$C^1$ Lipschitz function linear growth Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Approximate continous function with linear growth condition by Lipschitz functionWhy is this function not locally Lipschitz?Classify the growth of functions and find a more general growth functionLipschitz/Hölder continuity of cosine functionLinear functions $mathbbC^nlongrightarrowmathbbC^m$ are Lipschitz continousLipschitz Continuity For a Vector Function.Some questions about function's growth.Linear function: relation between linearity and continuityDifferentiable Manifolds, Metric Space Structure, Lipschitz ContinuityDifference between “Rate of growth of a function” and “Order of growth of a function”
What would you call this weird metallic apparatus that allows you to lift people?
Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?
How many serial port on PI3
How does light 'choose' between wave and particle behaviour?
What is the meaning of 'breadth' in breadth first search?
AppleTVs create a chatty alternate WiFi network
Should I use a zero-interest credit card for a large one-time purchase?
Triggering an ultrasonic sensor
"Lost his faith in humanity in the trenches of Verdun" — last line of an SF story
What is "gratricide"?
What do you call the main part of a joke?
Monolithic + MicroServices
Is it possible for SQL statements to execute concurrently within a single session in SQL Server?
Crossing US/Canada Border for less than 24 hours
Find 108 by using 3,4,6
Sum letters are not two different
A term for a woman complaining about things/begging in a cute/childish way
Maximum summed subsequences with non-adjacent items
Most bit efficient text communication method?
Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?
Project Euler #1 in C++
Can a new player join a group only when a new campaign starts?
How to improve on this Stylesheet Manipulation for Message Styling
Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?
$C^1$ Lipschitz function linear growth
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Approximate continous function with linear growth condition by Lipschitz functionWhy is this function not locally Lipschitz?Classify the growth of functions and find a more general growth functionLipschitz/Hölder continuity of cosine functionLinear functions $mathbbC^nlongrightarrowmathbbC^m$ are Lipschitz continousLipschitz Continuity For a Vector Function.Some questions about function's growth.Linear function: relation between linearity and continuityDifferentiable Manifolds, Metric Space Structure, Lipschitz ContinuityDifference between “Rate of growth of a function” and “Order of growth of a function”
$begingroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
edited Mar 27 at 19:42
Bernard
124k742117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
$endgroup$
add a comment |
$begingroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
edited Mar 27 at 19:42
Bernard
124k742117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
$endgroup$
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
add a comment |
$begingroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
edited Mar 27 at 19:42
Bernard
124k742117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
$endgroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
analysis
edited Mar 27 at 19:42
Bernard
124k742117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
edited Mar 27 at 19:42
Bernard
124k742117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
edited Mar 27 at 19:42
Bernard
124k742117
edited Mar 27 at 19:42
Bernard
124k742117
edited Mar 27 at 19:42
Bernard
124k742117
124k742117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
113
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
add a comment |
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No need for $C^1$:
$$fracf(x)1+lefracf(x)-f(0)1+lefrac) + 1+le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,441421
$endgroup$
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
Your Answer
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%2f3165027%2fc1-lipschitz-function-linear-growth%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
$begingroup$
No need for $C^1$:
$$fracf(x)1+lefracf(x)-f(0)1+lefrac) + 1+le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,441421
$endgroup$
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
$begingroup$
No need for $C^1$:
$$fracf(x)1+lefracf(x)-f(0)1+lefrac) + 1+le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,441421
$endgroup$
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
$begingroup$
No need for $C^1$:
$$fracf(x)1+lefracf(x)-f(0)1+lefrac) + 1+le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,441421
$endgroup$
No need for $C^1$:
$$fracf(x)1+lefracf(x)-f(0)1+lefrac) + 1+le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,441421
answered Mar 27 at 20:47
amsmathamsmath
3,441421
answered Mar 27 at 20:47
amsmathamsmath
3,441421
answered Mar 27 at 20:47
amsmathamsmath
3,441421
3,441421
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
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%2f3165027%2fc1-lipschitz-function-linear-growth%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
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45