. Tu's book.



But I have two problems:




  1. This bump function takes values from $0$ to $1$, but the function that I want should take values between $1$ and $f(epsilon+r_0)$ respectively $f(x_0-varepsilon-r_1)$ and $0$. I don't really know how to do this.


  2. I am concerned with smoothness. I have an idea of what these bump functions would look like, for instance, in the intervall $(varepsilon - r_0, varepsilon + r_0)$ it should look like the constant function $1$ near the left boundary and the constant function $f(varepsilon+r_0)$ near the right one. How can the transition from the bump function to my linear function $f$ be smooth?







real-analysis functional-analysis differential-equations analysis







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    up vote
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    down vote

    favorite












    Let $f : mathbb{R} to mathbb{R}$ be a linear function with $f(x_0)=0$, $f(epsilon)=1$, where $varepsilon >0$ fulfills $x_0-varepsilon > varepsilon>0$. I am trying to construct a smooth function $psi: mathbb{R} to [0,1]$, which fullfills $psi(x)=begin{cases} 1, text{ for } x le varepsilon - r_0 \ f(x), text{ for } varepsilon +r_0 le x le x_0-varepsilon -r_1 \ 0, text{ for } x ge x_0 -varepsilon + r_1 end{cases}$,



    where $r_0, r_1 >0$ are to be chosen in such a way that the open neighbourhoods $(varepsilon - r_0, varepsilon + r_0)$ and $(x_0 - varepsilon - r_1, x_0-varepsilon +r_1)$ are non-empty and disjoint.



    I am now looking for a way to construct two bump functions, which will make my function smooth (by which I mean $C^{infty}$) in those intervals. I know that I can construct a bump function by using the function
    $ f(x)=begin{cases}e^{-frac{1}{x}}&text{for $x ge 0$}\ 0&text{for $x le 0$.}end{cases}$
    and then taking $g(x)=frac{f(x)}{f(x) + f(1-x)}=begin{cases}1&text{for $x ge 1$} \ 0 &text{for $x le 0$ } end{cases}$.



    I know how to change the shape by shifting etc. In particular I am familiar with the construction in Loring W. Tu's book.



    But I have two problems:




    1. This bump function takes values from $0$ to $1$, but the function that I want should take values between $1$ and $f(epsilon+r_0)$ respectively $f(x_0-varepsilon-r_1)$ and $0$. I don't really know how to do this.


    2. I am concerned with smoothness. I have an idea of what these bump functions would look like, for instance, in the intervall $(varepsilon - r_0, varepsilon + r_0)$ it should look like the constant function $1$ near the left boundary and the constant function $f(varepsilon+r_0)$ near the right one. How can the transition from the bump function to my linear function $f$ be smooth?







    real-analysis functional-analysis differential-equations analysis







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $f : mathbb{R} to mathbb{R}$ be a linear function with $f(x_0)=0$, $f(epsilon)=1$, where $varepsilon >0$ fulfills $x_0-varepsilon > varepsilon>0$. I am trying to construct a smooth function $psi: mathbb{R} to [0,1]$, which fullfills $psi(x)=begin{cases} 1, text{ for } x le varepsilon - r_0 \ f(x), text{ for } varepsilon +r_0 le x le x_0-varepsilon -r_1 \ 0, text{ for } x ge x_0 -varepsilon + r_1 end{cases}$,



      where $r_0, r_1 >0$ are to be chosen in such a way that the open neighbourhoods $(varepsilon - r_0, varepsilon + r_0)$ and $(x_0 - varepsilon - r_1, x_0-varepsilon +r_1)$ are non-empty and disjoint.



      I am now looking for a way to construct two bump functions, which will make my function smooth (by which I mean $C^{infty}$) in those intervals. I know that I can construct a bump function by using the function
      $ f(x)=begin{cases}e^{-frac{1}{x}}&text{for $x ge 0$}\ 0&text{for $x le 0$.}end{cases}$
      and then taking $g(x)=frac{f(x)}{f(x) + f(1-x)}=begin{cases}1&text{for $x ge 1$} \ 0 &text{for $x le 0$ } end{cases}$.



      I know how to change the shape by shifting etc. In particular I am familiar with the construction in Loring W. Tu's book.



      But I have two problems:




      1. This bump function takes values from $0$ to $1$, but the function that I want should take values between $1$ and $f(epsilon+r_0)$ respectively $f(x_0-varepsilon-r_1)$ and $0$. I don't really know how to do this.


      2. I am concerned with smoothness. I have an idea of what these bump functions would look like, for instance, in the intervall $(varepsilon - r_0, varepsilon + r_0)$ it should look like the constant function $1$ near the left boundary and the constant function $f(varepsilon+r_0)$ near the right one. How can the transition from the bump function to my linear function $f$ be smooth?







      real-analysis functional-analysis differential-equations analysis







      share|cite|improve this question















      Let $f : mathbb{R} to mathbb{R}$ be a linear function with $f(x_0)=0$, $f(epsilon)=1$, where $varepsilon >0$ fulfills $x_0-varepsilon > varepsilon>0$. I am trying to construct a smooth function $psi: mathbb{R} to [0,1]$, which fullfills $psi(x)=begin{cases} 1, text{ for } x le varepsilon - r_0 \ f(x), text{ for } varepsilon +r_0 le x le x_0-varepsilon -r_1 \ 0, text{ for } x ge x_0 -varepsilon + r_1 end{cases}$,



      where $r_0, r_1 >0$ are to be chosen in such a way that the open neighbourhoods $(varepsilon - r_0, varepsilon + r_0)$ and $(x_0 - varepsilon - r_1, x_0-varepsilon +r_1)$ are non-empty and disjoint.



      I am now looking for a way to construct two bump functions, which will make my function smooth (by which I mean $C^{infty}$) in those intervals. I know that I can construct a bump function by using the function
      $ f(x)=begin{cases}e^{-frac{1}{x}}&text{for $x ge 0$}\ 0&text{for $x le 0$.}end{cases}$
      and then taking $g(x)=frac{f(x)}{f(x) + f(1-x)}=begin{cases}1&text{for $x ge 1$} \ 0 &text{for $x le 0$ } end{cases}$.



      I know how to change the shape by shifting etc. In particular I am familiar with the construction in Loring W. Tu's book.



      But I have two problems:




      1. This bump function takes values from $0$ to $1$, but the function that I want should take values between $1$ and $f(epsilon+r_0)$ respectively $f(x_0-varepsilon-r_1)$ and $0$. I don't really know how to do this.


      2. I am concerned with smoothness. I have an idea of what these bump functions would look like, for instance, in the intervall $(varepsilon - r_0, varepsilon + r_0)$ it should look like the constant function $1$ near the left boundary and the constant function $f(varepsilon+r_0)$ near the right one. How can the transition from the bump function to my linear function $f$ be smooth?







      real-analysis functional-analysis differential-equations analysis




      real-analysis functional-analysis differential-equations analysis






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 22 at 16:50

























      asked Nov 22 at 16:43









      rhodelta

      206




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          1 Answer
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          You should construct a bump function $h_{a,b,varepsilon}$ with $mathrm{supp} , h = [a-varepsilon,b+varepsilon]$ and $h(x) =1$ on $[a,b]$, see e.g. here. You can also construct such functions by using $g$: Define $$h(x) = gleft(frac{x-(a-varepsilon)}{varepsilon}right) cdot g left(frac{(b+varepsilon)-x}{varepsilon} right).$$
          Now you can easily construct your function by taking bump-functions which are supported on disjoint neighbourhoods.






          share|cite|improve this answer





















          • I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
            – rhodelta
            Nov 26 at 10:18












          • Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
            – rhodelta
            Nov 26 at 10:38










          • You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
            – p4sch
            Nov 26 at 10:46












          • Thank you so much, this really helped me a lot.
            – rhodelta
            Nov 26 at 21:58










          • I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
            – rhodelta
            Nov 29 at 19:02











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          1 Answer
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          up vote
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          accepted










          You should construct a bump function $h_{a,b,varepsilon}$ with $mathrm{supp} , h = [a-varepsilon,b+varepsilon]$ and $h(x) =1$ on $[a,b]$, see e.g. here. You can also construct such functions by using $g$: Define $$h(x) = gleft(frac{x-(a-varepsilon)}{varepsilon}right) cdot g left(frac{(b+varepsilon)-x}{varepsilon} right).$$
          Now you can easily construct your function by taking bump-functions which are supported on disjoint neighbourhoods.






          share|cite|improve this answer





















          • I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
            – rhodelta
            Nov 26 at 10:18












          • Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
            – rhodelta
            Nov 26 at 10:38










          • You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
            – p4sch
            Nov 26 at 10:46












          • Thank you so much, this really helped me a lot.
            – rhodelta
            Nov 26 at 21:58










          • I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
            – rhodelta
            Nov 29 at 19:02















          up vote
          1
          down vote



          accepted










          You should construct a bump function $h_{a,b,varepsilon}$ with $mathrm{supp} , h = [a-varepsilon,b+varepsilon]$ and $h(x) =1$ on $[a,b]$, see e.g. here. You can also construct such functions by using $g$: Define $$h(x) = gleft(frac{x-(a-varepsilon)}{varepsilon}right) cdot g left(frac{(b+varepsilon)-x}{varepsilon} right).$$
          Now you can easily construct your function by taking bump-functions which are supported on disjoint neighbourhoods.






          share|cite|improve this answer





















          • I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
            – rhodelta
            Nov 26 at 10:18












          • Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
            – rhodelta
            Nov 26 at 10:38










          • You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
            – p4sch
            Nov 26 at 10:46












          • Thank you so much, this really helped me a lot.
            – rhodelta
            Nov 26 at 21:58










          • I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
            – rhodelta
            Nov 29 at 19:02













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You should construct a bump function $h_{a,b,varepsilon}$ with $mathrm{supp} , h = [a-varepsilon,b+varepsilon]$ and $h(x) =1$ on $[a,b]$, see e.g. here. You can also construct such functions by using $g$: Define $$h(x) = gleft(frac{x-(a-varepsilon)}{varepsilon}right) cdot g left(frac{(b+varepsilon)-x}{varepsilon} right).$$
          Now you can easily construct your function by taking bump-functions which are supported on disjoint neighbourhoods.






          share|cite|improve this answer












          You should construct a bump function $h_{a,b,varepsilon}$ with $mathrm{supp} , h = [a-varepsilon,b+varepsilon]$ and $h(x) =1$ on $[a,b]$, see e.g. here. You can also construct such functions by using $g$: Define $$h(x) = gleft(frac{x-(a-varepsilon)}{varepsilon}right) cdot g left(frac{(b+varepsilon)-x}{varepsilon} right).$$
          Now you can easily construct your function by taking bump-functions which are supported on disjoint neighbourhoods.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 17:16









          p4sch

          4,800217




          4,800217












          • I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
            – rhodelta
            Nov 26 at 10:18












          • Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
            – rhodelta
            Nov 26 at 10:38










          • You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
            – p4sch
            Nov 26 at 10:46












          • Thank you so much, this really helped me a lot.
            – rhodelta
            Nov 26 at 21:58










          • I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
            – rhodelta
            Nov 29 at 19:02


















          • I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
            – rhodelta
            Nov 26 at 10:18












          • Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
            – rhodelta
            Nov 26 at 10:38










          • You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
            – p4sch
            Nov 26 at 10:46












          • Thank you so much, this really helped me a lot.
            – rhodelta
            Nov 26 at 21:58










          • I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
            – rhodelta
            Nov 29 at 19:02
















          I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
          – rhodelta
          Nov 26 at 10:18






          I see that if I take for instance $h(x)=1-g(frac{x-(varepsilon-r_0)}{2r_0})$, that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \0, text{ for } x ge varepsilon+r_0 end{cases}$. But, what I want is instead that $h(x)=begin{cases} 1, text{ for } x le varepsilon -r_0, \ f(varepsilon+r_0), text{ for } x ge varepsilon+r_0 end{cases}$. If I just shift my function, this will also change the value where $h$ was 1. How can I change my $h$ in such a way, that this condition is fulfilled?
          – rhodelta
          Nov 26 at 10:18














          Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
          – rhodelta
          Nov 26 at 10:38




          Okay, I think I could take instead $1-(f(varepsilon+r_0)+1) g(frac{x-(varepsilon-r_0)}{2r_0})=begin{cases} 1, text{ if } x le varepsilon-r_0, \ f(varepsilon+r_0), text{ if } x ge varepsilon+r_0 end{cases}$. Will this glue smoothly?
          – rhodelta
          Nov 26 at 10:38












          You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
          – p4sch
          Nov 26 at 10:46






          You can take a function, say $h_1$, with $h_1(x) =1$ for $x le varepsilon-r_0$ and $h_1(x)=0$ for $x ge varepsilon-r_0/2$. (Just modifiy your function $g$ by rescaling!) Now take $h_2$ as in my answer with $a= varepsilon+r_0$ and $b=x_0-varepsilon -r_1$ and $widetilde{epsilon} = min {r_0,r_1}$. Then $h_2$ is supported on $[varepsilon,x_0-varepsilon]$ and is constant one on $[varepsilon+r_0,x_0-varepsilon-r_1]$. Define $psi = h_1 + h_2 f$. (This choice works!)
          – p4sch
          Nov 26 at 10:46














          Thank you so much, this really helped me a lot.
          – rhodelta
          Nov 26 at 21:58




          Thank you so much, this really helped me a lot.
          – rhodelta
          Nov 26 at 21:58












          I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
          – rhodelta
          Nov 29 at 19:02




          I don't know if this is easy to see, but I have been wondering if it is possible to not have to restrict to $varepsilon=min{r_0, r_1}$. I feel like this is very restrictive, for instance if I want my function to be 1 in 0, then I wouldn't be able to use this construction.
          – rhodelta
          Nov 29 at 19:02


















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