Prove that $G$ is differentiable on $U$ and $G'(x)=int^b_aD_2f(s,x)ds,;;xin U$

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Let $U$ be open in $BbbR^n$ and beginalignf:[a,b]times Uto BbbR^mendalign
be continuous. Let beginalignG(x)=int^b_af(s,x)ds,;;xin U.endalign
We assume that $D_2 f$ is continuous on $[a,b]times U.$ I want to prove that $G$ is differentiable on $U$ and
beginalignG'(x)=int^b_aD_2f(s,x)ds,;;xin U.endalign.



The case of differentiability:



I picked $hin U.$ Then,
beginalignG(x+h)-G(x)=int^b_af(s,x+h)ds-int^b_af(s,x)ds,;;xin U.endalign
beginalign=int^b_abig[f(s,x+h)-f(s,x)big]ds.endalign
So, I got stuck here and couldn't proceed. Please, can anyone please help out?







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  • I need full detail about this proof!
    – Mike
    Aug 24 at 5:58














up vote
0
down vote

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Let $U$ be open in $BbbR^n$ and beginalignf:[a,b]times Uto BbbR^mendalign
be continuous. Let beginalignG(x)=int^b_af(s,x)ds,;;xin U.endalign
We assume that $D_2 f$ is continuous on $[a,b]times U.$ I want to prove that $G$ is differentiable on $U$ and
beginalignG'(x)=int^b_aD_2f(s,x)ds,;;xin U.endalign.



The case of differentiability:



I picked $hin U.$ Then,
beginalignG(x+h)-G(x)=int^b_af(s,x+h)ds-int^b_af(s,x)ds,;;xin U.endalign
beginalign=int^b_abig[f(s,x+h)-f(s,x)big]ds.endalign
So, I got stuck here and couldn't proceed. Please, can anyone please help out?







share|cite|improve this question






















  • I need full detail about this proof!
    – Mike
    Aug 24 at 5:58












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $U$ be open in $BbbR^n$ and beginalignf:[a,b]times Uto BbbR^mendalign
be continuous. Let beginalignG(x)=int^b_af(s,x)ds,;;xin U.endalign
We assume that $D_2 f$ is continuous on $[a,b]times U.$ I want to prove that $G$ is differentiable on $U$ and
beginalignG'(x)=int^b_aD_2f(s,x)ds,;;xin U.endalign.



The case of differentiability:



I picked $hin U.$ Then,
beginalignG(x+h)-G(x)=int^b_af(s,x+h)ds-int^b_af(s,x)ds,;;xin U.endalign
beginalign=int^b_abig[f(s,x+h)-f(s,x)big]ds.endalign
So, I got stuck here and couldn't proceed. Please, can anyone please help out?







share|cite|improve this question














Let $U$ be open in $BbbR^n$ and beginalignf:[a,b]times Uto BbbR^mendalign
be continuous. Let beginalignG(x)=int^b_af(s,x)ds,;;xin U.endalign
We assume that $D_2 f$ is continuous on $[a,b]times U.$ I want to prove that $G$ is differentiable on $U$ and
beginalignG'(x)=int^b_aD_2f(s,x)ds,;;xin U.endalign.



The case of differentiability:



I picked $hin U.$ Then,
beginalignG(x+h)-G(x)=int^b_af(s,x+h)ds-int^b_af(s,x)ds,;;xin U.endalign
beginalign=int^b_abig[f(s,x+h)-f(s,x)big]ds.endalign
So, I got stuck here and couldn't proceed. Please, can anyone please help out?









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edited Aug 24 at 6:00

























asked Aug 21 at 3:33









Mike

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  • I need full detail about this proof!
    – Mike
    Aug 24 at 5:58
















  • I need full detail about this proof!
    – Mike
    Aug 24 at 5:58















I need full detail about this proof!
– Mike
Aug 24 at 5:58




I need full detail about this proof!
– Mike
Aug 24 at 5:58










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Your formula for $G(x+h)-G(x)$ is wrong. The correct expression is $int_a^b f(s,x+h), ds -int_a^b f(s,x), ds$. To prove the two statement about $G$ it is enough restrict your attention to a compact subset of $U$. When you do this $D_2f$ becomes uniformly bounded. By MVT $f(s,x+h) - f(s,x)=hD_2f(s,t)$ for some $t$between $x$ and $x+h$. Can you now complete the argument using the fact that $D_2f(s,t) to D_2f(s,x)$?






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    1 Answer
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    active

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    up vote
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    down vote



    +50










    Your formula for $G(x+h)-G(x)$ is wrong. The correct expression is $int_a^b f(s,x+h), ds -int_a^b f(s,x), ds$. To prove the two statement about $G$ it is enough restrict your attention to a compact subset of $U$. When you do this $D_2f$ becomes uniformly bounded. By MVT $f(s,x+h) - f(s,x)=hD_2f(s,t)$ for some $t$between $x$ and $x+h$. Can you now complete the argument using the fact that $D_2f(s,t) to D_2f(s,x)$?






    share|cite|improve this answer
























      up vote
      1
      down vote



      +50










      Your formula for $G(x+h)-G(x)$ is wrong. The correct expression is $int_a^b f(s,x+h), ds -int_a^b f(s,x), ds$. To prove the two statement about $G$ it is enough restrict your attention to a compact subset of $U$. When you do this $D_2f$ becomes uniformly bounded. By MVT $f(s,x+h) - f(s,x)=hD_2f(s,t)$ for some $t$between $x$ and $x+h$. Can you now complete the argument using the fact that $D_2f(s,t) to D_2f(s,x)$?






      share|cite|improve this answer






















        up vote
        1
        down vote



        +50







        up vote
        1
        down vote



        +50




        +50




        Your formula for $G(x+h)-G(x)$ is wrong. The correct expression is $int_a^b f(s,x+h), ds -int_a^b f(s,x), ds$. To prove the two statement about $G$ it is enough restrict your attention to a compact subset of $U$. When you do this $D_2f$ becomes uniformly bounded. By MVT $f(s,x+h) - f(s,x)=hD_2f(s,t)$ for some $t$between $x$ and $x+h$. Can you now complete the argument using the fact that $D_2f(s,t) to D_2f(s,x)$?






        share|cite|improve this answer












        Your formula for $G(x+h)-G(x)$ is wrong. The correct expression is $int_a^b f(s,x+h), ds -int_a^b f(s,x), ds$. To prove the two statement about $G$ it is enough restrict your attention to a compact subset of $U$. When you do this $D_2f$ becomes uniformly bounded. By MVT $f(s,x+h) - f(s,x)=hD_2f(s,t)$ for some $t$between $x$ and $x+h$. Can you now complete the argument using the fact that $D_2f(s,t) to D_2f(s,x)$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 21 at 6:06









        Kavi Rama Murthy

        23.4k2933




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