Existence of a function under certain given conditions.
$begingroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
$endgroup$
add a comment |
$begingroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
$endgroup$
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
add a comment |
$begingroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
$endgroup$
Is there exists a function $f$ such that $f(-1)=-1$ and $f(1)=1$ and $|f(x)-f(y)| leq |x-y|^{3/2}$ for all $x,y in mathbb{R}$
How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.
real-analysis
real-analysis
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
asked Dec 7 '18 at 5:13
blue boyblue boy
1,236613
1,236613
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
add a comment |
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
1
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
1766
$endgroup$
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%2f3029490%2fexistence-of-a-function-under-certain-given-conditions%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
$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
1766
$endgroup$
add a comment |
$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
1766
$endgroup$
add a comment |
$begingroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
1766
$endgroup$
Note that when $xnot = y$, we have that $|frac{f(x)-f(y)}{x-y}|le sqrt{|x-y|}$. If we take the limit as $yto x$ on both sides, we get information about the derivative of $f$ at any point $x$.
answered Dec 7 '18 at 5:32
John BJohn B
1766
answered Dec 7 '18 at 5:32
John BJohn B
1766
answered Dec 7 '18 at 5:32
John BJohn B
1766
answered Dec 7 '18 at 5:32
John BJohn B
1766
1766
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%2f3029490%2fexistence-of-a-function-under-certain-given-conditions%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
1
$begingroup$
Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality?
$endgroup$
– John B
Dec 7 '18 at 5:16
$begingroup$
$f(x)=x$? {}{}{}{}
$endgroup$
– blue boy
Dec 7 '18 at 5:23
$begingroup$
Oh nevermind, that only works when |x-y| is greater than 1
$endgroup$
– John B
Dec 7 '18 at 5:24
$begingroup$
I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$
$endgroup$
– user25959
Dec 7 '18 at 5:25