Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$.
0
Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 at 1:48
Rebuild
31
add a comment |
0
Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 at 1:48
Rebuild
31
add a comment |
0
0
0
Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 at 1:48
Rebuild
31
Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 at 1:48
Rebuild
31
asked Dec 2 at 1:48
Rebuild
31
asked Dec 2 at 1:48
Rebuild
31
asked Dec 2 at 1:48
Rebuild
31
asked Dec 2 at 1:48
Rebuild
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
0
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 at 1:52
Anupam
2,3701823
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
draft saved
draft discarded
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%2f3022117%2ffind-a-sequence-of-lipschitz-continuous-functions-on-0-1-whose-uniform-limit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 at 1:52
Anupam
2,3701823
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
add a comment |
0
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 at 1:52
Anupam
2,3701823
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
add a comment |
0
0
0
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 at 1:52
Anupam
2,3701823
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 at 1:52
Anupam
2,3701823
answered Dec 2 at 1:52
Anupam
2,3701823
answered Dec 2 at 1:52
Anupam
2,3701823
answered Dec 2 at 1:52
Anupam
2,3701823
2,3701823
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
add a comment |
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_n$ are not Lipschitz continuous functions, take $n=1$.
– Rebuild
Dec 2 at 2:04
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
– Anupam
Dec 2 at 2:08
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
– Rebuild
Dec 2 at 2:11
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
– Rebuild
Dec 2 at 2:18
add a comment |
draft saved
draft discarded
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
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%2f3022117%2ffind-a-sequence-of-lipschitz-continuous-functions-on-0-1-whose-uniform-limit%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
