What is the main difference between pointwise and uniform convergence as defined here?
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
asked 2 hours ago
Mike
1,498321
add a comment |
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
asked 2 hours ago
Mike
1,498321
2
Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago
add a comment |
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
asked 2 hours ago
Mike
1,498321
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
asked 2 hours ago
Mike
1,498321
asked 2 hours ago
Mike
1,498321
edited 2 hours ago
asked 2 hours ago
Mike
1,498321
asked 2 hours ago
Mike
1,498321
asked 2 hours ago
Mike
1,498321
1,498321
2
Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago
add a comment |
2
Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago
2
2
Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago
Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$f_n$ 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 $x$.
answered 2 hours ago
Tsemo Aristide
56.2k11444
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
That's so true.
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
|
show 1 more comment
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.
answered 1 hour ago
xbh
5,7451522
1
Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
– Matt A Pelto
49 mins ago
add a comment |
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.
answered 35 mins ago
zipirovich
11.1k11631
add a comment |
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3 Answers
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3 Answers
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$f_n$ 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 $x$.
answered 2 hours ago
Tsemo Aristide
56.2k11444
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
That's so true.
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
|
show 1 more comment
$f_n$ 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 $x$.
answered 2 hours ago
Tsemo Aristide
56.2k11444
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
That's so true.
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
|
show 1 more comment
$f_n$ 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 $x$.
answered 2 hours ago
Tsemo Aristide
56.2k11444
$f_n$ 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 $x$.
answered 2 hours ago
Tsemo Aristide
56.2k11444
answered 2 hours ago
Tsemo Aristide
56.2k11444
answered 2 hours ago
Tsemo Aristide
56.2k11444
answered 2 hours ago
Tsemo Aristide
56.2k11444
56.2k11444
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
That's so true.
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
|
show 1 more comment
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
That's so true.
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
(+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
– Mike
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
– Tsemo Aristide
2 hours ago
That's so true.
– Mike
2 hours ago
That's so true.
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
– Mike
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
$f_n(n)=1, f_n(x)=0$ if $xneq n$
– Tsemo Aristide
2 hours ago
|
show 1 more comment
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.
answered 1 hour ago
xbh
5,7451522
1
Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
– Matt A Pelto
49 mins ago
add a comment |
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.
answered 1 hour ago
xbh
5,7451522
1
Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
– Matt A Pelto
49 mins ago
add a comment |
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.
answered 1 hour ago
xbh
5,7451522
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.
answered 1 hour ago
xbh
5,7451522
answered 1 hour ago
xbh
5,7451522
answered 1 hour ago
xbh
5,7451522
answered 1 hour ago
xbh
5,7451522
5,7451522
1
Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
– Matt A Pelto
49 mins ago
add a comment |
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
49 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
49 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
49 mins ago
add a comment |
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.
answered 35 mins ago
zipirovich
11.1k11631
add a comment |
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.
answered 35 mins ago
zipirovich
11.1k11631
add a comment |
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.
answered 35 mins ago
zipirovich
11.1k11631
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.
answered 35 mins ago
zipirovich
11.1k11631
answered 35 mins ago
zipirovich
11.1k11631
answered 35 mins ago
zipirovich
11.1k11631
answered 35 mins ago
zipirovich
11.1k11631
11.1k11631
add a comment |
add a comment |
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2
Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago