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?










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  • 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?










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?










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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edited 1 hour ago

























asked 1 hour 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










2 Answers
2






active

oldest

votes


















2














$f_n$ 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





















2














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





















  • 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
    11 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














$f_n$ 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


















2














$f_n$ 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
















2












2








2






$f_n$ 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













2














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





















  • 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
    11 mins ago
















2














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





















  • 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
    11 mins ago














2












2








2






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

5,7251522




5,7251522












  • 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
    11 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
    11 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
11 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
11 mins ago


















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