$G'(0) = int_{[0,1] backslash Z(g)} h(x) cdot text{sign}(g(x)) mathrm{d}x$











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Today during an exam I got the following exercise :




Let $h, g in C^0([0,1], | cdot |_1)$ such that the set : $Z(g) = {x in [0,1] mid g(x) = 0}$ is a finite union of intervals. Then let's defined :
$$ G : t mapsto | g + th |_1$$



If the function $G$ has a derivative at $0$ prove that :
$$G'(0) = int_{[0,1] backslash Z(g)} h(x) cdot sign(g(x)) mathrm{d}x$$




First of all I don't understand how $G$ is defined because do we consider that :
$$G(t) = int_{[0,1]} mid g(x) + th(x) mid mathrm{d}x$$



Or we consider that :



$$G(t) = int_{[0,1]} mid g(t) + th(t) mid mathrm{d}t$$
?



Then I tried considering :
$$frac{G(h) - G(0)}{h}$$
In order to find the derivative. Yet the $mid cdot mid$ make the task not so easy.



So I tried the to split the integrand and study the part : $int_{Z(g)}$
I get (using the second interpretation of $G$) :



$$frac{int_{Z(g)} mid th(t) mid mathrm{d}t}{t} $$



But it doesn't seem to help...










share|cite|improve this question




























    up vote
    2
    down vote

    favorite












    Today during an exam I got the following exercise :




    Let $h, g in C^0([0,1], | cdot |_1)$ such that the set : $Z(g) = {x in [0,1] mid g(x) = 0}$ is a finite union of intervals. Then let's defined :
    $$ G : t mapsto | g + th |_1$$



    If the function $G$ has a derivative at $0$ prove that :
    $$G'(0) = int_{[0,1] backslash Z(g)} h(x) cdot sign(g(x)) mathrm{d}x$$




    First of all I don't understand how $G$ is defined because do we consider that :
    $$G(t) = int_{[0,1]} mid g(x) + th(x) mid mathrm{d}x$$



    Or we consider that :



    $$G(t) = int_{[0,1]} mid g(t) + th(t) mid mathrm{d}t$$
    ?



    Then I tried considering :
    $$frac{G(h) - G(0)}{h}$$
    In order to find the derivative. Yet the $mid cdot mid$ make the task not so easy.



    So I tried the to split the integrand and study the part : $int_{Z(g)}$
    I get (using the second interpretation of $G$) :



    $$frac{int_{Z(g)} mid th(t) mid mathrm{d}t}{t} $$



    But it doesn't seem to help...










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Today during an exam I got the following exercise :




      Let $h, g in C^0([0,1], | cdot |_1)$ such that the set : $Z(g) = {x in [0,1] mid g(x) = 0}$ is a finite union of intervals. Then let's defined :
      $$ G : t mapsto | g + th |_1$$



      If the function $G$ has a derivative at $0$ prove that :
      $$G'(0) = int_{[0,1] backslash Z(g)} h(x) cdot sign(g(x)) mathrm{d}x$$




      First of all I don't understand how $G$ is defined because do we consider that :
      $$G(t) = int_{[0,1]} mid g(x) + th(x) mid mathrm{d}x$$



      Or we consider that :



      $$G(t) = int_{[0,1]} mid g(t) + th(t) mid mathrm{d}t$$
      ?



      Then I tried considering :
      $$frac{G(h) - G(0)}{h}$$
      In order to find the derivative. Yet the $mid cdot mid$ make the task not so easy.



      So I tried the to split the integrand and study the part : $int_{Z(g)}$
      I get (using the second interpretation of $G$) :



      $$frac{int_{Z(g)} mid th(t) mid mathrm{d}t}{t} $$



      But it doesn't seem to help...










      share|cite|improve this question















      Today during an exam I got the following exercise :




      Let $h, g in C^0([0,1], | cdot |_1)$ such that the set : $Z(g) = {x in [0,1] mid g(x) = 0}$ is a finite union of intervals. Then let's defined :
      $$ G : t mapsto | g + th |_1$$



      If the function $G$ has a derivative at $0$ prove that :
      $$G'(0) = int_{[0,1] backslash Z(g)} h(x) cdot sign(g(x)) mathrm{d}x$$




      First of all I don't understand how $G$ is defined because do we consider that :
      $$G(t) = int_{[0,1]} mid g(x) + th(x) mid mathrm{d}x$$



      Or we consider that :



      $$G(t) = int_{[0,1]} mid g(t) + th(t) mid mathrm{d}t$$
      ?



      Then I tried considering :
      $$frac{G(h) - G(0)}{h}$$
      In order to find the derivative. Yet the $mid cdot mid$ make the task not so easy.



      So I tried the to split the integrand and study the part : $int_{Z(g)}$
      I get (using the second interpretation of $G$) :



      $$frac{int_{Z(g)} mid th(t) mid mathrm{d}t}{t} $$



      But it doesn't seem to help...







      calculus real-analysis integration sequences-and-series derivatives






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 at 21:36









      Henry Lee

      1,684218




      1,684218










      asked Nov 25 at 19:27









      DP_q

      856




      856






















          1 Answer
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          It has to be
          $$G(t)=int_0^1|g(x)+th(x)|,dx $$
          otherwise $G(t)$ does not depend on $t$. We have
          $$frac{G(t)-G(0)}{t}=int_0^1frac{|g(x)+th(x)|-|g(x)|}{t},dx= $$
          By triangle inequality,
          $$frac{|g(x)+th(x)|-|g(x)|}{t}leq |h(x)|in L^1(0,1) $$
          Thus by dominated convergence, if $G'(0)$ exists, then
          $$G'(0)=int_0^1lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t},dx$$
          Now we distinsguish three cases.




          • If $g(x)=0$, i.e. $xin Z_g$, then $$ lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0}frac{|t|h(x)}{t}$$


          Notice that the above limit exists if and only if $h(x)=0$, and in this case, it is equal to $0$. Recall that we are assuming that $G'(0)$ exists. Now suppose by contradiction that there existed an $x_0in [0,1]$ such that $g(x_0)=0$ but $h(x_0)neq 0$. Since $Z_g$ is a finite union of intervals, we have $x_0in [a,b]$ with $g(x)=0$ on $[a,b]$. Moreover, since $h(x_0)neq 0$ and $h$ is continuous, we have $h(x)neq 0$ on $(x_0-varepsilon,x_0+varepsilon)subset [a,b]$. Thus the above limit does not exist on a whole interval $(x_0-varepsilon,x_0+varepsilon)$ (not just on $x_0$!). This implies that $G'(0)$ is not defined, contradicting our assumption. Therefore the above limit always exists and is equal to $0$.





          • If $g(x)>0$, then for $t$ small enough we have $g(x)+th(x)>0$. Thus



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{g(x)+th(x)-g(x)}{t}=h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          • Similarly, if $g(x)<0$ then



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{-g(x)-th(x)+g(x)}{t}=-h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          Hence
          $$G'(0)=int_{[0,1]setminus Z_g}h(x)cdot operatorname{sign}(g(x)),dx $$






          share|cite|improve this answer























          • This is very clear, thank you very much for your answer.
            – DP_q
            Nov 25 at 20:31











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          up vote
          2
          down vote



          accepted










          It has to be
          $$G(t)=int_0^1|g(x)+th(x)|,dx $$
          otherwise $G(t)$ does not depend on $t$. We have
          $$frac{G(t)-G(0)}{t}=int_0^1frac{|g(x)+th(x)|-|g(x)|}{t},dx= $$
          By triangle inequality,
          $$frac{|g(x)+th(x)|-|g(x)|}{t}leq |h(x)|in L^1(0,1) $$
          Thus by dominated convergence, if $G'(0)$ exists, then
          $$G'(0)=int_0^1lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t},dx$$
          Now we distinsguish three cases.




          • If $g(x)=0$, i.e. $xin Z_g$, then $$ lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0}frac{|t|h(x)}{t}$$


          Notice that the above limit exists if and only if $h(x)=0$, and in this case, it is equal to $0$. Recall that we are assuming that $G'(0)$ exists. Now suppose by contradiction that there existed an $x_0in [0,1]$ such that $g(x_0)=0$ but $h(x_0)neq 0$. Since $Z_g$ is a finite union of intervals, we have $x_0in [a,b]$ with $g(x)=0$ on $[a,b]$. Moreover, since $h(x_0)neq 0$ and $h$ is continuous, we have $h(x)neq 0$ on $(x_0-varepsilon,x_0+varepsilon)subset [a,b]$. Thus the above limit does not exist on a whole interval $(x_0-varepsilon,x_0+varepsilon)$ (not just on $x_0$!). This implies that $G'(0)$ is not defined, contradicting our assumption. Therefore the above limit always exists and is equal to $0$.





          • If $g(x)>0$, then for $t$ small enough we have $g(x)+th(x)>0$. Thus



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{g(x)+th(x)-g(x)}{t}=h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          • Similarly, if $g(x)<0$ then



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{-g(x)-th(x)+g(x)}{t}=-h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          Hence
          $$G'(0)=int_{[0,1]setminus Z_g}h(x)cdot operatorname{sign}(g(x)),dx $$






          share|cite|improve this answer























          • This is very clear, thank you very much for your answer.
            – DP_q
            Nov 25 at 20:31















          up vote
          2
          down vote



          accepted










          It has to be
          $$G(t)=int_0^1|g(x)+th(x)|,dx $$
          otherwise $G(t)$ does not depend on $t$. We have
          $$frac{G(t)-G(0)}{t}=int_0^1frac{|g(x)+th(x)|-|g(x)|}{t},dx= $$
          By triangle inequality,
          $$frac{|g(x)+th(x)|-|g(x)|}{t}leq |h(x)|in L^1(0,1) $$
          Thus by dominated convergence, if $G'(0)$ exists, then
          $$G'(0)=int_0^1lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t},dx$$
          Now we distinsguish three cases.




          • If $g(x)=0$, i.e. $xin Z_g$, then $$ lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0}frac{|t|h(x)}{t}$$


          Notice that the above limit exists if and only if $h(x)=0$, and in this case, it is equal to $0$. Recall that we are assuming that $G'(0)$ exists. Now suppose by contradiction that there existed an $x_0in [0,1]$ such that $g(x_0)=0$ but $h(x_0)neq 0$. Since $Z_g$ is a finite union of intervals, we have $x_0in [a,b]$ with $g(x)=0$ on $[a,b]$. Moreover, since $h(x_0)neq 0$ and $h$ is continuous, we have $h(x)neq 0$ on $(x_0-varepsilon,x_0+varepsilon)subset [a,b]$. Thus the above limit does not exist on a whole interval $(x_0-varepsilon,x_0+varepsilon)$ (not just on $x_0$!). This implies that $G'(0)$ is not defined, contradicting our assumption. Therefore the above limit always exists and is equal to $0$.





          • If $g(x)>0$, then for $t$ small enough we have $g(x)+th(x)>0$. Thus



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{g(x)+th(x)-g(x)}{t}=h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          • Similarly, if $g(x)<0$ then



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{-g(x)-th(x)+g(x)}{t}=-h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          Hence
          $$G'(0)=int_{[0,1]setminus Z_g}h(x)cdot operatorname{sign}(g(x)),dx $$






          share|cite|improve this answer























          • This is very clear, thank you very much for your answer.
            – DP_q
            Nov 25 at 20:31













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          It has to be
          $$G(t)=int_0^1|g(x)+th(x)|,dx $$
          otherwise $G(t)$ does not depend on $t$. We have
          $$frac{G(t)-G(0)}{t}=int_0^1frac{|g(x)+th(x)|-|g(x)|}{t},dx= $$
          By triangle inequality,
          $$frac{|g(x)+th(x)|-|g(x)|}{t}leq |h(x)|in L^1(0,1) $$
          Thus by dominated convergence, if $G'(0)$ exists, then
          $$G'(0)=int_0^1lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t},dx$$
          Now we distinsguish three cases.




          • If $g(x)=0$, i.e. $xin Z_g$, then $$ lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0}frac{|t|h(x)}{t}$$


          Notice that the above limit exists if and only if $h(x)=0$, and in this case, it is equal to $0$. Recall that we are assuming that $G'(0)$ exists. Now suppose by contradiction that there existed an $x_0in [0,1]$ such that $g(x_0)=0$ but $h(x_0)neq 0$. Since $Z_g$ is a finite union of intervals, we have $x_0in [a,b]$ with $g(x)=0$ on $[a,b]$. Moreover, since $h(x_0)neq 0$ and $h$ is continuous, we have $h(x)neq 0$ on $(x_0-varepsilon,x_0+varepsilon)subset [a,b]$. Thus the above limit does not exist on a whole interval $(x_0-varepsilon,x_0+varepsilon)$ (not just on $x_0$!). This implies that $G'(0)$ is not defined, contradicting our assumption. Therefore the above limit always exists and is equal to $0$.





          • If $g(x)>0$, then for $t$ small enough we have $g(x)+th(x)>0$. Thus



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{g(x)+th(x)-g(x)}{t}=h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          • Similarly, if $g(x)<0$ then



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{-g(x)-th(x)+g(x)}{t}=-h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          Hence
          $$G'(0)=int_{[0,1]setminus Z_g}h(x)cdot operatorname{sign}(g(x)),dx $$






          share|cite|improve this answer














          It has to be
          $$G(t)=int_0^1|g(x)+th(x)|,dx $$
          otherwise $G(t)$ does not depend on $t$. We have
          $$frac{G(t)-G(0)}{t}=int_0^1frac{|g(x)+th(x)|-|g(x)|}{t},dx= $$
          By triangle inequality,
          $$frac{|g(x)+th(x)|-|g(x)|}{t}leq |h(x)|in L^1(0,1) $$
          Thus by dominated convergence, if $G'(0)$ exists, then
          $$G'(0)=int_0^1lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t},dx$$
          Now we distinsguish three cases.




          • If $g(x)=0$, i.e. $xin Z_g$, then $$ lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0}frac{|t|h(x)}{t}$$


          Notice that the above limit exists if and only if $h(x)=0$, and in this case, it is equal to $0$. Recall that we are assuming that $G'(0)$ exists. Now suppose by contradiction that there existed an $x_0in [0,1]$ such that $g(x_0)=0$ but $h(x_0)neq 0$. Since $Z_g$ is a finite union of intervals, we have $x_0in [a,b]$ with $g(x)=0$ on $[a,b]$. Moreover, since $h(x_0)neq 0$ and $h$ is continuous, we have $h(x)neq 0$ on $(x_0-varepsilon,x_0+varepsilon)subset [a,b]$. Thus the above limit does not exist on a whole interval $(x_0-varepsilon,x_0+varepsilon)$ (not just on $x_0$!). This implies that $G'(0)$ is not defined, contradicting our assumption. Therefore the above limit always exists and is equal to $0$.





          • If $g(x)>0$, then for $t$ small enough we have $g(x)+th(x)>0$. Thus



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{g(x)+th(x)-g(x)}{t}=h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          • Similarly, if $g(x)<0$ then



            $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{-g(x)-th(x)+g(x)}{t}=-h(x)=h(x)cdotoperatorname{sign}(g(x))$$




          Hence
          $$G'(0)=int_{[0,1]setminus Z_g}h(x)cdot operatorname{sign}(g(x)),dx $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 20:03

























          answered Nov 25 at 19:54









          Lorenzo Quarisa

          3,121316




          3,121316












          • This is very clear, thank you very much for your answer.
            – DP_q
            Nov 25 at 20:31


















          • This is very clear, thank you very much for your answer.
            – DP_q
            Nov 25 at 20:31
















          This is very clear, thank you very much for your answer.
          – DP_q
          Nov 25 at 20:31




          This is very clear, thank you very much for your answer.
          – DP_q
          Nov 25 at 20:31


















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