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?
real-analysis analysis multivariable-calculus derivatives partial-derivative
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up 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?
real-analysis analysis multivariable-calculus derivatives partial-derivative
I need full detail about this proof!
â Mike
Aug 24 at 5:58
add a comment |Â
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?
real-analysis analysis multivariable-calculus derivatives partial-derivative
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?
real-analysis analysis multivariable-calculus derivatives partial-derivative
edited Aug 24 at 6:00
asked Aug 21 at 3:33
Mike
75615
75615
I need full detail about this proof!
â Mike
Aug 24 at 5:58
add a comment |Â
I need full detail about this proof!
â Mike
Aug 24 at 5:58
I need full detail about this proof!
â Mike
Aug 24 at 5:58
I need full detail about this proof!
â Mike
Aug 24 at 5:58
add a comment |Â
1 Answer
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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
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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 $D_2f(s,t) to D_2f(s,x)$?
add a comment |Â
up vote
1
down vote
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 $D_2f(s,t) to D_2f(s,x)$?
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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 $D_2f(s,t) to D_2f(s,x)$?
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 $D_2f(s,t) to D_2f(s,x)$?
answered Aug 21 at 6:06


Kavi Rama Murthy
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23.4k2933
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I need full detail about this proof!
â Mike
Aug 24 at 5:58