Alternative proof. There is $f_0 in C([0,1], K)$ such that $ell(f_0) leq ell(f)quad forall f$

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Let $a,b in K$ where $K subset mathbbR^n$ is compact. Let $C$ the set of continuous paths $f: [0,1] to K$, rectifiable, with $f(0)=a$ and $f(1)=b$. If $C neq emptyset$ then there is $f_0 in C$ such that $ell(f_0) leq ell(f)$ for all $f in C$.




Notation. $ell(f)$ denotes the length of the path $f$.



I know a proof of this result using the Ascoli-Arzelá Theorem, but I'm trying to get a proof without using this theorem. Does anyone know (if any)?




real-analysis continuity




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  • I suspect it might be difficult to avoid a compactness argument?
    – copper.hat
    Jul 31 at 4:59














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Let $a,b in K$ where $K subset mathbbR^n$ is compact. Let $C$ the set of continuous paths $f: [0,1] to K$, rectifiable, with $f(0)=a$ and $f(1)=b$. If $C neq emptyset$ then there is $f_0 in C$ such that $ell(f_0) leq ell(f)$ for all $f in C$.




Notation. $ell(f)$ denotes the length of the path $f$.



I know a proof of this result using the Ascoli-Arzelá Theorem, but I'm trying to get a proof without using this theorem. Does anyone know (if any)?




real-analysis continuity




share|cite|improve this question



















  • I suspect it might be difficult to avoid a compactness argument?
    – copper.hat
    Jul 31 at 4:59












up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2






Let $a,b in K$ where $K subset mathbbR^n$ is compact. Let $C$ the set of continuous paths $f: [0,1] to K$, rectifiable, with $f(0)=a$ and $f(1)=b$. If $C neq emptyset$ then there is $f_0 in C$ such that $ell(f_0) leq ell(f)$ for all $f in C$.




Notation. $ell(f)$ denotes the length of the path $f$.



I know a proof of this result using the Ascoli-Arzelá Theorem, but I'm trying to get a proof without using this theorem. Does anyone know (if any)?




real-analysis continuity




share|cite|improve this question












Let $a,b in K$ where $K subset mathbbR^n$ is compact. Let $C$ the set of continuous paths $f: [0,1] to K$, rectifiable, with $f(0)=a$ and $f(1)=b$. If $C neq emptyset$ then there is $f_0 in C$ such that $ell(f_0) leq ell(f)$ for all $f in C$.




Notation. $ell(f)$ denotes the length of the path $f$.



I know a proof of this result using the Ascoli-Arzelá Theorem, but I'm trying to get a proof without using this theorem. Does anyone know (if any)?





real-analysis continuity





share|cite|improve this question










share|cite|improve this question




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asked Jul 31 at 3:34









Lucas Corrêa

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  • I suspect it might be difficult to avoid a compactness argument?
    – copper.hat
    Jul 31 at 4:59
















  • I suspect it might be difficult to avoid a compactness argument?
    – copper.hat
    Jul 31 at 4:59















I suspect it might be difficult to avoid a compactness argument?
– copper.hat
Jul 31 at 4:59




I suspect it might be difficult to avoid a compactness argument?
– copper.hat
Jul 31 at 4:59















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