Pointwise convergence of uniformly continuous functions to zero, but not uniformlyUniform 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?
When does Haskell complain about incorrect typing in functions?
Request for a Latin phrase as motto "God is highest/supreme"
Is this photo showing a woman standing in the nude before teenagers real?
Is a topological space considered to be a class in set theory?
Pointwise convergence of uniformly continuous functions to zero, but not uniformly
How much were the LMs maneuvered to their landing points?
Isolated audio without a transformer
Are there any examples of technologies have been lost over time?
Use cases for M-0 & C-0?
Correlation length anisotropy in the 2D Ising model
Defining a Function programmatically
The best place for swimming in Arctic Ocean
What is the use of で in this sentence?
Melee or Ranged attacks by Monsters, no distinction in modifiers?
Assuring luggage isn't lost with short layover
Is a fighting a fallen friend with the help of a redeemed villain story too much for one book
What is the difference between 1/3, 1/2, and full casters?
Where to find an interactive PDF or HTML version of the tex.web documentation?
Why force the nose of 737 Max down in the first place?
Decreasing star size
Checking if an integer is a member of an integer list
If a 2019 UA artificer has the Repeating Shot infusion on two hand crossbows, can they use two-weapon fighting?
Unethical behavior : should I report it?
Did the IBM PC use the 8088's NMI line?
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?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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 8 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
$endgroup$
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 8 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
8 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
8 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 8 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 8 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 8 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 8 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
asked 8 hours ago
Jannik PittJannik Pitt
6416 silver badges19 bronze badges
asked 8 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
8 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
8 hours ago
add a comment |
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
8 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
8 hours ago
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
8 hours ago
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
8 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
8 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
8 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 8 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 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 8 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 8 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3306545%2fpointwise-convergence-of-uniformly-continuous-functions-to-zero-but-not-uniform%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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 8 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 8 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 8 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 8 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
edited 8 hours ago
answered 8 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
answered 8 hours ago
Robert ZRobert Z
107k10 gold badges76 silver badges149 bronze badges
answered 8 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 8 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 8 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 8 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 8 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
edited 8 hours ago
answered 8 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
answered 8 hours ago
Chinnapparaj RChinnapparaj R
8,6912 gold badges10 silver badges32 bronze badges
answered 8 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 8 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 8 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 8 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 8 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
edited 8 hours ago
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
203k25 gold badges159 silver badges280 bronze badges
answered 8 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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3306545%2fpointwise-convergence-of-uniformly-continuous-functions-to-zero-but-not-uniform%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example.
$endgroup$
– Suzet
8 hours ago
$begingroup$
@Suzet I forgot, I want the domain to be compact.
$endgroup$
– Jannik Pitt
8 hours ago