An example of function that is not constant but its derivative is 0
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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 yesterday
Rob Arthan
28.4k42865
asked yesterday
Gabi G
1687
add a comment |
up vote
0
down vote
favorite
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 yesterday
Rob Arthan
28.4k42865
asked yesterday
Gabi G
1687
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
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 yesterday
Rob Arthan
28.4k42865
asked yesterday
Gabi G
1687
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 yesterday
Rob Arthan
28.4k42865
asked yesterday
Gabi G
1687
edited yesterday
Rob Arthan
28.4k42865
asked yesterday
Gabi G
1687
edited yesterday
Rob Arthan
28.4k42865
edited yesterday
Rob Arthan
28.4k42865
edited yesterday
Rob Arthan
28.4k42865
28.4k42865
asked yesterday
Gabi G
1687
asked yesterday
Gabi G
1687
asked yesterday
Gabi G
1687
1687
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
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 yesterday
Will M.
1,949213
add a comment |
up vote
1
down vote
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 yesterday
Nicolas
315
Thanks Nicolas!
– Gabi G
yesterday
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
yesterday
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
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 yesterday
Will M.
1,949213
add a comment |
up vote
2
down vote
accepted
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 yesterday
Will M.
1,949213
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 yesterday
Will M.
1,949213
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 yesterday
Will M.
1,949213
answered yesterday
Will M.
1,949213
answered yesterday
Will M.
1,949213
answered yesterday
Will M.
1,949213
1,949213
add a comment |
add a comment |
up vote
1
down vote
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 yesterday
Nicolas
315
Thanks Nicolas!
– Gabi G
yesterday
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
yesterday
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
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
add a comment |
up vote
1
down vote
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 yesterday
Nicolas
315
Thanks Nicolas!
– Gabi G
yesterday
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
yesterday
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
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
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 yesterday
Nicolas
315
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 yesterday
Nicolas
315
answered yesterday
Nicolas
315
answered yesterday
Nicolas
315
answered yesterday
Nicolas
315
315
Thanks Nicolas!
– Gabi G
yesterday
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
yesterday
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
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
add a comment |
Thanks Nicolas!
– Gabi G
yesterday
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
yesterday
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
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
Thanks Nicolas!
– Gabi G
yesterday
Thanks Nicolas!
– Gabi G
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
Could you explain what are the coordinate projections? And how can you use this theorem from a multi dimensional domain?
– Gabi G
yesterday
add a comment |
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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
yesterday