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Convergence of uniformly continous functions to a uniformly continous function
Clash Royale CLAN TAG#URR8PPP
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Say $f_n,f:mathbbR to mathbbR$ are uniformly continous functions a $f_nto f$ pointwise.
Is the convergence nessecarily uniform?
I couldn't find an appropriate counter-example for this case so I tried proving it :
We have $|f_n(x)-f_n(y)|<epsilon/3$ for all $|x-y|<delta_1$, also $|f(x)-f(y)|<epsilon/3$ for $|x-y|<delta_2$ and for all $n>N_x$ we have $|f_n(x)-f(x)|<epsilon /3$.
Overall for any $|x-y|<delta=min(delta_1,delta_2)$ and any $n>N_x$ we have $|f_n(y)-f(y)|leq |f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(x)-f(y)|<epsilon $ for any $yin (x-delta,x+delta)$ and for any $xin AsubsetmathbbR$.
Now if $A$ is compact, we have a cover of $A$ and we can choose a finite subcover $U$, and take $N=maxxin U. $ and for all $n>N$ we have $|f_n(y)-f(y)|<epsilon$ uniformly .
However, if my proof is correct I have only proven uniform convergence in compact sets, and not on the entire number line. Can my proof be modified so we can get uniform convergence on $mathbbR$? Or can someone provide a counter example where it fails in this case?
Edit: Thanks to the counter-examples in the answers I can see my proof is false. Can anyone point out where my proof fails?
real-analysis sequences-and-series proof-verification uniform-convergence uniform-continuity
asked 2 days ago
Sar
3529
add a comment |Â
up vote
5
down vote
favorite
Say $f_n,f:mathbbR to mathbbR$ are uniformly continous functions a $f_nto f$ pointwise.
Is the convergence nessecarily uniform?
I couldn't find an appropriate counter-example for this case so I tried proving it :
We have $|f_n(x)-f_n(y)|<epsilon/3$ for all $|x-y|<delta_1$, also $|f(x)-f(y)|<epsilon/3$ for $|x-y|<delta_2$ and for all $n>N_x$ we have $|f_n(x)-f(x)|<epsilon /3$.
Overall for any $|x-y|<delta=min(delta_1,delta_2)$ and any $n>N_x$ we have $|f_n(y)-f(y)|leq |f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(x)-f(y)|<epsilon $ for any $yin (x-delta,x+delta)$ and for any $xin AsubsetmathbbR$.
Now if $A$ is compact, we have a cover of $A$ and we can choose a finite subcover $U$, and take $N=maxxin U. $ and for all $n>N$ we have $|f_n(y)-f(y)|<epsilon$ uniformly .
However, if my proof is correct I have only proven uniform convergence in compact sets, and not on the entire number line. Can my proof be modified so we can get uniform convergence on $mathbbR$? Or can someone provide a counter example where it fails in this case?
Edit: Thanks to the counter-examples in the answers I can see my proof is false. Can anyone point out where my proof fails?
real-analysis sequences-and-series proof-verification uniform-convergence uniform-continuity
asked 2 days ago
Sar
3529
This kind of like the Dini's theorem, in which the convergence is monotonic
– xbh
2 days ago
This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions.
– Sar
2 days ago
1
A related concept is equicontinuity, which says that the same $delta$ works for all values of $n$.
– Teepeemm
2 days ago
Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for.
– Sar
2 days ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Say $f_n,f:mathbbR to mathbbR$ are uniformly continous functions a $f_nto f$ pointwise.
Is the convergence nessecarily uniform?
I couldn't find an appropriate counter-example for this case so I tried proving it :
We have $|f_n(x)-f_n(y)|<epsilon/3$ for all $|x-y|<delta_1$, also $|f(x)-f(y)|<epsilon/3$ for $|x-y|<delta_2$ and for all $n>N_x$ we have $|f_n(x)-f(x)|<epsilon /3$.
Overall for any $|x-y|<delta=min(delta_1,delta_2)$ and any $n>N_x$ we have $|f_n(y)-f(y)|leq |f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(x)-f(y)|<epsilon $ for any $yin (x-delta,x+delta)$ and for any $xin AsubsetmathbbR$.
Now if $A$ is compact, we have a cover of $A$ and we can choose a finite subcover $U$, and take $N=maxxin U. $ and for all $n>N$ we have $|f_n(y)-f(y)|<epsilon$ uniformly .
However, if my proof is correct I have only proven uniform convergence in compact sets, and not on the entire number line. Can my proof be modified so we can get uniform convergence on $mathbbR$? Or can someone provide a counter example where it fails in this case?
Edit: Thanks to the counter-examples in the answers I can see my proof is false. Can anyone point out where my proof fails?
real-analysis sequences-and-series proof-verification uniform-convergence uniform-continuity
asked 2 days ago
Sar
3529
Say $f_n,f:mathbbR to mathbbR$ are uniformly continous functions a $f_nto f$ pointwise.
Is the convergence nessecarily uniform?
I couldn't find an appropriate counter-example for this case so I tried proving it :
We have $|f_n(x)-f_n(y)|<epsilon/3$ for all $|x-y|<delta_1$, also $|f(x)-f(y)|<epsilon/3$ for $|x-y|<delta_2$ and for all $n>N_x$ we have $|f_n(x)-f(x)|<epsilon /3$.
Overall for any $|x-y|<delta=min(delta_1,delta_2)$ and any $n>N_x$ we have $|f_n(y)-f(y)|leq |f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(x)-f(y)|<epsilon $ for any $yin (x-delta,x+delta)$ and for any $xin AsubsetmathbbR$.
Now if $A$ is compact, we have a cover of $A$ and we can choose a finite subcover $U$, and take $N=maxxin U. $ and for all $n>N$ we have $|f_n(y)-f(y)|<epsilon$ uniformly .
However, if my proof is correct I have only proven uniform convergence in compact sets, and not on the entire number line. Can my proof be modified so we can get uniform convergence on $mathbbR$? Or can someone provide a counter example where it fails in this case?
Edit: Thanks to the counter-examples in the answers I can see my proof is false. Can anyone point out where my proof fails?
real-analysis sequences-and-series proof-verification uniform-convergence uniform-continuity
asked 2 days ago
Sar
3529
edited 2 days ago
asked 2 days ago
Sar
3529
asked 2 days ago
Sar
3529
asked 2 days ago
Sar
3529
3529
This kind of like the Dini's theorem, in which the convergence is monotonic
– xbh
2 days ago
This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions.
– Sar
2 days ago
1
A related concept is equicontinuity, which says that the same $delta$ works for all values of $n$.
– Teepeemm
2 days ago
Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for.
– Sar
2 days ago
add a comment |Â
This kind of like the Dini's theorem, in which the convergence is monotonic
– xbh
2 days ago
This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions.
– Sar
2 days ago
1
A related concept is equicontinuity, which says that the same $delta$ works for all values of $n$.
– Teepeemm
2 days ago
Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for.
– Sar
2 days ago
This kind of like the Dini's theorem, in which the convergence is monotonic
– xbh
2 days ago
This kind of like the Dini's theorem, in which the convergence is monotonic
– xbh
2 days ago
This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions.
– Sar
2 days ago
This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions.
– Sar
2 days ago
1
1
A related concept is equicontinuity, which says that the same $delta$ works for all values of $n$.
– Teepeemm
2 days ago
A related concept is equicontinuity, which says that the same $delta$ works for all values of $n$.
– Teepeemm
2 days ago
Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for.
– Sar
2 days ago
Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for.
– Sar
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
The answer is no. To construct a counter-example, we consider $f_n(x)=fracnx1+(nx)^2$ and $f(x)=0$ on interval [0,1]. Obviously they are all continuous functions defined on a compact subset of real axis, hence they are all uniform continuous. It’s also easy to verify that the series fn(x) converge point-wise to f(x). However, since for each n, fn(x) attains its maximum 1/2 at x=1/n, so the convergence is not a uniform one.
answered 2 days ago
Legolas Hu
562
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
2
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
1
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
add a comment |Â
up vote
6
down vote
I'll give an example on $mathbbR$. Take the sequence $f_n(x)=fracxn$. All the functions are uniformly continuous, the limit function is the zero function which is also uniformly continuous. But I say there is no uniform convergence here. Take $epsilon=1$. For any index $n_oin mathbbN$ you can take $n=n_0+1$ and $x=2n$ and you will get $|fracxn-0|=2 geq 1=epsilon$. Hence there is no uniform convergence.
answered 2 days ago
Mark
5849
add a comment |Â
up vote
2
down vote
No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $textbf1_1$, but the convergence is not uniform.
answered 2 days ago
Daniel Xiang
1,788413
2
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
1
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
 |Â
show 4 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The answer is no. To construct a counter-example, we consider $f_n(x)=fracnx1+(nx)^2$ and $f(x)=0$ on interval [0,1]. Obviously they are all continuous functions defined on a compact subset of real axis, hence they are all uniform continuous. It’s also easy to verify that the series fn(x) converge point-wise to f(x). However, since for each n, fn(x) attains its maximum 1/2 at x=1/n, so the convergence is not a uniform one.
answered 2 days ago
Legolas Hu
562
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
2
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
1
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
add a comment |Â
up vote
4
down vote
accepted
The answer is no. To construct a counter-example, we consider $f_n(x)=fracnx1+(nx)^2$ and $f(x)=0$ on interval [0,1]. Obviously they are all continuous functions defined on a compact subset of real axis, hence they are all uniform continuous. It’s also easy to verify that the series fn(x) converge point-wise to f(x). However, since for each n, fn(x) attains its maximum 1/2 at x=1/n, so the convergence is not a uniform one.
answered 2 days ago
Legolas Hu
562
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
2
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
1
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The answer is no. To construct a counter-example, we consider $f_n(x)=fracnx1+(nx)^2$ and $f(x)=0$ on interval [0,1]. Obviously they are all continuous functions defined on a compact subset of real axis, hence they are all uniform continuous. It’s also easy to verify that the series fn(x) converge point-wise to f(x). However, since for each n, fn(x) attains its maximum 1/2 at x=1/n, so the convergence is not a uniform one.
answered 2 days ago
Legolas Hu
562
The answer is no. To construct a counter-example, we consider $f_n(x)=fracnx1+(nx)^2$ and $f(x)=0$ on interval [0,1]. Obviously they are all continuous functions defined on a compact subset of real axis, hence they are all uniform continuous. It’s also easy to verify that the series fn(x) converge point-wise to f(x). However, since for each n, fn(x) attains its maximum 1/2 at x=1/n, so the convergence is not a uniform one.
answered 2 days ago
Legolas Hu
562
answered 2 days ago
Legolas Hu
562
answered 2 days ago
Legolas Hu
562
answered 2 days ago
Legolas Hu
562
562
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
2
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
1
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
add a comment |Â
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
2
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
1
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails?
– Sar
2 days ago
2
2
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
One of the problems I see in your proof is that you wrote that for a given $epsilon$ there exists a unique $delta_1$ that works for all $n in mathbbN$. This is not true in general. All the functions $f_n$ are uniformly continuous but $delta$ is dependent on $n$.
– Mark
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
@Mark Thank you for clarifying the same question for me at the same time!
– xbh
2 days ago
1
1
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence.
– Mark
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
@Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $delta$ and complete the proof. Thanks again!
– xbh
2 days ago
add a comment |Â
up vote
6
down vote
I'll give an example on $mathbbR$. Take the sequence $f_n(x)=fracxn$. All the functions are uniformly continuous, the limit function is the zero function which is also uniformly continuous. But I say there is no uniform convergence here. Take $epsilon=1$. For any index $n_oin mathbbN$ you can take $n=n_0+1$ and $x=2n$ and you will get $|fracxn-0|=2 geq 1=epsilon$. Hence there is no uniform convergence.
answered 2 days ago
Mark
5849
add a comment |Â
up vote
6
down vote
I'll give an example on $mathbbR$. Take the sequence $f_n(x)=fracxn$. All the functions are uniformly continuous, the limit function is the zero function which is also uniformly continuous. But I say there is no uniform convergence here. Take $epsilon=1$. For any index $n_oin mathbbN$ you can take $n=n_0+1$ and $x=2n$ and you will get $|fracxn-0|=2 geq 1=epsilon$. Hence there is no uniform convergence.
answered 2 days ago
Mark
5849
add a comment |Â
up vote
6
down vote
up vote
6
down vote
I'll give an example on $mathbbR$. Take the sequence $f_n(x)=fracxn$. All the functions are uniformly continuous, the limit function is the zero function which is also uniformly continuous. But I say there is no uniform convergence here. Take $epsilon=1$. For any index $n_oin mathbbN$ you can take $n=n_0+1$ and $x=2n$ and you will get $|fracxn-0|=2 geq 1=epsilon$. Hence there is no uniform convergence.
answered 2 days ago
Mark
5849
I'll give an example on $mathbbR$. Take the sequence $f_n(x)=fracxn$. All the functions are uniformly continuous, the limit function is the zero function which is also uniformly continuous. But I say there is no uniform convergence here. Take $epsilon=1$. For any index $n_oin mathbbN$ you can take $n=n_0+1$ and $x=2n$ and you will get $|fracxn-0|=2 geq 1=epsilon$. Hence there is no uniform convergence.
answered 2 days ago
Mark
5849
edited 2 days ago
answered 2 days ago
Mark
5849
answered 2 days ago
Mark
5849
answered 2 days ago
Mark
5849
5849
add a comment |Â
add a comment |Â
up vote
2
down vote
No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $textbf1_1$, but the convergence is not uniform.
answered 2 days ago
Daniel Xiang
1,788413
2
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
1
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
 |Â
show 4 more comments
up vote
2
down vote
No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $textbf1_1$, but the convergence is not uniform.
answered 2 days ago
Daniel Xiang
1,788413
2
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
1
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $textbf1_1$, but the convergence is not uniform.
answered 2 days ago
Daniel Xiang
1,788413
No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $textbf1_1$, but the convergence is not uniform.
answered 2 days ago
Daniel Xiang
1,788413
answered 2 days ago
Daniel Xiang
1,788413
answered 2 days ago
Daniel Xiang
1,788413
answered 2 days ago
Daniel Xiang
1,788413
1,788413
2
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
1
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
 |Â
show 4 more comments
2
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
1
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
2
2
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$.
– Mark
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
The limit function is $f=begincases 1spacespace, x=1 \ 0 spacespace,xin[0,1)endcases$
– Sar
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$
– xbh
2 days ago
1
1
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit.
– Rob Arthan
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
@RobArthan Okay, thanks.
– xbh
2 days ago
 |Â
show 4 more comments
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This kind of like the Dini's theorem, in which the convergence is monotonic
– xbh
2 days ago
This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions.
– Sar
2 days ago
1
A related concept is equicontinuity, which says that the same $delta$ works for all values of $n$.
– Teepeemm
2 days ago
Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for.
– Sar
2 days ago