What does this notation means? (proof of Picard–Lindelöf theorem in wikipedia)












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$begingroup$


I was read a detailed proof of Picard–Lindelöf theorem. And, in this article of Wikipedia I found a notation some weird.
I saw that:
$$overline {I_{a}(t_{0})} = [t_0-a,t_0+a]$$
$$overline {B_{b}(y_{0})} = [y_0-b,y_0+b]$$
This is al normal, it is a definition.
But after, I saw:



$$displaystyle Gamma : {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )longrightarrow {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )$$



So, What does this line over the letter mean? It is an error? Because I had the idea that it is:



$$displaystyle Gamma : {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)longrightarrow {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)$$



The other thing is, according to me, the theorem should have a proof with $overline {B_{b}(y_{0})} subseteq mathbb{R}^n$, but $overline {B_{b}(y_{0})}$ is a interval in this proof.



Sorry for my english.










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$endgroup$












  • $begingroup$
    I don't see any 'hat' anywhere.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:56










  • $begingroup$
    I wanted to say "line over the letter", thank you
    $endgroup$
    – El borito
    Dec 30 '18 at 7:36










  • $begingroup$
    An overline is usually to denote closure. For example if $I = (a, b)$ is an open interval, then it's closure is $overline{I} = [a, b]$.
    $endgroup$
    – Pratyush Sarkar
    Dec 30 '18 at 7:51












  • $begingroup$
    So, there is a possibility that it is an error?
    $endgroup$
    – El borito
    Dec 30 '18 at 8:23










  • $begingroup$
    The proof is for $n=1$, so a closed ball reduces to a compact interval.
    $endgroup$
    – user539887
    Dec 30 '18 at 9:04


















0












$begingroup$


I was read a detailed proof of Picard–Lindelöf theorem. And, in this article of Wikipedia I found a notation some weird.
I saw that:
$$overline {I_{a}(t_{0})} = [t_0-a,t_0+a]$$
$$overline {B_{b}(y_{0})} = [y_0-b,y_0+b]$$
This is al normal, it is a definition.
But after, I saw:



$$displaystyle Gamma : {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )longrightarrow {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )$$



So, What does this line over the letter mean? It is an error? Because I had the idea that it is:



$$displaystyle Gamma : {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)longrightarrow {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)$$



The other thing is, according to me, the theorem should have a proof with $overline {B_{b}(y_{0})} subseteq mathbb{R}^n$, but $overline {B_{b}(y_{0})}$ is a interval in this proof.



Sorry for my english.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see any 'hat' anywhere.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:56










  • $begingroup$
    I wanted to say "line over the letter", thank you
    $endgroup$
    – El borito
    Dec 30 '18 at 7:36










  • $begingroup$
    An overline is usually to denote closure. For example if $I = (a, b)$ is an open interval, then it's closure is $overline{I} = [a, b]$.
    $endgroup$
    – Pratyush Sarkar
    Dec 30 '18 at 7:51












  • $begingroup$
    So, there is a possibility that it is an error?
    $endgroup$
    – El borito
    Dec 30 '18 at 8:23










  • $begingroup$
    The proof is for $n=1$, so a closed ball reduces to a compact interval.
    $endgroup$
    – user539887
    Dec 30 '18 at 9:04
















0












0








0





$begingroup$


I was read a detailed proof of Picard–Lindelöf theorem. And, in this article of Wikipedia I found a notation some weird.
I saw that:
$$overline {I_{a}(t_{0})} = [t_0-a,t_0+a]$$
$$overline {B_{b}(y_{0})} = [y_0-b,y_0+b]$$
This is al normal, it is a definition.
But after, I saw:



$$displaystyle Gamma : {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )longrightarrow {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )$$



So, What does this line over the letter mean? It is an error? Because I had the idea that it is:



$$displaystyle Gamma : {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)longrightarrow {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)$$



The other thing is, according to me, the theorem should have a proof with $overline {B_{b}(y_{0})} subseteq mathbb{R}^n$, but $overline {B_{b}(y_{0})}$ is a interval in this proof.



Sorry for my english.










share|cite|improve this question











$endgroup$




I was read a detailed proof of Picard–Lindelöf theorem. And, in this article of Wikipedia I found a notation some weird.
I saw that:
$$overline {I_{a}(t_{0})} = [t_0-a,t_0+a]$$
$$overline {B_{b}(y_{0})} = [y_0-b,y_0+b]$$
This is al normal, it is a definition.
But after, I saw:



$$displaystyle Gamma : {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )longrightarrow {mathcal {C}} ( I_{a}(t_{0}) , B_{b}(y_{0}) )$$



So, What does this line over the letter mean? It is an error? Because I had the idea that it is:



$$displaystyle Gamma : {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)longrightarrow {mathcal {C}} left( overline {I_{a}(t_{0})} , overline {B_{b}(y_{0})} right)$$



The other thing is, according to me, the theorem should have a proof with $overline {B_{b}(y_{0})} subseteq mathbb{R}^n$, but $overline {B_{b}(y_{0})}$ is a interval in this proof.



Sorry for my english.







general-topology ordinary-differential-equations notation






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edited Dec 30 '18 at 7:34







El borito

















asked Dec 29 '18 at 23:09









El boritoEl borito

664216




664216












  • $begingroup$
    I don't see any 'hat' anywhere.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:56










  • $begingroup$
    I wanted to say "line over the letter", thank you
    $endgroup$
    – El borito
    Dec 30 '18 at 7:36










  • $begingroup$
    An overline is usually to denote closure. For example if $I = (a, b)$ is an open interval, then it's closure is $overline{I} = [a, b]$.
    $endgroup$
    – Pratyush Sarkar
    Dec 30 '18 at 7:51












  • $begingroup$
    So, there is a possibility that it is an error?
    $endgroup$
    – El borito
    Dec 30 '18 at 8:23










  • $begingroup$
    The proof is for $n=1$, so a closed ball reduces to a compact interval.
    $endgroup$
    – user539887
    Dec 30 '18 at 9:04




















  • $begingroup$
    I don't see any 'hat' anywhere.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:56










  • $begingroup$
    I wanted to say "line over the letter", thank you
    $endgroup$
    – El borito
    Dec 30 '18 at 7:36










  • $begingroup$
    An overline is usually to denote closure. For example if $I = (a, b)$ is an open interval, then it's closure is $overline{I} = [a, b]$.
    $endgroup$
    – Pratyush Sarkar
    Dec 30 '18 at 7:51












  • $begingroup$
    So, there is a possibility that it is an error?
    $endgroup$
    – El borito
    Dec 30 '18 at 8:23










  • $begingroup$
    The proof is for $n=1$, so a closed ball reduces to a compact interval.
    $endgroup$
    – user539887
    Dec 30 '18 at 9:04


















$begingroup$
I don't see any 'hat' anywhere.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:56




$begingroup$
I don't see any 'hat' anywhere.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:56












$begingroup$
I wanted to say "line over the letter", thank you
$endgroup$
– El borito
Dec 30 '18 at 7:36




$begingroup$
I wanted to say "line over the letter", thank you
$endgroup$
– El borito
Dec 30 '18 at 7:36












$begingroup$
An overline is usually to denote closure. For example if $I = (a, b)$ is an open interval, then it's closure is $overline{I} = [a, b]$.
$endgroup$
– Pratyush Sarkar
Dec 30 '18 at 7:51






$begingroup$
An overline is usually to denote closure. For example if $I = (a, b)$ is an open interval, then it's closure is $overline{I} = [a, b]$.
$endgroup$
– Pratyush Sarkar
Dec 30 '18 at 7:51














$begingroup$
So, there is a possibility that it is an error?
$endgroup$
– El borito
Dec 30 '18 at 8:23




$begingroup$
So, there is a possibility that it is an error?
$endgroup$
– El borito
Dec 30 '18 at 8:23












$begingroup$
The proof is for $n=1$, so a closed ball reduces to a compact interval.
$endgroup$
– user539887
Dec 30 '18 at 9:04






$begingroup$
The proof is for $n=1$, so a closed ball reduces to a compact interval.
$endgroup$
– user539887
Dec 30 '18 at 9:04












1 Answer
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Yes, the wikipedia entry is sub-optimal. There should be mention of a domain $cal D$ of $f$ in the first statement in the intro, later that the cylinder $bar I_atimes bar B_b$ has to be a sub-set of $cal D$.



And you need the closed sets, first for compactness to have a maximum $M$ of $f$ over the cylinder and later for completeness of the function space as you apply the Banach fixed-point theorem.



(Btw., there is no need to invoke Grönwall, as the BFS already supplies the uniqueness of the fixed point.)





In short, as always with the things written in wikipedia, look for other sources (more than one). For this topic, preferably find text books, as the proofs there are usually more carefully composed.






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    $begingroup$

    Yes, the wikipedia entry is sub-optimal. There should be mention of a domain $cal D$ of $f$ in the first statement in the intro, later that the cylinder $bar I_atimes bar B_b$ has to be a sub-set of $cal D$.



    And you need the closed sets, first for compactness to have a maximum $M$ of $f$ over the cylinder and later for completeness of the function space as you apply the Banach fixed-point theorem.



    (Btw., there is no need to invoke Grönwall, as the BFS already supplies the uniqueness of the fixed point.)





    In short, as always with the things written in wikipedia, look for other sources (more than one). For this topic, preferably find text books, as the proofs there are usually more carefully composed.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Yes, the wikipedia entry is sub-optimal. There should be mention of a domain $cal D$ of $f$ in the first statement in the intro, later that the cylinder $bar I_atimes bar B_b$ has to be a sub-set of $cal D$.



      And you need the closed sets, first for compactness to have a maximum $M$ of $f$ over the cylinder and later for completeness of the function space as you apply the Banach fixed-point theorem.



      (Btw., there is no need to invoke Grönwall, as the BFS already supplies the uniqueness of the fixed point.)





      In short, as always with the things written in wikipedia, look for other sources (more than one). For this topic, preferably find text books, as the proofs there are usually more carefully composed.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Yes, the wikipedia entry is sub-optimal. There should be mention of a domain $cal D$ of $f$ in the first statement in the intro, later that the cylinder $bar I_atimes bar B_b$ has to be a sub-set of $cal D$.



        And you need the closed sets, first for compactness to have a maximum $M$ of $f$ over the cylinder and later for completeness of the function space as you apply the Banach fixed-point theorem.



        (Btw., there is no need to invoke Grönwall, as the BFS already supplies the uniqueness of the fixed point.)





        In short, as always with the things written in wikipedia, look for other sources (more than one). For this topic, preferably find text books, as the proofs there are usually more carefully composed.






        share|cite|improve this answer









        $endgroup$



        Yes, the wikipedia entry is sub-optimal. There should be mention of a domain $cal D$ of $f$ in the first statement in the intro, later that the cylinder $bar I_atimes bar B_b$ has to be a sub-set of $cal D$.



        And you need the closed sets, first for compactness to have a maximum $M$ of $f$ over the cylinder and later for completeness of the function space as you apply the Banach fixed-point theorem.



        (Btw., there is no need to invoke Grönwall, as the BFS already supplies the uniqueness of the fixed point.)





        In short, as always with the things written in wikipedia, look for other sources (more than one). For this topic, preferably find text books, as the proofs there are usually more carefully composed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 10:31









        LutzLLutzL

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