Mixing
Prove that $f(x)=xsin(1/x)$ for $xne0$, $f(0)=0$, is not Lipschitz on $[0,1]$
$begingroup$
Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$
MY TRIAL
My idea is to show that $f$ does not have a bounded derivative.
So, suppose for contradiction that there exists $K>0$ such that
$$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
As $xto 0$,
begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
Please, I'm I right? Any other way of showing this is also accepted.
real-analysis analysis uniform-continuity lipschitz-functions
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
$endgroup$
add a comment |
$begingroup$
Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$
MY TRIAL
My idea is to show that $f$ does not have a bounded derivative.
So, suppose for contradiction that there exists $K>0$ such that
$$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
As $xto 0$,
begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
Please, I'm I right? Any other way of showing this is also accepted.
real-analysis analysis uniform-continuity lipschitz-functions
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
$endgroup$
add a comment |
$begingroup$
Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$
MY TRIAL
My idea is to show that $f$ does not have a bounded derivative.
So, suppose for contradiction that there exists $K>0$ such that
$$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
As $xto 0$,
begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
Please, I'm I right? Any other way of showing this is also accepted.
real-analysis analysis uniform-continuity lipschitz-functions
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
$endgroup$
Prove that $f(x)=cases{0& if $x=0$\xsin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$
MY TRIAL
My idea is to show that $f$ does not have a bounded derivative.
So, suppose for contradiction that there exists $K>0$ such that
$$left| f'(x)-f'(0)right|=left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq K |x|,;forall;xin [0,1].$$
As $xto 0$,
begin{align}limlimits_{xto 0}left| f'(x)-f'(0)right|&=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|,end{align}
which implies that begin{align}+infty=limlimits_{xto 0}left| sinleft(frac{1}{x}right)-frac{1}{x}cosleft(frac{1}{x}right)right|leq Klimlimits_{xto 0} |x|=0,;;text{contradiction.}end{align}
Please, I'm I right? Any other way of showing this is also accepted.
real-analysis analysis uniform-continuity lipschitz-functions
real-analysis analysis uniform-continuity lipschitz-functions
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
edited Jan 5 at 14:58
Omojola Micheal
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
asked Jan 5 at 14:29
Omojola MichealOmojola Micheal
1,831324
1,831324
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$
for $n>C.$ So, $f$ can't be Lipschitz.
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
$endgroup$
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
|
show 4 more comments
$begingroup$
$f'(0)$ does not exist so your approach won't work. But the idea is correct:
You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$
Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
$endgroup$
1
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
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%2f3062770%2fprove-that-fx-x-sin1-x-for-x-ne0-f0-0-is-not-lipschitz-on-0-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$
for $n>C.$ So, $f$ can't be Lipschitz.
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
$endgroup$
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
|
show 4 more comments
$begingroup$
$fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$
for $n>C.$ So, $f$ can't be Lipschitz.
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
$endgroup$
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
|
show 4 more comments
$begingroup$
$fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$
for $n>C.$ So, $f$ can't be Lipschitz.
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
$endgroup$
$fleft(frac{2}{(2n+1)pi}right)=frac{2}{(2n+1)pi}(-1)^n.$ So, for any $C>0$ $left|fleft(frac{2}{(2n+1)pi}right)-fleft(frac{2}{(2n-1)pi}right)right|=frac{2}{(2n-1)pi}+frac{2}{(2n+1)pi}=frac{8n}{(2n+1)(2n-1)pi}> Cfrac{4}{(2n+1)(2n-1)pi}=Cleft|frac{2}{(2n-1)pi}-frac{2}{(2n+1)pi}right|$
for $n>C.$ So, $f$ can't be Lipschitz.
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
edited Jan 6 at 4:24
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
answered Jan 5 at 14:42
John_WickJohn_Wick
1,486111
1,486111
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
|
show 4 more comments
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
(+1) for that. I'm I right too?
$endgroup$
– Omojola Micheal
Jan 5 at 14:51
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Your approach was right, but the reasoning was a little bit shaky.
$endgroup$
– John_Wick
Jan 5 at 14:53
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
Oh, at what point?
$endgroup$
– Omojola Micheal
Jan 5 at 14:54
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
You showed that $f'$ is not Lipschitz. But why does that imply $f$ is not Lipschitz is not very clear.
$endgroup$
– John_Wick
Jan 5 at 14:56
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
$begingroup$
Alright, let me edit it.
$endgroup$
– Omojola Micheal
Jan 5 at 14:57
|
show 4 more comments
$begingroup$
$f'(0)$ does not exist so your approach won't work. But the idea is correct:
You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$
Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
$endgroup$
1
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
add a comment |
$begingroup$
$f'(0)$ does not exist so your approach won't work. But the idea is correct:
You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$
Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
$endgroup$
1
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
add a comment |
$begingroup$
$f'(0)$ does not exist so your approach won't work. But the idea is correct:
You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$
Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
$endgroup$
$f'(0)$ does not exist so your approach won't work. But the idea is correct:
You want to show that $frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$
Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_nto 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $frac{|xsin(1/x)-ysin(1/y)|}{|x - y|}$, we might try $x_n = frac{1}{2 pi n + pi/2}\$ because then the sines are equal to $1$. I'll leave it to you to find $y_nin [0,1]$ that gives us what we want.
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
answered Jan 5 at 16:17
MatematletaMatematleta
10.3k2918
10.3k2918
1
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
add a comment |
1
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
1
1
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
Thanks for that, Matematleta!
$endgroup$
– Omojola Micheal
Jan 5 at 16:23
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
$begingroup$
You are welcome!
$endgroup$
– Matematleta
Jan 5 at 16:25
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%2f3062770%2fprove-that-fx-x-sin1-x-for-x-ne0-f0-0-is-not-lipschitz-on-0-1%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
