Mixing
An example of function that is not constant but its derivative is 0
0
Let $U subset R^{n}$ be and open set. Let $f:U to R^n$ be a differentiable function.
I wonder if there is such a function that is not constant but with $Df=0$ at every point. (The derivative is 0).
calculus real-analysis multivariable-calculus
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
asked Nov 20 '18 at 22:59
Gabi G
36218
add a comment |
0
Let $U subset R^{n}$ be and open set. Let $f:U to R^n$ be a differentiable function.
I wonder if there is such a function that is not constant but with $Df=0$ at every point. (The derivative is 0).
calculus real-analysis multivariable-calculus
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
asked Nov 20 '18 at 22:59
Gabi G
36218
3
Well, if $U$ is not connected you could define a function to be 1 on one of the connected components and 0 on the others. But if $U$ is connected that implies $f$ is constant; and in general, $f$ has to be locally constant.
– Daniel Schepler
Nov 20 '18 at 23:04
add a comment |
0
0
0
Let $U subset R^{n}$ be and open set. Let $f:U to R^n$ be a differentiable function.
I wonder if there is such a function that is not constant but with $Df=0$ at every point. (The derivative is 0).
calculus real-analysis multivariable-calculus
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
asked Nov 20 '18 at 22:59
Gabi G
36218
Let $U subset R^{n}$ be and open set. Let $f:U to R^n$ be a differentiable function.
I wonder if there is such a function that is not constant but with $Df=0$ at every point. (The derivative is 0).
calculus real-analysis multivariable-calculus
calculus real-analysis multivariable-calculus
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
asked Nov 20 '18 at 22:59
Gabi G
36218
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
asked Nov 20 '18 at 22:59
Gabi G
36218
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
edited Nov 20 '18 at 23:44
Rob Arthan
29.1k42866
29.1k42866
asked Nov 20 '18 at 22:59
Gabi G
36218
asked Nov 20 '18 at 22:59
Gabi G
36218
asked Nov 20 '18 at 22:59
Gabi G
36218
36218
3
Well, if $U$ is not connected you could define a function to be 1 on one of the connected components and 0 on the others. But if $U$ is connected that implies $f$ is constant; and in general, $f$ has to be locally constant.
– Daniel Schepler
Nov 20 '18 at 23:04
add a comment |
3
Well, if $U$ is not connected you could define a function to be 1 on one of the connected components and 0 on the others. But if $U$ is connected that implies $f$ is constant; and in general, $f$ has to be locally constant.
– Daniel Schepler
Nov 20 '18 at 23:04
3
3
Well, if $U$ is not connected you could define a function to be 1 on one of the connected components and 0 on the others. But if $U$ is connected that implies $f$ is constant; and in general, $f$ has to be locally constant.
– Daniel Schepler
Nov 20 '18 at 23:04
Well, if $U$ is not connected you could define a function to be 1 on one of the connected components and 0 on the others. But if $U$ is connected that implies $f$ is constant; and in general, $f$ has to be locally constant.
– Daniel Schepler
Nov 20 '18 at 23:04
add a comment |
2 Answers
2
active
oldest
votes
2
Consider $f$ defined on $mathrm{U} = mathbf{R}^2 setminus{(x,0) : xin mathbf{R}}.$ Here $mathrm{U}$ is the plane minus the $x$-axis. Define $f = (0,0)$ on the upper part of $mathrm{U}$ and $f = (-1,0)$ on the lower part.
answered Nov 20 '18 at 23:04
Will M.
2,387314
add a comment |
1
Let $U = (0,1)cup (2,3)$ and $f: U rightarrow mathbb{R}$ and $f(x) = 0$ for $x in (0,1)$ and $f(x) = 1$ for $x in (2,3)$. For open and connected sets $U$, $Df equiv 0$ implies that $f$ is constant (I can provide a proof if you want).
answered Nov 20 '18 at 23:08
Nicolas
415
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
1
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
|
show 1 more 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
});
}
});
draft saved
draft discarded
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%2f3007022%2fan-example-of-function-that-is-not-constant-but-its-derivative-is-0%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
2
Consider $f$ defined on $mathrm{U} = mathbf{R}^2 setminus{(x,0) : xin mathbf{R}}.$ Here $mathrm{U}$ is the plane minus the $x$-axis. Define $f = (0,0)$ on the upper part of $mathrm{U}$ and $f = (-1,0)$ on the lower part.
answered Nov 20 '18 at 23:04
Will M.
2,387314
add a comment |
2
Consider $f$ defined on $mathrm{U} = mathbf{R}^2 setminus{(x,0) : xin mathbf{R}}.$ Here $mathrm{U}$ is the plane minus the $x$-axis. Define $f = (0,0)$ on the upper part of $mathrm{U}$ and $f = (-1,0)$ on the lower part.
answered Nov 20 '18 at 23:04
Will M.
2,387314
add a comment |
2
2
2
Consider $f$ defined on $mathrm{U} = mathbf{R}^2 setminus{(x,0) : xin mathbf{R}}.$ Here $mathrm{U}$ is the plane minus the $x$-axis. Define $f = (0,0)$ on the upper part of $mathrm{U}$ and $f = (-1,0)$ on the lower part.
answered Nov 20 '18 at 23:04
Will M.
2,387314
Consider $f$ defined on $mathrm{U} = mathbf{R}^2 setminus{(x,0) : xin mathbf{R}}.$ Here $mathrm{U}$ is the plane minus the $x$-axis. Define $f = (0,0)$ on the upper part of $mathrm{U}$ and $f = (-1,0)$ on the lower part.
answered Nov 20 '18 at 23:04
Will M.
2,387314
answered Nov 20 '18 at 23:04
Will M.
2,387314
answered Nov 20 '18 at 23:04
Will M.
2,387314
answered Nov 20 '18 at 23:04
Will M.
2,387314
2,387314
add a comment |
add a comment |
1
Let $U = (0,1)cup (2,3)$ and $f: U rightarrow mathbb{R}$ and $f(x) = 0$ for $x in (0,1)$ and $f(x) = 1$ for $x in (2,3)$. For open and connected sets $U$, $Df equiv 0$ implies that $f$ is constant (I can provide a proof if you want).
answered Nov 20 '18 at 23:08
Nicolas
415
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
1
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
|
show 1 more comment
1
Let $U = (0,1)cup (2,3)$ and $f: U rightarrow mathbb{R}$ and $f(x) = 0$ for $x in (0,1)$ and $f(x) = 1$ for $x in (2,3)$. For open and connected sets $U$, $Df equiv 0$ implies that $f$ is constant (I can provide a proof if you want).
answered Nov 20 '18 at 23:08
Nicolas
415
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
1
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
|
show 1 more comment
1
1
1
Let $U = (0,1)cup (2,3)$ and $f: U rightarrow mathbb{R}$ and $f(x) = 0$ for $x in (0,1)$ and $f(x) = 1$ for $x in (2,3)$. For open and connected sets $U$, $Df equiv 0$ implies that $f$ is constant (I can provide a proof if you want).
answered Nov 20 '18 at 23:08
Nicolas
415
Let $U = (0,1)cup (2,3)$ and $f: U rightarrow mathbb{R}$ and $f(x) = 0$ for $x in (0,1)$ and $f(x) = 1$ for $x in (2,3)$. For open and connected sets $U$, $Df equiv 0$ implies that $f$ is constant (I can provide a proof if you want).
answered Nov 20 '18 at 23:08
Nicolas
415
answered Nov 20 '18 at 23:08
Nicolas
415
answered Nov 20 '18 at 23:08
Nicolas
415
answered Nov 20 '18 at 23:08
Nicolas
415
415
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
1
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
|
show 1 more comment
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
1
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
Thanks Nicolas!
– Gabi G
Nov 20 '18 at 23:24
1
1
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
About the case where $U$ is connected and open, can I prove it with the multi variable version of the Newton Leibniz fundamental thorem?
– Gabi G
Nov 20 '18 at 23:27
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
If $U$ is connected and open and $x, y in U$, you can compose a diferentiable path from $x$ to $y$ with each of the coordinate projections of $Bbb{R}^n$ onto $Bbb{R}$ and then deduce the result from the case when $n = 1$ (where it follows from the mean value theorem).
– Rob Arthan
Nov 20 '18 at 23:51
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
Nov 21 '18 at 0:04
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
Sorry for the late reply. I proved it by fixing $x_0 in U$ and considering the set $D := {x in U: f(x) = f(x_0)}$. Since $U$ is connected, it suffices to show that $D$ is open, closed and non-empty. Since $x_0 in D$, $D$ is non-empty, and since $D = f^{-1}(x^0)$ and $f$ is differentiable (and thus continuous), $D$ is closed. Show $D$ is open by considering $x in D$, then there exists $epsilon > 0$ such that $B_epsilon(x) subset U$. Use $Df equiv 0$ and consider $phi(t) = f(xt+(1-t)y)$ for $t in [0,1], y in B_epsilon(x)$ to show $B_epsilon(x) subset D$.
– Nicolas
Dec 4 '18 at 23:29
|
show 1 more comment
draft saved
draft discarded
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
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%2f3007022%2fan-example-of-function-that-is-not-constant-but-its-derivative-is-0%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
3
Well, if $U$ is not connected you could define a function to be 1 on one of the connected components and 0 on the others. But if $U$ is connected that implies $f$ is constant; and in general, $f$ has to be locally constant.
– Daniel Schepler
Nov 20 '18 at 23:04