If $xmapsto F(x)=int^x_af(s)ds$, then $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b]$

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Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?







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  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02















up vote
1
down vote

favorite












Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?







share|cite|improve this question
















  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?







share|cite|improve this question












Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 28 at 15:20









Micheal

25710




25710







  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02













  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02








1




1




Mean value theorem for integrals...
– amsmath
Aug 28 at 15:27




Mean value theorem for integrals...
– amsmath
Aug 28 at 15:27




2




2




@amsmath Doesn't the Mean value theorem fail for vector-valued functions?
– Sobi
Aug 28 at 15:29





@amsmath Doesn't the Mean value theorem fail for vector-valued functions?
– Sobi
Aug 28 at 15:29





1




1




@Sobi Sure it does. I did not read carefully enough. My bad.
– amsmath
Aug 28 at 15:30




@Sobi Sure it does. I did not read carefully enough. My bad.
– amsmath
Aug 28 at 15:30












@amsmath Happened to me a million times! :)
– Sobi
Aug 28 at 15:31




@amsmath Happened to me a million times! :)
– Sobi
Aug 28 at 15:31




1




1




@Sobi That happens because we wanna be the first to help, right? ;-)
– amsmath
Aug 28 at 18:02





@Sobi That happens because we wanna be the first to help, right? ;-)
– amsmath
Aug 28 at 18:02











1 Answer
1






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oldest

votes

















up vote
4
down vote



accepted










Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer




















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer




















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55














up vote
4
down vote



accepted










Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer




















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55












up vote
4
down vote



accepted







up vote
4
down vote



accepted






Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer












Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 28 at 15:27









LutzL

50.1k31849




50.1k31849











  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55
















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55















Wow, I love this!
– Micheal
Aug 28 at 15:28




Wow, I love this!
– Micheal
Aug 28 at 15:28












Now, use uniform continuity of $f$ on $[a,b]$.
– amsmath
Aug 28 at 15:29




Now, use uniform continuity of $f$ on $[a,b]$.
– amsmath
Aug 28 at 15:29












@amsmath: Done!
– Micheal
Aug 28 at 18:55




@amsmath: Done!
– Micheal
Aug 28 at 18:55

















 

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