$f$ is a $k$-Lipschitz continuous function & continuously differentiable $forall x in mathbb{R}^n$....
$begingroup$
EDIT - generalization for vector analysis
for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
$k$-Lipschitz continuous function.
ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$
$ $
previous Version:
I was asking the same question but for $f: mathbb{R} to mathbb{R}$
Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.
Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?
Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$
The only intuition I have in mind is that:
$$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$
Therefore
$$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$
But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.
vector-analysis lipschitz-functions
asked Jan 23 at 11:40
JnevenJneven
906322
$endgroup$
add a comment |
$begingroup$
EDIT - generalization for vector analysis
for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
$k$-Lipschitz continuous function.
ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$
$ $
previous Version:
I was asking the same question but for $f: mathbb{R} to mathbb{R}$
Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.
Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?
Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$
The only intuition I have in mind is that:
$$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$
Therefore
$$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$
But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.
vector-analysis lipschitz-functions
asked Jan 23 at 11:40
JnevenJneven
906322
$endgroup$
add a comment |
$begingroup$
EDIT - generalization for vector analysis
for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
$k$-Lipschitz continuous function.
ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$
$ $
previous Version:
I was asking the same question but for $f: mathbb{R} to mathbb{R}$
Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.
Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?
Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$
The only intuition I have in mind is that:
$$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$
Therefore
$$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$
But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.
vector-analysis lipschitz-functions
asked Jan 23 at 11:40
JnevenJneven
906322
$endgroup$
EDIT - generalization for vector analysis
for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
$k$-Lipschitz continuous function.
ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$
$ $
previous Version:
I was asking the same question but for $f: mathbb{R} to mathbb{R}$
Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.
Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?
Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$
The only intuition I have in mind is that:
$$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$
Therefore
$$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$
But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.
vector-analysis lipschitz-functions
vector-analysis lipschitz-functions
asked Jan 23 at 11:40
JnevenJneven
906322
asked Jan 23 at 11:40
JnevenJneven
906322
edited Jan 23 at 13:14
Jneven
asked Jan 23 at 11:40
JnevenJneven
906322
asked Jan 23 at 11:40
JnevenJneven
906322
asked Jan 23 at 11:40
JnevenJneven
906322
906322
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We have
$$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
But by the lipschitz condition we get
$$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
And thus, we have
$$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
implying that
$$ Vert D_f(x) Vert_{op} leq k. $$
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
$endgroup$
add a comment |
$begingroup$
If $f$ is k -Lipschitz continuous, you have
$$
left| frac{f(y)-f(x)}{y-x} right| le k
$$
for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
You should isolate instead $f'(x)$, in that case:
$f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$
Then:
$vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
$endgroup$
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%2f3084358%2ff-is-a-k-lipschitz-continuous-function-continuously-differentiable-foral%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
$begingroup$
We have
$$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
But by the lipschitz condition we get
$$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
And thus, we have
$$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
implying that
$$ Vert D_f(x) Vert_{op} leq k. $$
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
$endgroup$
add a comment |
$begingroup$
We have
$$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
But by the lipschitz condition we get
$$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
And thus, we have
$$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
implying that
$$ Vert D_f(x) Vert_{op} leq k. $$
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
$endgroup$
add a comment |
$begingroup$
We have
$$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
But by the lipschitz condition we get
$$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
And thus, we have
$$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
implying that
$$ Vert D_f(x) Vert_{op} leq k. $$
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
$endgroup$
We have
$$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
But by the lipschitz condition we get
$$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
And thus, we have
$$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
implying that
$$ Vert D_f(x) Vert_{op} leq k. $$
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
answered Jan 23 at 13:30
Severin SchravenSeverin Schraven
6,3381934
6,3381934
add a comment |
add a comment |
$begingroup$
If $f$ is k -Lipschitz continuous, you have
$$
left| frac{f(y)-f(x)}{y-x} right| le k
$$
for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
If $f$ is k -Lipschitz continuous, you have
$$
left| frac{f(y)-f(x)}{y-x} right| le k
$$
for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
If $f$ is k -Lipschitz continuous, you have
$$
left| frac{f(y)-f(x)}{y-x} right| le k
$$
for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
$endgroup$
If $f$ is k -Lipschitz continuous, you have
$$
left| frac{f(y)-f(x)}{y-x} right| le k
$$
for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
answered Jan 23 at 11:46
Reiner MartinReiner Martin
3,509414
3,509414
add a comment |
add a comment |
$begingroup$
You should isolate instead $f'(x)$, in that case:
$f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$
Then:
$vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
$endgroup$
add a comment |
$begingroup$
You should isolate instead $f'(x)$, in that case:
$f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$
Then:
$vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
$endgroup$
add a comment |
$begingroup$
You should isolate instead $f'(x)$, in that case:
$f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$
Then:
$vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
$endgroup$
You should isolate instead $f'(x)$, in that case:
$f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$
Then:
$vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
answered Jan 23 at 11:49
Keen-ameteurKeen-ameteur
1,477416
1,477416
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%2f3084358%2ff-is-a-k-lipschitz-continuous-function-continuously-differentiable-foral%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
