Mixing
Showing function is not Lipschitz continuous
$begingroup$
For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.
Any help is appreciated.
real-analysis lipschitz-functions
asked Jan 14 at 19:54
Tyler6Tyler6
838413
$endgroup$
add a comment |
$begingroup$
For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.
Any help is appreciated.
real-analysis lipschitz-functions
asked Jan 14 at 19:54
Tyler6Tyler6
838413
$endgroup$
$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58
1
$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02
$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04
add a comment |
$begingroup$
For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.
Any help is appreciated.
real-analysis lipschitz-functions
asked Jan 14 at 19:54
Tyler6Tyler6
838413
$endgroup$
For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.
Any help is appreciated.
real-analysis lipschitz-functions
real-analysis lipschitz-functions
asked Jan 14 at 19:54
Tyler6Tyler6
838413
asked Jan 14 at 19:54
Tyler6Tyler6
838413
asked Jan 14 at 19:54
Tyler6Tyler6
838413
asked Jan 14 at 19:54
Tyler6Tyler6
838413
asked Jan 14 at 19:54
Tyler6Tyler6
838413
838413
$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58
1
$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02
$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04
add a comment |
$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58
1
$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02
$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04
$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58
$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58
1
1
$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02
$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02
$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04
$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
$$
However,
$$
limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
$$
hence no $K$ satisfies $(1)$.
answered Jan 14 at 20:15
HaykHayk
2,6271214
$endgroup$
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
});
}
});
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%2f3073665%2fshowing-function-is-not-lipschitz-continuous%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
$begingroup$
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
$$
However,
$$
limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
$$
hence no $K$ satisfies $(1)$.
answered Jan 14 at 20:15
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
$$
However,
$$
limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
$$
hence no $K$ satisfies $(1)$.
answered Jan 14 at 20:15
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
$$
However,
$$
limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
$$
hence no $K$ satisfies $(1)$.
answered Jan 14 at 20:15
HaykHayk
2,6271214
$endgroup$
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
$$
However,
$$
limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
$$
hence no $K$ satisfies $(1)$.
answered Jan 14 at 20:15
HaykHayk
2,6271214
answered Jan 14 at 20:15
HaykHayk
2,6271214
answered Jan 14 at 20:15
HaykHayk
2,6271214
answered Jan 14 at 20:15
HaykHayk
2,6271214
2,6271214
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%2f3073665%2fshowing-function-is-not-lipschitz-continuous%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
$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58
1
$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02
$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04