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Pointwise convergence of uniformly continuous functions to zero, but not uniformly
Uniform convergence on interval.Pointwise but not Uniformly ConvergentPointwise but not uniform convergence of continuous functions on $[0,1]$Uniformly continuous functions sequence $f_n(x)$ converges uniformly to a uniformly continuous function $f(x)$?Pointwise convergence to infinity implies uniformly convergence?A function that converges pointwise but not uniformly?Pointwise Convergence. Uniform ConvergencePointwise convergence to a uniform continuous functionContinuous function as pointwise limit but not as uniform limit of a sequence of continuous functions on $[0,1]$Does pointwise convergence to a continuous function in compact set imply uniform convergence?Sequence of integrable functions on [a,b] converges pointwise but not uniformly?
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$begingroup$
What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $0$, but not uniformly?
real-analysis examples-counterexamples uniform-convergence uniform-continuity pointwise-convergence
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
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$endgroup$
add a comment |
$begingroup$
What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $0$, but not uniformly?
real-analysis examples-counterexamples uniform-convergence uniform-continuity pointwise-convergence
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
$endgroup$
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
9 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
9 hours ago
add a comment |
$begingroup$
What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $0$, but not uniformly?
real-analysis examples-counterexamples uniform-convergence uniform-continuity pointwise-convergence
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
$endgroup$
What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $0$, but not uniformly?
real-analysis examples-counterexamples uniform-convergence uniform-continuity pointwise-convergence
real-analysis examples-counterexamples uniform-convergence uniform-continuity pointwise-convergence
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
edited 8 hours ago
José Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
203k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
asked 9 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
6416 silver badges19 bronze badges
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
9 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
9 hours ago
add a comment |
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
9 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
9 hours ago
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
9 hours ago
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
9 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
9 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider
$$f_n(x)=1-min(1,n|x-1/n|)=begincases
nx&text if x<frac1n\2-nx&text if frac1nleq xleqfrac2n\0&text otherwise.endcases$$
Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover, $f_n(x)to 0$ for any $xin [0,1]$, but the convergence is not uniform on $[0,1]$ because $max_xin[0,1]|f_n(x)|=f(1/n))=1$.
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
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add a comment |
$begingroup$
The another standard one is the growing steeple on $[0,1]$:
$$f_n(x)=begincasesn^2 x &textif;0 leq x leq frac1n\ 2n-n^2 x &textif;frac1n leq x leq frac2n\ 0 &textif;frac2n leq x leq 1 endcases$$
Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height $n$ attained at $1/n$
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&nx^n(1-x).endarray$$The sequence $(f_n)_ninmathbb N$ converges pointwise to the null function, but not uniformly, since$$(forall ninmathbb N):f_nleft(frac nn+1right)=left(frac nn+1right)^n+1$$and $lim_ntoinftyleft(frac nn+1right)^n+1=e^-1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
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add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
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active
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$begingroup$
Consider
$$f_n(x)=1-min(1,n|x-1/n|)=begincases
nx&text if x<frac1n\2-nx&text if frac1nleq xleqfrac2n\0&text otherwise.endcases$$
Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover, $f_n(x)to 0$ for any $xin [0,1]$, but the convergence is not uniform on $[0,1]$ because $max_xin[0,1]|f_n(x)|=f(1/n))=1$.
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
$endgroup$
add a comment |
$begingroup$
Consider
$$f_n(x)=1-min(1,n|x-1/n|)=begincases
nx&text if x<frac1n\2-nx&text if frac1nleq xleqfrac2n\0&text otherwise.endcases$$
Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover, $f_n(x)to 0$ for any $xin [0,1]$, but the convergence is not uniform on $[0,1]$ because $max_xin[0,1]|f_n(x)|=f(1/n))=1$.
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
$endgroup$
add a comment |
$begingroup$
Consider
$$f_n(x)=1-min(1,n|x-1/n|)=begincases
nx&text if x<frac1n\2-nx&text if frac1nleq xleqfrac2n\0&text otherwise.endcases$$
Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover, $f_n(x)to 0$ for any $xin [0,1]$, but the convergence is not uniform on $[0,1]$ because $max_xin[0,1]|f_n(x)|=f(1/n))=1$.
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
$endgroup$
Consider
$$f_n(x)=1-min(1,n|x-1/n|)=begincases
nx&text if x<frac1n\2-nx&text if frac1nleq xleqfrac2n\0&text otherwise.endcases$$
Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover, $f_n(x)to 0$ for any $xin [0,1]$, but the convergence is not uniform on $[0,1]$ because $max_xin[0,1]|f_n(x)|=f(1/n))=1$.
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
edited 9 hours ago
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
answered 9 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
107k10 gold badges76 silver badges149 bronze badges
add a comment |
add a comment |
$begingroup$
The another standard one is the growing steeple on $[0,1]$:
$$f_n(x)=begincasesn^2 x &textif;0 leq x leq frac1n\ 2n-n^2 x &textif;frac1n leq x leq frac2n\ 0 &textif;frac2n leq x leq 1 endcases$$
Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height $n$ attained at $1/n$
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
The another standard one is the growing steeple on $[0,1]$:
$$f_n(x)=begincasesn^2 x &textif;0 leq x leq frac1n\ 2n-n^2 x &textif;frac1n leq x leq frac2n\ 0 &textif;frac2n leq x leq 1 endcases$$
Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height $n$ attained at $1/n$
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
The another standard one is the growing steeple on $[0,1]$:
$$f_n(x)=begincasesn^2 x &textif;0 leq x leq frac1n\ 2n-n^2 x &textif;frac1n leq x leq frac2n\ 0 &textif;frac2n leq x leq 1 endcases$$
Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height $n$ attained at $1/n$
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
$endgroup$
The another standard one is the growing steeple on $[0,1]$:
$$f_n(x)=begincasesn^2 x &textif;0 leq x leq frac1n\ 2n-n^2 x &textif;frac1n leq x leq frac2n\ 0 &textif;frac2n leq x leq 1 endcases$$
Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height $n$ attained at $1/n$
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
edited 9 hours ago
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
answered 9 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
8,6912 gold badges10 silver badges32 bronze badges
add a comment |
add a comment |
$begingroup$
Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&nx^n(1-x).endarray$$The sequence $(f_n)_ninmathbb N$ converges pointwise to the null function, but not uniformly, since$$(forall ninmathbb N):f_nleft(frac nn+1right)=left(frac nn+1right)^n+1$$and $lim_ntoinftyleft(frac nn+1right)^n+1=e^-1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
$endgroup$
add a comment |
$begingroup$
Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&nx^n(1-x).endarray$$The sequence $(f_n)_ninmathbb N$ converges pointwise to the null function, but not uniformly, since$$(forall ninmathbb N):f_nleft(frac nn+1right)=left(frac nn+1right)^n+1$$and $lim_ntoinftyleft(frac nn+1right)^n+1=e^-1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
$endgroup$
add a comment |
$begingroup$
Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&nx^n(1-x).endarray$$The sequence $(f_n)_ninmathbb N$ converges pointwise to the null function, but not uniformly, since$$(forall ninmathbb N):f_nleft(frac nn+1right)=left(frac nn+1right)^n+1$$and $lim_ntoinftyleft(frac nn+1right)^n+1=e^-1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
$endgroup$
Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&nx^n(1-x).endarray$$The sequence $(f_n)_ninmathbb N$ converges pointwise to the null function, but not uniformly, since$$(forall ninmathbb N):f_nleft(frac nn+1right)=left(frac nn+1right)^n+1$$and $lim_ntoinftyleft(frac nn+1right)^n+1=e^-1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
edited 9 hours ago
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
203k25 gold badges159 silver badges280 bronze badges
add a comment |
add a comment |
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$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
9 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
9 hours ago