Integration in $mathbb R^n$ [on hold]

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Let $C([0,a],mathbbR^n)$ be the set of continuous functions defined from $[0,a]$ to $mathbbR^n$, where $a>0$ is a real number. Let $f:[0,a] times mathbbR^n rightarrow mathbbR^n$ be a continuous function.For $x in mathbbR^n$ define $Vert x Vert= Vert x_1,x_2,ldots , x_n Vert= max vert x_1 vert, vert x_2 vert,ldots , vert x_n vert $.



Prove that $$ BigVert intlimits_0^a [f(s,u(s))-f(s,v(s))]ds BigVert leq intlimits_0^a Vert f(s,u(s))-f(s,v(s)) Vert ds,$$ where $u,v in C([0,a],mathbbR^n)$.




calculus functional-analysis




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put on hold as off-topic by José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants Aug 3 at 14:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.












  • What are your thoughts on this problem? For example, are you familiar with Minkowski's inequality?
    – Strants
    Aug 3 at 14:24










  • Yes I"m familiar
    – Manu Rohilla
    Aug 3 at 15:10














up vote
0
down vote

favorite












Let $C([0,a],mathbbR^n)$ be the set of continuous functions defined from $[0,a]$ to $mathbbR^n$, where $a>0$ is a real number. Let $f:[0,a] times mathbbR^n rightarrow mathbbR^n$ be a continuous function.For $x in mathbbR^n$ define $Vert x Vert= Vert x_1,x_2,ldots , x_n Vert= max vert x_1 vert, vert x_2 vert,ldots , vert x_n vert $.



Prove that $$ BigVert intlimits_0^a [f(s,u(s))-f(s,v(s))]ds BigVert leq intlimits_0^a Vert f(s,u(s))-f(s,v(s)) Vert ds,$$ where $u,v in C([0,a],mathbbR^n)$.




calculus functional-analysis




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put on hold as off-topic by José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants Aug 3 at 14:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.












  • What are your thoughts on this problem? For example, are you familiar with Minkowski's inequality?
    – Strants
    Aug 3 at 14:24










  • Yes I"m familiar
    – Manu Rohilla
    Aug 3 at 15:10












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $C([0,a],mathbbR^n)$ be the set of continuous functions defined from $[0,a]$ to $mathbbR^n$, where $a>0$ is a real number. Let $f:[0,a] times mathbbR^n rightarrow mathbbR^n$ be a continuous function.For $x in mathbbR^n$ define $Vert x Vert= Vert x_1,x_2,ldots , x_n Vert= max vert x_1 vert, vert x_2 vert,ldots , vert x_n vert $.



Prove that $$ BigVert intlimits_0^a [f(s,u(s))-f(s,v(s))]ds BigVert leq intlimits_0^a Vert f(s,u(s))-f(s,v(s)) Vert ds,$$ where $u,v in C([0,a],mathbbR^n)$.




calculus functional-analysis




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Let $C([0,a],mathbbR^n)$ be the set of continuous functions defined from $[0,a]$ to $mathbbR^n$, where $a>0$ is a real number. Let $f:[0,a] times mathbbR^n rightarrow mathbbR^n$ be a continuous function.For $x in mathbbR^n$ define $Vert x Vert= Vert x_1,x_2,ldots , x_n Vert= max vert x_1 vert, vert x_2 vert,ldots , vert x_n vert $.



Prove that $$ BigVert intlimits_0^a [f(s,u(s))-f(s,v(s))]ds BigVert leq intlimits_0^a Vert f(s,u(s))-f(s,v(s)) Vert ds,$$ where $u,v in C([0,a],mathbbR^n)$.





calculus functional-analysis





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edited Aug 3 at 11:01









amWhy

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189k25219431









asked Aug 3 at 7:41









Manu Rohilla

1209




1209




put on hold as off-topic by José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants Aug 3 at 14:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants Aug 3 at 14:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, amWhy, Claude Leibovici, Micah, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.











  • What are your thoughts on this problem? For example, are you familiar with Minkowski's inequality?
    – Strants
    Aug 3 at 14:24










  • Yes I"m familiar
    – Manu Rohilla
    Aug 3 at 15:10
















  • What are your thoughts on this problem? For example, are you familiar with Minkowski's inequality?
    – Strants
    Aug 3 at 14:24










  • Yes I"m familiar
    – Manu Rohilla
    Aug 3 at 15:10















What are your thoughts on this problem? For example, are you familiar with Minkowski's inequality?
– Strants
Aug 3 at 14:24




What are your thoughts on this problem? For example, are you familiar with Minkowski's inequality?
– Strants
Aug 3 at 14:24












Yes I"m familiar
– Manu Rohilla
Aug 3 at 15:10




Yes I"m familiar
– Manu Rohilla
Aug 3 at 15:10










1 Answer
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If $u$ is a vector with $sum_i=1^n |u_i|=1$ and $langle u, v rangle$ denotes $sum_i=1^n u_iv_i$ then $langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle=int_0^a langle u, [f(s,u(s)-f(s,v(s)] rangle , dsleq RHS$. Take sup over all $u$. Details: let $g(s)=f(s,u(s))-f(s,v(s))$. Then $g$ maps $[0,a]$ into $mathbb R^n$. Write $$g=(g_1,g_2,...,g_n)$$ . [$g_i$'s are real valued functions]. By definition $int_0^ag(s)ds$ is the vector $$(int_0^ag_1(s)ds,int_0^ag_2(s)ds,...,int_0^ag_n(s)ds)$$. If $u=(u_1,u_2,...,u_n)$ is as above then $$langle u, int_0^a g(s) , ds rangle= sum_i=1^n u_iint_0^ag_i(s)ds=int_0^asum_i=1^nu_ig_i(s)ds$$. But $sum_i=1^nu_ig_i(s) leq |g(s)|$ so we get $$langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle leq int_0^a |g(s)| , ds$$ and the last quantity is the right hand side of your inequality. it is easy to see that for any vector $x$, $sup (u_i)=|x|$. Hence, if you take sup over all $u$ you will get the desired inequality.






share|cite|improve this answer























  • Please explain a bit more. I'm not able to get it.
    – Manu Rohilla
    Aug 3 at 7:50










  • This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
    – Strants
    Aug 3 at 15:14










  • @Strants Thanks. I have revised the answer.
    – Kavi Rama Murthy
    Aug 3 at 23:50

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










If $u$ is a vector with $sum_i=1^n |u_i|=1$ and $langle u, v rangle$ denotes $sum_i=1^n u_iv_i$ then $langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle=int_0^a langle u, [f(s,u(s)-f(s,v(s)] rangle , dsleq RHS$. Take sup over all $u$. Details: let $g(s)=f(s,u(s))-f(s,v(s))$. Then $g$ maps $[0,a]$ into $mathbb R^n$. Write $$g=(g_1,g_2,...,g_n)$$ . [$g_i$'s are real valued functions]. By definition $int_0^ag(s)ds$ is the vector $$(int_0^ag_1(s)ds,int_0^ag_2(s)ds,...,int_0^ag_n(s)ds)$$. If $u=(u_1,u_2,...,u_n)$ is as above then $$langle u, int_0^a g(s) , ds rangle= sum_i=1^n u_iint_0^ag_i(s)ds=int_0^asum_i=1^nu_ig_i(s)ds$$. But $sum_i=1^nu_ig_i(s) leq |g(s)|$ so we get $$langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle leq int_0^a |g(s)| , ds$$ and the last quantity is the right hand side of your inequality. it is easy to see that for any vector $x$, $sup (u_i)=|x|$. Hence, if you take sup over all $u$ you will get the desired inequality.






share|cite|improve this answer























  • Please explain a bit more. I'm not able to get it.
    – Manu Rohilla
    Aug 3 at 7:50










  • This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
    – Strants
    Aug 3 at 15:14










  • @Strants Thanks. I have revised the answer.
    – Kavi Rama Murthy
    Aug 3 at 23:50














up vote
1
down vote



accepted










If $u$ is a vector with $sum_i=1^n |u_i|=1$ and $langle u, v rangle$ denotes $sum_i=1^n u_iv_i$ then $langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle=int_0^a langle u, [f(s,u(s)-f(s,v(s)] rangle , dsleq RHS$. Take sup over all $u$. Details: let $g(s)=f(s,u(s))-f(s,v(s))$. Then $g$ maps $[0,a]$ into $mathbb R^n$. Write $$g=(g_1,g_2,...,g_n)$$ . [$g_i$'s are real valued functions]. By definition $int_0^ag(s)ds$ is the vector $$(int_0^ag_1(s)ds,int_0^ag_2(s)ds,...,int_0^ag_n(s)ds)$$. If $u=(u_1,u_2,...,u_n)$ is as above then $$langle u, int_0^a g(s) , ds rangle= sum_i=1^n u_iint_0^ag_i(s)ds=int_0^asum_i=1^nu_ig_i(s)ds$$. But $sum_i=1^nu_ig_i(s) leq |g(s)|$ so we get $$langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle leq int_0^a |g(s)| , ds$$ and the last quantity is the right hand side of your inequality. it is easy to see that for any vector $x$, $sup (u_i)=|x|$. Hence, if you take sup over all $u$ you will get the desired inequality.






share|cite|improve this answer























  • Please explain a bit more. I'm not able to get it.
    – Manu Rohilla
    Aug 3 at 7:50










  • This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
    – Strants
    Aug 3 at 15:14










  • @Strants Thanks. I have revised the answer.
    – Kavi Rama Murthy
    Aug 3 at 23:50












up vote
1
down vote



accepted







up vote
1
down vote



accepted






If $u$ is a vector with $sum_i=1^n |u_i|=1$ and $langle u, v rangle$ denotes $sum_i=1^n u_iv_i$ then $langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle=int_0^a langle u, [f(s,u(s)-f(s,v(s)] rangle , dsleq RHS$. Take sup over all $u$. Details: let $g(s)=f(s,u(s))-f(s,v(s))$. Then $g$ maps $[0,a]$ into $mathbb R^n$. Write $$g=(g_1,g_2,...,g_n)$$ . [$g_i$'s are real valued functions]. By definition $int_0^ag(s)ds$ is the vector $$(int_0^ag_1(s)ds,int_0^ag_2(s)ds,...,int_0^ag_n(s)ds)$$. If $u=(u_1,u_2,...,u_n)$ is as above then $$langle u, int_0^a g(s) , ds rangle= sum_i=1^n u_iint_0^ag_i(s)ds=int_0^asum_i=1^nu_ig_i(s)ds$$. But $sum_i=1^nu_ig_i(s) leq |g(s)|$ so we get $$langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle leq int_0^a |g(s)| , ds$$ and the last quantity is the right hand side of your inequality. it is easy to see that for any vector $x$, $sup (u_i)=|x|$. Hence, if you take sup over all $u$ you will get the desired inequality.






share|cite|improve this answer















If $u$ is a vector with $sum_i=1^n |u_i|=1$ and $langle u, v rangle$ denotes $sum_i=1^n u_iv_i$ then $langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle=int_0^a langle u, [f(s,u(s)-f(s,v(s)] rangle , dsleq RHS$. Take sup over all $u$. Details: let $g(s)=f(s,u(s))-f(s,v(s))$. Then $g$ maps $[0,a]$ into $mathbb R^n$. Write $$g=(g_1,g_2,...,g_n)$$ . [$g_i$'s are real valued functions]. By definition $int_0^ag(s)ds$ is the vector $$(int_0^ag_1(s)ds,int_0^ag_2(s)ds,...,int_0^ag_n(s)ds)$$. If $u=(u_1,u_2,...,u_n)$ is as above then $$langle u, int_0^a g(s) , ds rangle= sum_i=1^n u_iint_0^ag_i(s)ds=int_0^asum_i=1^nu_ig_i(s)ds$$. But $sum_i=1^nu_ig_i(s) leq |g(s)|$ so we get $$langle u, int_0^a [f(s,u(s))-f(s,v(s)] , ds rangle leq int_0^a |g(s)| , ds$$ and the last quantity is the right hand side of your inequality. it is easy to see that for any vector $x$, $sup (u_i)=|x|$. Hence, if you take sup over all $u$ you will get the desired inequality.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 3 at 23:49


























answered Aug 3 at 7:47









Kavi Rama Murthy

19.2k2829




19.2k2829











  • Please explain a bit more. I'm not able to get it.
    – Manu Rohilla
    Aug 3 at 7:50










  • This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
    – Strants
    Aug 3 at 15:14










  • @Strants Thanks. I have revised the answer.
    – Kavi Rama Murthy
    Aug 3 at 23:50
















  • Please explain a bit more. I'm not able to get it.
    – Manu Rohilla
    Aug 3 at 7:50










  • This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
    – Strants
    Aug 3 at 15:14










  • @Strants Thanks. I have revised the answer.
    – Kavi Rama Murthy
    Aug 3 at 23:50















Please explain a bit more. I'm not able to get it.
– Manu Rohilla
Aug 3 at 7:50




Please explain a bit more. I'm not able to get it.
– Manu Rohilla
Aug 3 at 7:50












This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
– Strants
Aug 3 at 15:14




This answer isn't quite correct: the norm $lVert x rVert$ given in the question is not generated by an inner product, so Cauchy-Schwarz does not apply.
– Strants
Aug 3 at 15:14












@Strants Thanks. I have revised the answer.
– Kavi Rama Murthy
Aug 3 at 23:50




@Strants Thanks. I have revised the answer.
– Kavi Rama Murthy
Aug 3 at 23:50


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