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
    20 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
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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    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
      20 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
      20 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
      20 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
      20 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
    20 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
    20 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 6 mins ago









        zipirovich

        11.1k11631




        11.1k11631






























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            Zagreb

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            This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
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            Understanding the information contained in the Deep Space Network XML data?

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            up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
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