What is the main difference between pointwise and uniform convergence as defined here?












3














I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence







share|cite|improve this question




















  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    2 hours ago
















3














I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence







share|cite|improve this question




















  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    2 hours ago














3












3








3


1





I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence







share|cite|improve this question















I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence




real-analysis analysis definition uniform-convergence pointwise-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago

























asked 2 hours ago









Mike

1,498321




1,498321








  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    2 hours ago














  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    2 hours ago








2




2




Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
2 hours ago




Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
2 hours ago










3 Answers
3






active

oldest

votes


















5














Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
$$
f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
$$
This is strictly stronger than pointwise convergence.



Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






share|cite|improve this answer

















  • 1




    Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
    – Matt A Pelto
    1 hour ago



















4














$f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



$f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






share|cite|improve this answer





















  • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
    – Mike
    2 hours ago










  • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
    – Tsemo Aristide
    2 hours ago










  • That's so true.
    – Mike
    2 hours ago










  • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
    – Mike
    2 hours ago












  • $f_n(n)=1, f_n(x)=0$ if $xneq n$
    – Tsemo Aristide
    2 hours ago





















2














Sorry, but yes, you probably are missing something important, because the second statement in your post




On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
$$f_n(x)to f(x),;forall,xin E.$$




is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



Your best bet is to check the original source to find out what exactly it says there.






share|cite|improve this answer





















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    3 Answers
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    3 Answers
    3






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    oldest

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    active

    oldest

    votes






    active

    oldest

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    5














    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






    share|cite|improve this answer

















    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      1 hour ago
















    5














    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






    share|cite|improve this answer

















    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      1 hour ago














    5












    5








    5






    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






    share|cite|improve this answer












    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    xbh

    5,7551522




    5,7551522








    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      1 hour ago














    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      1 hour ago








    1




    1




    Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
    – Matt A Pelto
    1 hour ago




    Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
    – Matt A Pelto
    1 hour ago











    4














    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






    share|cite|improve this answer





















    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago


















    4














    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






    share|cite|improve this answer





















    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago
















    4












    4








    4






    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






    share|cite|improve this answer












    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Tsemo Aristide

    56.2k11444




    56.2k11444












    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago




















    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago


















    (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
    – Mike
    2 hours ago




    (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
    – Mike
    2 hours ago












    If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
    – Tsemo Aristide
    2 hours ago




    If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
    – Tsemo Aristide
    2 hours ago












    That's so true.
    – Mike
    2 hours ago




    That's so true.
    – Mike
    2 hours ago












    Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
    – Mike
    2 hours ago






    Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
    – Mike
    2 hours ago














    $f_n(n)=1, f_n(x)=0$ if $xneq n$
    – Tsemo Aristide
    2 hours ago






    $f_n(n)=1, f_n(x)=0$ if $xneq n$
    – Tsemo Aristide
    2 hours ago













    2














    Sorry, but yes, you probably are missing something important, because the second statement in your post




    On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
    $$f_n(x)to f(x),;forall,xin E.$$




    is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



    Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



    Your best bet is to check the original source to find out what exactly it says there.






    share|cite|improve this answer


























      2














      Sorry, but yes, you probably are missing something important, because the second statement in your post




      On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
      $$f_n(x)to f(x),;forall,xin E.$$




      is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



      Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



      Your best bet is to check the original source to find out what exactly it says there.






      share|cite|improve this answer
























        2












        2








        2






        Sorry, but yes, you probably are missing something important, because the second statement in your post




        On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
        $$f_n(x)to f(x),;forall,xin E.$$




        is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



        Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



        Your best bet is to check the original source to find out what exactly it says there.






        share|cite|improve this answer












        Sorry, but yes, you probably are missing something important, because the second statement in your post




        On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
        $$f_n(x)to f(x),;forall,xin E.$$




        is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



        Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



        Your best bet is to check the original source to find out what exactly it says there.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 58 mins ago









        zipirovich

        11.1k11631




        11.1k11631






























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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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