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












4














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?










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  • 7




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
















4














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?










share|cite|improve this question




















  • 7




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














4












4








4


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?










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






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share|cite|improve this question













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edited yesterday

























asked yesterday









Mike

1,504321




1,504321








  • 7




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














  • 7




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








7




7




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




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










5 Answers
5






active

oldest

votes


















5














$f_n$ 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
    yesterday










  • 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
    yesterday










  • That's so true.
    – Mike
    yesterday










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












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





















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
    yesterday



















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














    What you wrote is actually ill-formed, and that is why it is completely wrong. Nowhere did you specify or quantify $n$ in "$f_n(x) → f(x)$, $∀x∈E$". Furthermore, do not ever use symbols like "$∀$" unless you use them properly (quantifiers must always be put in front of the quantified statement).




    $f_n → f$ pointwise as $n→infty$ iff ( $f_n(x)-f(x) → 0$ as $n→∞$ ) for every $x∈E$.



    $f_n → f$ uniformly as $n→infty$ iff $sup_{x∈E} |f_n(x)-f(x)| → 0$ as $n→∞$.




    In one case the convergence may proceed differently for different $x∈E$. In the other case the sup norm between $f_n$ and $f$ (over all $x∈E$) must tend to zero, which intuitively means the convergence must proceed uniformly for all $x∈E$.



    In no way is it possible to correctly express uniform convergence using the expression you used, so it is most likely that you did not actually copy the definition or notation given to you.





    In purely logical form:
    $
    defnn{mathbb{N}}
    defrr{mathbb{R}}
    $




    $f_n → f$ pointwise as $n→infty$ iff $∀x∈E ∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ( |f_n(x)-f(x)| < ε )$.



    $f_n → f$ uniformly as $n→infty$ iff $∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ∀x∈E ( |f_n(x)-f(x)| < ε )$.




    This is an instance of a (restricted) quantifier swap. It is a basic logic fact that one implies the other, easily summarized as "$∃∀ ⇒ ∀∃$", but the reverse implication may not hold.






    share|cite|improve this answer































      1














      @xbh has pointed out that those are not the formal definitions of pointwise and uniform convergence. You can look up those definitions, so I hope to illustrate their difference with an intutive example: polynomial interpolation.



      Suppose we want to approximate $displaystyle f(x) = frac{1}{1 + 25 x^2}$ with a polynomial.



      Let $p_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ equally-spaced nodes in [–1, 1]. The plot below suggests that $p_n(x)$ converges pointwise to $f(x)$ in the interval (–1, 1) but diverges at the endpoints –1 and 1.





      Let $q_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ Chebyshev nodes in [–1, 1]. The plot below suggests that $q_n(x)$ converges uniformly to $f(x)$ in the interval [–1, 1].





      You can generate the plot above with more nodes:



      import numpy as np
      from numpy.polynomial.polynomial import polyfit, Polynomial
      import matplotlib.pyplot as plt
      xs = np.linspace(-1, 1, 256)
      def f(x): return 1.0 / (1 + 25 * x**2)

      N = 16
      nodes = np.cos(np.linspace(0, N, N) * np.pi / N)
      interp = Polynomial(polyfit(nodes, f(nodes), N - 1))
      plt.plot(xs, f(xs), "--", label = "$f(x) = (1 + 25 x^2)^{-1}$")
      plt.plot(xs, interp(xs), "-", label = "$q_n(x)$")
      plt.plot(nodes, f(nodes), "o")
      plt.legend(frameon = False)
      plt.show()





      share|cite|improve this answer










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        5 Answers
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        5














        $f_n$ 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
          yesterday










        • 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
          yesterday










        • That's so true.
          – Mike
          yesterday










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












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


















        5














        $f_n$ 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
          yesterday










        • 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
          yesterday










        • That's so true.
          – Mike
          yesterday










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












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
















        5












        5








        5






        $f_n$ 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 yesterday









        Tsemo Aristide

        56.3k11444




        56.3k11444












        • (+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
          yesterday










        • 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
          yesterday










        • That's so true.
          – Mike
          yesterday










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












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




















        • (+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
          yesterday










        • 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
          yesterday










        • That's so true.
          – Mike
          yesterday










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












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


















        (+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
        yesterday




        (+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
        yesterday












        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
        yesterday




        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
        yesterday












        That's so true.
        – Mike
        yesterday




        That's so true.
        – Mike
        yesterday












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






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














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






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













        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
          yesterday
















        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
          yesterday














        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 yesterday









        xbh

        5,7651522




        5,7651522








        • 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
          yesterday














        • 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
          yesterday








        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
        yesterday




        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
        yesterday











        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 yesterday









            zipirovich

            11.1k11631




            11.1k11631























                2














                What you wrote is actually ill-formed, and that is why it is completely wrong. Nowhere did you specify or quantify $n$ in "$f_n(x) → f(x)$, $∀x∈E$". Furthermore, do not ever use symbols like "$∀$" unless you use them properly (quantifiers must always be put in front of the quantified statement).




                $f_n → f$ pointwise as $n→infty$ iff ( $f_n(x)-f(x) → 0$ as $n→∞$ ) for every $x∈E$.



                $f_n → f$ uniformly as $n→infty$ iff $sup_{x∈E} |f_n(x)-f(x)| → 0$ as $n→∞$.




                In one case the convergence may proceed differently for different $x∈E$. In the other case the sup norm between $f_n$ and $f$ (over all $x∈E$) must tend to zero, which intuitively means the convergence must proceed uniformly for all $x∈E$.



                In no way is it possible to correctly express uniform convergence using the expression you used, so it is most likely that you did not actually copy the definition or notation given to you.





                In purely logical form:
                $
                defnn{mathbb{N}}
                defrr{mathbb{R}}
                $




                $f_n → f$ pointwise as $n→infty$ iff $∀x∈E ∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ( |f_n(x)-f(x)| < ε )$.



                $f_n → f$ uniformly as $n→infty$ iff $∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ∀x∈E ( |f_n(x)-f(x)| < ε )$.




                This is an instance of a (restricted) quantifier swap. It is a basic logic fact that one implies the other, easily summarized as "$∃∀ ⇒ ∀∃$", but the reverse implication may not hold.






                share|cite|improve this answer




























                  2














                  What you wrote is actually ill-formed, and that is why it is completely wrong. Nowhere did you specify or quantify $n$ in "$f_n(x) → f(x)$, $∀x∈E$". Furthermore, do not ever use symbols like "$∀$" unless you use them properly (quantifiers must always be put in front of the quantified statement).




                  $f_n → f$ pointwise as $n→infty$ iff ( $f_n(x)-f(x) → 0$ as $n→∞$ ) for every $x∈E$.



                  $f_n → f$ uniformly as $n→infty$ iff $sup_{x∈E} |f_n(x)-f(x)| → 0$ as $n→∞$.




                  In one case the convergence may proceed differently for different $x∈E$. In the other case the sup norm between $f_n$ and $f$ (over all $x∈E$) must tend to zero, which intuitively means the convergence must proceed uniformly for all $x∈E$.



                  In no way is it possible to correctly express uniform convergence using the expression you used, so it is most likely that you did not actually copy the definition or notation given to you.





                  In purely logical form:
                  $
                  defnn{mathbb{N}}
                  defrr{mathbb{R}}
                  $




                  $f_n → f$ pointwise as $n→infty$ iff $∀x∈E ∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ( |f_n(x)-f(x)| < ε )$.



                  $f_n → f$ uniformly as $n→infty$ iff $∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ∀x∈E ( |f_n(x)-f(x)| < ε )$.




                  This is an instance of a (restricted) quantifier swap. It is a basic logic fact that one implies the other, easily summarized as "$∃∀ ⇒ ∀∃$", but the reverse implication may not hold.






                  share|cite|improve this answer


























                    2












                    2








                    2






                    What you wrote is actually ill-formed, and that is why it is completely wrong. Nowhere did you specify or quantify $n$ in "$f_n(x) → f(x)$, $∀x∈E$". Furthermore, do not ever use symbols like "$∀$" unless you use them properly (quantifiers must always be put in front of the quantified statement).




                    $f_n → f$ pointwise as $n→infty$ iff ( $f_n(x)-f(x) → 0$ as $n→∞$ ) for every $x∈E$.



                    $f_n → f$ uniformly as $n→infty$ iff $sup_{x∈E} |f_n(x)-f(x)| → 0$ as $n→∞$.




                    In one case the convergence may proceed differently for different $x∈E$. In the other case the sup norm between $f_n$ and $f$ (over all $x∈E$) must tend to zero, which intuitively means the convergence must proceed uniformly for all $x∈E$.



                    In no way is it possible to correctly express uniform convergence using the expression you used, so it is most likely that you did not actually copy the definition or notation given to you.





                    In purely logical form:
                    $
                    defnn{mathbb{N}}
                    defrr{mathbb{R}}
                    $




                    $f_n → f$ pointwise as $n→infty$ iff $∀x∈E ∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ( |f_n(x)-f(x)| < ε )$.



                    $f_n → f$ uniformly as $n→infty$ iff $∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ∀x∈E ( |f_n(x)-f(x)| < ε )$.




                    This is an instance of a (restricted) quantifier swap. It is a basic logic fact that one implies the other, easily summarized as "$∃∀ ⇒ ∀∃$", but the reverse implication may not hold.






                    share|cite|improve this answer














                    What you wrote is actually ill-formed, and that is why it is completely wrong. Nowhere did you specify or quantify $n$ in "$f_n(x) → f(x)$, $∀x∈E$". Furthermore, do not ever use symbols like "$∀$" unless you use them properly (quantifiers must always be put in front of the quantified statement).




                    $f_n → f$ pointwise as $n→infty$ iff ( $f_n(x)-f(x) → 0$ as $n→∞$ ) for every $x∈E$.



                    $f_n → f$ uniformly as $n→infty$ iff $sup_{x∈E} |f_n(x)-f(x)| → 0$ as $n→∞$.




                    In one case the convergence may proceed differently for different $x∈E$. In the other case the sup norm between $f_n$ and $f$ (over all $x∈E$) must tend to zero, which intuitively means the convergence must proceed uniformly for all $x∈E$.



                    In no way is it possible to correctly express uniform convergence using the expression you used, so it is most likely that you did not actually copy the definition or notation given to you.





                    In purely logical form:
                    $
                    defnn{mathbb{N}}
                    defrr{mathbb{R}}
                    $




                    $f_n → f$ pointwise as $n→infty$ iff $∀x∈E ∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ( |f_n(x)-f(x)| < ε )$.



                    $f_n → f$ uniformly as $n→infty$ iff $∀ε∈rr_{>0} ∃k∈nn ∀n∈nn_{≥k} ∀x∈E ( |f_n(x)-f(x)| < ε )$.




                    This is an instance of a (restricted) quantifier swap. It is a basic logic fact that one implies the other, easily summarized as "$∃∀ ⇒ ∀∃$", but the reverse implication may not hold.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 22 hours ago

























                    answered 23 hours ago









                    user21820

                    38.7k543153




                    38.7k543153























                        1














                        @xbh has pointed out that those are not the formal definitions of pointwise and uniform convergence. You can look up those definitions, so I hope to illustrate their difference with an intutive example: polynomial interpolation.



                        Suppose we want to approximate $displaystyle f(x) = frac{1}{1 + 25 x^2}$ with a polynomial.



                        Let $p_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ equally-spaced nodes in [–1, 1]. The plot below suggests that $p_n(x)$ converges pointwise to $f(x)$ in the interval (–1, 1) but diverges at the endpoints –1 and 1.





                        Let $q_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ Chebyshev nodes in [–1, 1]. The plot below suggests that $q_n(x)$ converges uniformly to $f(x)$ in the interval [–1, 1].





                        You can generate the plot above with more nodes:



                        import numpy as np
                        from numpy.polynomial.polynomial import polyfit, Polynomial
                        import matplotlib.pyplot as plt
                        xs = np.linspace(-1, 1, 256)
                        def f(x): return 1.0 / (1 + 25 * x**2)

                        N = 16
                        nodes = np.cos(np.linspace(0, N, N) * np.pi / N)
                        interp = Polynomial(polyfit(nodes, f(nodes), N - 1))
                        plt.plot(xs, f(xs), "--", label = "$f(x) = (1 + 25 x^2)^{-1}$")
                        plt.plot(xs, interp(xs), "-", label = "$q_n(x)$")
                        plt.plot(nodes, f(nodes), "o")
                        plt.legend(frameon = False)
                        plt.show()





                        share|cite|improve this answer










                        New contributor




                        farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.























                          1














                          @xbh has pointed out that those are not the formal definitions of pointwise and uniform convergence. You can look up those definitions, so I hope to illustrate their difference with an intutive example: polynomial interpolation.



                          Suppose we want to approximate $displaystyle f(x) = frac{1}{1 + 25 x^2}$ with a polynomial.



                          Let $p_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ equally-spaced nodes in [–1, 1]. The plot below suggests that $p_n(x)$ converges pointwise to $f(x)$ in the interval (–1, 1) but diverges at the endpoints –1 and 1.





                          Let $q_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ Chebyshev nodes in [–1, 1]. The plot below suggests that $q_n(x)$ converges uniformly to $f(x)$ in the interval [–1, 1].





                          You can generate the plot above with more nodes:



                          import numpy as np
                          from numpy.polynomial.polynomial import polyfit, Polynomial
                          import matplotlib.pyplot as plt
                          xs = np.linspace(-1, 1, 256)
                          def f(x): return 1.0 / (1 + 25 * x**2)

                          N = 16
                          nodes = np.cos(np.linspace(0, N, N) * np.pi / N)
                          interp = Polynomial(polyfit(nodes, f(nodes), N - 1))
                          plt.plot(xs, f(xs), "--", label = "$f(x) = (1 + 25 x^2)^{-1}$")
                          plt.plot(xs, interp(xs), "-", label = "$q_n(x)$")
                          plt.plot(nodes, f(nodes), "o")
                          plt.legend(frameon = False)
                          plt.show()





                          share|cite|improve this answer










                          New contributor




                          farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





















                            1












                            1








                            1






                            @xbh has pointed out that those are not the formal definitions of pointwise and uniform convergence. You can look up those definitions, so I hope to illustrate their difference with an intutive example: polynomial interpolation.



                            Suppose we want to approximate $displaystyle f(x) = frac{1}{1 + 25 x^2}$ with a polynomial.



                            Let $p_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ equally-spaced nodes in [–1, 1]. The plot below suggests that $p_n(x)$ converges pointwise to $f(x)$ in the interval (–1, 1) but diverges at the endpoints –1 and 1.





                            Let $q_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ Chebyshev nodes in [–1, 1]. The plot below suggests that $q_n(x)$ converges uniformly to $f(x)$ in the interval [–1, 1].





                            You can generate the plot above with more nodes:



                            import numpy as np
                            from numpy.polynomial.polynomial import polyfit, Polynomial
                            import matplotlib.pyplot as plt
                            xs = np.linspace(-1, 1, 256)
                            def f(x): return 1.0 / (1 + 25 * x**2)

                            N = 16
                            nodes = np.cos(np.linspace(0, N, N) * np.pi / N)
                            interp = Polynomial(polyfit(nodes, f(nodes), N - 1))
                            plt.plot(xs, f(xs), "--", label = "$f(x) = (1 + 25 x^2)^{-1}$")
                            plt.plot(xs, interp(xs), "-", label = "$q_n(x)$")
                            plt.plot(nodes, f(nodes), "o")
                            plt.legend(frameon = False)
                            plt.show()





                            share|cite|improve this answer










                            New contributor




                            farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            @xbh has pointed out that those are not the formal definitions of pointwise and uniform convergence. You can look up those definitions, so I hope to illustrate their difference with an intutive example: polynomial interpolation.



                            Suppose we want to approximate $displaystyle f(x) = frac{1}{1 + 25 x^2}$ with a polynomial.



                            Let $p_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ equally-spaced nodes in [–1, 1]. The plot below suggests that $p_n(x)$ converges pointwise to $f(x)$ in the interval (–1, 1) but diverges at the endpoints –1 and 1.





                            Let $q_n(x)$ be the polynomial of degree $n-1$ that interpolates $f(x)$ on $n$ Chebyshev nodes in [–1, 1]. The plot below suggests that $q_n(x)$ converges uniformly to $f(x)$ in the interval [–1, 1].





                            You can generate the plot above with more nodes:



                            import numpy as np
                            from numpy.polynomial.polynomial import polyfit, Polynomial
                            import matplotlib.pyplot as plt
                            xs = np.linspace(-1, 1, 256)
                            def f(x): return 1.0 / (1 + 25 * x**2)

                            N = 16
                            nodes = np.cos(np.linspace(0, N, N) * np.pi / N)
                            interp = Polynomial(polyfit(nodes, f(nodes), N - 1))
                            plt.plot(xs, f(xs), "--", label = "$f(x) = (1 + 25 x^2)^{-1}$")
                            plt.plot(xs, interp(xs), "-", label = "$q_n(x)$")
                            plt.plot(nodes, f(nodes), "o")
                            plt.legend(frameon = False)
                            plt.show()






                            share|cite|improve this answer










                            New contributor




                            farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 20 hours ago





















                            New contributor




                            farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 23 hours ago









                            farmer

                            112




                            112




                            New contributor




                            farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            farmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






























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