a non-zero continuous function with compact support and its derivatives are zero at zero












3












$begingroup$


I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?






real-analysis derivatives continuity







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    3












    $begingroup$


    I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
    Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?






    real-analysis derivatives continuity







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
      Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?






      real-analysis derivatives continuity







      share|cite|improve this question









      $endgroup$




      I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
      Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?






      real-analysis derivatives continuity




      real-analysis derivatives continuity






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 23:20









      David LlerenaDavid Llerena

      254




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          2 Answers
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          4












          $begingroup$

          Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.



          Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.



          Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
            $$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              your function is not differentiable everywhere.
              $endgroup$
              – ersh
              Jan 15 at 0:15










            • $begingroup$
              I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
              $endgroup$
              – Matt A Pelto
              Jan 15 at 0:31














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            2 Answers
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            active

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            active

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            active

            oldest

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            4












            $begingroup$

            Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.



            Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.



            Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.



              Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.



              Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.



                Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.



                Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.






                share|cite|improve this answer









                $endgroup$



                Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.



                Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.



                Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 23:42









                MindlackMindlack

                4,910211




                4,910211























                    2












                    $begingroup$

                    A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
                    $$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      your function is not differentiable everywhere.
                      $endgroup$
                      – ersh
                      Jan 15 at 0:15










                    • $begingroup$
                      I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
                      $endgroup$
                      – Matt A Pelto
                      Jan 15 at 0:31


















                    2












                    $begingroup$

                    A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
                    $$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      your function is not differentiable everywhere.
                      $endgroup$
                      – ersh
                      Jan 15 at 0:15










                    • $begingroup$
                      I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
                      $endgroup$
                      – Matt A Pelto
                      Jan 15 at 0:31
















                    2












                    2








                    2





                    $begingroup$

                    A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
                    $$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$






                    share|cite|improve this answer









                    $endgroup$



                    A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
                    $$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 14 at 23:37









                    Matt A PeltoMatt A Pelto

                    2,707621




                    2,707621








                    • 1




                      $begingroup$
                      your function is not differentiable everywhere.
                      $endgroup$
                      – ersh
                      Jan 15 at 0:15










                    • $begingroup$
                      I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
                      $endgroup$
                      – Matt A Pelto
                      Jan 15 at 0:31
















                    • 1




                      $begingroup$
                      your function is not differentiable everywhere.
                      $endgroup$
                      – ersh
                      Jan 15 at 0:15










                    • $begingroup$
                      I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
                      $endgroup$
                      – Matt A Pelto
                      Jan 15 at 0:31










                    1




                    1




                    $begingroup$
                    your function is not differentiable everywhere.
                    $endgroup$
                    – ersh
                    Jan 15 at 0:15




                    $begingroup$
                    your function is not differentiable everywhere.
                    $endgroup$
                    – ersh
                    Jan 15 at 0:15












                    $begingroup$
                    I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
                    $endgroup$
                    – Matt A Pelto
                    Jan 15 at 0:31






                    $begingroup$
                    I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
                    $endgroup$
                    – Matt A Pelto
                    Jan 15 at 0:31




















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