Proving the chain rule with upper derivative












0














Let $f$ be a function on $[a,b]$, and $g$ is continuous on $[alpha,beta]$ that is differentiable at $gammain (alpha,beta)$ with $g(gamma)=c in (a,b).$ Show that



(i) If $g'(gamma)>0,$ then $overline{D}(f circ g)(gamma)=overline{D}f(c).g'(gamma).$



(ii) If $g'(gamma)=0$, and $overline{D}f(c)$, $underline{D}f(c)$ finite, then $overline{D}(f circ g)(gamma)=0.$



By the definition we have that $$overline{D}(f circ g)(gamma)=lim_{h rightarrow 0}supfrac{(f circ g)(gamma+h)-(f circ g)(gamma)}{h}=lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}.$$
Since $g$ is differantiable at $gamma$ then, $overline{D}g(gamma)=g'(gamma)>0$, and by the definition of the differentiability existence $overline{D}g(gamma)=g'(gamma)$ is finite, so we can multiply and divide (1) by $g'(gamma)$, so
begin{align*}
begin{split}
& lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}right).frac{g'(gamma)}{g'(gamma)}
\ &= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}. frac{g(gamma+h)-g(gamma)}{g(gamma+h)-g(gamma)} right)=left(lim_{h rightarrow 0}supfrac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)
end{split}
end{align*}
where we used the limit properties for the quantity before the last quantity. Now, define $u:=frac{g(gamma+h)-g(gamma)}{h}-g'(gamma)$, and note that $u$ depends on $h$. Moreover, we see $u rightarrow 0$ as $h rightarrow 0$, and we have $g(gamma+h)=(u+g'(gamma))h+g(gamma)=(u+g'(gamma))h+c,$ so $g(gamma+h)-c=g(gamma+h)-g(gamma)=(u+g'(gamma))h.$ Thus, $$left(lim_{h rightarrow 0}sup frac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)=left(lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}right).g'(gamma) $$



Then, I am saying that $$lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=lim_{(u+g'(gamma)h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=overline{D}f(c)$$
which is I am not sure about it, and I dont know how to use the continuity of $g$ here.



Thanks for any help.






measure-theory lebesgue-measure







share|cite|improve this question



























    0














    Let $f$ be a function on $[a,b]$, and $g$ is continuous on $[alpha,beta]$ that is differentiable at $gammain (alpha,beta)$ with $g(gamma)=c in (a,b).$ Show that



    (i) If $g'(gamma)>0,$ then $overline{D}(f circ g)(gamma)=overline{D}f(c).g'(gamma).$



    (ii) If $g'(gamma)=0$, and $overline{D}f(c)$, $underline{D}f(c)$ finite, then $overline{D}(f circ g)(gamma)=0.$



    By the definition we have that $$overline{D}(f circ g)(gamma)=lim_{h rightarrow 0}supfrac{(f circ g)(gamma+h)-(f circ g)(gamma)}{h}=lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}.$$
    Since $g$ is differantiable at $gamma$ then, $overline{D}g(gamma)=g'(gamma)>0$, and by the definition of the differentiability existence $overline{D}g(gamma)=g'(gamma)$ is finite, so we can multiply and divide (1) by $g'(gamma)$, so
    begin{align*}
    begin{split}
    & lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}right).frac{g'(gamma)}{g'(gamma)}
    \ &= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}. frac{g(gamma+h)-g(gamma)}{g(gamma+h)-g(gamma)} right)=left(lim_{h rightarrow 0}supfrac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)
    end{split}
    end{align*}
    where we used the limit properties for the quantity before the last quantity. Now, define $u:=frac{g(gamma+h)-g(gamma)}{h}-g'(gamma)$, and note that $u$ depends on $h$. Moreover, we see $u rightarrow 0$ as $h rightarrow 0$, and we have $g(gamma+h)=(u+g'(gamma))h+g(gamma)=(u+g'(gamma))h+c,$ so $g(gamma+h)-c=g(gamma+h)-g(gamma)=(u+g'(gamma))h.$ Thus, $$left(lim_{h rightarrow 0}sup frac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)=left(lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}right).g'(gamma) $$



    Then, I am saying that $$lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=lim_{(u+g'(gamma)h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=overline{D}f(c)$$
    which is I am not sure about it, and I dont know how to use the continuity of $g$ here.



    Thanks for any help.






    measure-theory lebesgue-measure







    share|cite|improve this question

























      0












      0








      0







      Let $f$ be a function on $[a,b]$, and $g$ is continuous on $[alpha,beta]$ that is differentiable at $gammain (alpha,beta)$ with $g(gamma)=c in (a,b).$ Show that



      (i) If $g'(gamma)>0,$ then $overline{D}(f circ g)(gamma)=overline{D}f(c).g'(gamma).$



      (ii) If $g'(gamma)=0$, and $overline{D}f(c)$, $underline{D}f(c)$ finite, then $overline{D}(f circ g)(gamma)=0.$



      By the definition we have that $$overline{D}(f circ g)(gamma)=lim_{h rightarrow 0}supfrac{(f circ g)(gamma+h)-(f circ g)(gamma)}{h}=lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}.$$
      Since $g$ is differantiable at $gamma$ then, $overline{D}g(gamma)=g'(gamma)>0$, and by the definition of the differentiability existence $overline{D}g(gamma)=g'(gamma)$ is finite, so we can multiply and divide (1) by $g'(gamma)$, so
      begin{align*}
      begin{split}
      & lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}right).frac{g'(gamma)}{g'(gamma)}
      \ &= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}. frac{g(gamma+h)-g(gamma)}{g(gamma+h)-g(gamma)} right)=left(lim_{h rightarrow 0}supfrac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)
      end{split}
      end{align*}
      where we used the limit properties for the quantity before the last quantity. Now, define $u:=frac{g(gamma+h)-g(gamma)}{h}-g'(gamma)$, and note that $u$ depends on $h$. Moreover, we see $u rightarrow 0$ as $h rightarrow 0$, and we have $g(gamma+h)=(u+g'(gamma))h+g(gamma)=(u+g'(gamma))h+c,$ so $g(gamma+h)-c=g(gamma+h)-g(gamma)=(u+g'(gamma))h.$ Thus, $$left(lim_{h rightarrow 0}sup frac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)=left(lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}right).g'(gamma) $$



      Then, I am saying that $$lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=lim_{(u+g'(gamma)h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=overline{D}f(c)$$
      which is I am not sure about it, and I dont know how to use the continuity of $g$ here.



      Thanks for any help.






      measure-theory lebesgue-measure







      share|cite|improve this question













      Let $f$ be a function on $[a,b]$, and $g$ is continuous on $[alpha,beta]$ that is differentiable at $gammain (alpha,beta)$ with $g(gamma)=c in (a,b).$ Show that



      (i) If $g'(gamma)>0,$ then $overline{D}(f circ g)(gamma)=overline{D}f(c).g'(gamma).$



      (ii) If $g'(gamma)=0$, and $overline{D}f(c)$, $underline{D}f(c)$ finite, then $overline{D}(f circ g)(gamma)=0.$



      By the definition we have that $$overline{D}(f circ g)(gamma)=lim_{h rightarrow 0}supfrac{(f circ g)(gamma+h)-(f circ g)(gamma)}{h}=lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}.$$
      Since $g$ is differantiable at $gamma$ then, $overline{D}g(gamma)=g'(gamma)>0$, and by the definition of the differentiability existence $overline{D}g(gamma)=g'(gamma)$ is finite, so we can multiply and divide (1) by $g'(gamma)$, so
      begin{align*}
      begin{split}
      & lim_{h rightarrow 0}supfrac{f (g(gamma+h))-f (g(gamma))}{h}= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}right).frac{g'(gamma)}{g'(gamma)}
      \ &= lim_{h rightarrow 0}sup left(frac{f (g(gamma+h))-f (g(gamma))}{h}. frac{g(gamma+h)-g(gamma)}{g(gamma+h)-g(gamma)} right)=left(lim_{h rightarrow 0}supfrac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)
      end{split}
      end{align*}
      where we used the limit properties for the quantity before the last quantity. Now, define $u:=frac{g(gamma+h)-g(gamma)}{h}-g'(gamma)$, and note that $u$ depends on $h$. Moreover, we see $u rightarrow 0$ as $h rightarrow 0$, and we have $g(gamma+h)=(u+g'(gamma))h+g(gamma)=(u+g'(gamma))h+c,$ so $g(gamma+h)-c=g(gamma+h)-g(gamma)=(u+g'(gamma))h.$ Thus, $$left(lim_{h rightarrow 0}sup frac{f(g(gamma+h))-f(g(gamma))}{g(gamma+h)-g(gamma)}right).g'(gamma)=left(lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}right).g'(gamma) $$



      Then, I am saying that $$lim_{h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=lim_{(u+g'(gamma)h rightarrow 0} sup frac{f((u+g'(gamma)h+c)-f(c)}{(u+g'(gamma))h}=overline{D}f(c)$$
      which is I am not sure about it, and I dont know how to use the continuity of $g$ here.



      Thanks for any help.






      measure-theory lebesgue-measure




      measure-theory lebesgue-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 19 '18 at 1:47









      Ahmed

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