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




















  • 1




    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




















  • 1




    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 1 hour ago

























asked 2 hours ago









Mike

1,498321




1,498321








  • 1




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














  • 1




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








1




1




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


















3














$f_n$ 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
    1 hour 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
    1 hour ago










  • That's so true.
    – Mike
    1 hour ago










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












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





















3














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
    17 mins ago



















0














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





















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    3 Answers
    3






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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    $f_n$ 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
      1 hour 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
      1 hour ago










    • That's so true.
      – Mike
      1 hour ago










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












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


















    3














    $f_n$ 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
      1 hour 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
      1 hour ago










    • That's so true.
      – Mike
      1 hour ago










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












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
















    3












    3








    3






    $f_n$ 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 1 hour 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
      1 hour 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
      1 hour ago










    • That's so true.
      – Mike
      1 hour ago










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












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










    • That's so true.
      – Mike
      1 hour ago










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












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












    That's so true.
    – Mike
    1 hour ago




    That's so true.
    – Mike
    1 hour ago












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






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














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






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













    3














    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
      17 mins ago
















    3














    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
      17 mins ago














    3












    3








    3






    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

    5,7351522




    5,7351522








    • 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
      17 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
      17 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
    17 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
    17 mins ago











    0














    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


























      0














      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
























        0












        0








        0






        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












        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












        share|cite



        share|cite










        answered 3 mins ago









        zipirovich

        11.1k11631




        11.1k11631






























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            數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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            When can things happen in Etherscan, such as the picture below?

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            2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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            格利澤436b

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            格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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