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
    1 hour 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
    1 hour 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
    1 hour ago














  • 2




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








2




2




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




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










3 Answers
3






active

oldest

votes


















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














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
    43 mins ago



















1














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






    active

    oldest

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    active

    oldest

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    active

    oldest

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    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













    4














    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
      43 mins ago
















    4














    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
      43 mins ago














    4












    4








    4






    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 1 hour ago









    xbh

    5,7451522




    5,7451522








    • 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
      43 mins 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
      43 mins 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
    43 mins 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
    43 mins ago











    1














    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


























      1














      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
























        1












        1








        1






        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 29 mins ago









        zipirovich

        11.1k11631




        11.1k11631






























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            Alexandru Averescu

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            Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center
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            Trompette piccolo

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            Ne doit pas être confondu avec trompette de poche. Trompette piccolo en si bémol, avec des branches d'embouchure différentes pour le si bémol (la plus courte) ou la (la plus longue) Trompette piccolo La trompette piccolo est la plus petite de la famille des trompettes. Elle joue une octave plus haut que la trompette standard en si bémol. La plupart des trompettes piccolo sont conçues pour jouer dans les deux tonalités de si bémol ou la, en utilisant une branche d'embouchure différente pour chaque tonalité. Le tube de la trompette piccolo en si bémol a une longueur moitié de celle de la trompette normale en si bémol. Des trompettes piccolo en sol, fa, ut sont également fabriquées, mais elles sont rares. La trompette soprano en ré, aussi connu comme la « trompette Bach », a été inventée vers 1890 par le facteur belge Victor-Charles Mahillon pour jouer les parties hautes de la trompette dans la musique de Bach et Haendel. La trompette piccolo moderne permet aux
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