if derivative vanishes in a path connected set, then $f$ is constant on that set?
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Let $f: mathbb{R}^d to mathbb{R}^m $ be a function sucht that $Df(a) = 0 $ for all $a in U subseteq mathbb{R}^d$. Also, suppose $U$ is open and path-connected.
Question: IS $f$ constant on $U$ ??
calculus real-analysis
add a comment |
up vote
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Let $f: mathbb{R}^d to mathbb{R}^m $ be a function sucht that $Df(a) = 0 $ for all $a in U subseteq mathbb{R}^d$. Also, suppose $U$ is open and path-connected.
Question: IS $f$ constant on $U$ ??
calculus real-analysis
I think you don't need $U$ to be open
– dani_s
Mar 6 '14 at 10:13
@dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:mathbb R^2tomathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake.
– user127096
Mar 8 '14 at 4:35
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f: mathbb{R}^d to mathbb{R}^m $ be a function sucht that $Df(a) = 0 $ for all $a in U subseteq mathbb{R}^d$. Also, suppose $U$ is open and path-connected.
Question: IS $f$ constant on $U$ ??
calculus real-analysis
Let $f: mathbb{R}^d to mathbb{R}^m $ be a function sucht that $Df(a) = 0 $ for all $a in U subseteq mathbb{R}^d$. Also, suppose $U$ is open and path-connected.
Question: IS $f$ constant on $U$ ??
calculus real-analysis
calculus real-analysis
asked Mar 3 '14 at 10:44
user130448
I think you don't need $U$ to be open
– dani_s
Mar 6 '14 at 10:13
@dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:mathbb R^2tomathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake.
– user127096
Mar 8 '14 at 4:35
add a comment |
I think you don't need $U$ to be open
– dani_s
Mar 6 '14 at 10:13
@dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:mathbb R^2tomathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake.
– user127096
Mar 8 '14 at 4:35
I think you don't need $U$ to be open
– dani_s
Mar 6 '14 at 10:13
I think you don't need $U$ to be open
– dani_s
Mar 6 '14 at 10:13
@dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:mathbb R^2tomathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake.
– user127096
Mar 8 '14 at 4:35
@dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:mathbb R^2tomathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake.
– user127096
Mar 8 '14 at 4:35
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
Here is little more precise answer: Let $a, b in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $gamma :[0, 1] to U$ from $a$ to $b$. Then
begin{align*}
f(b) -f(a) &= f(gamma(1))- f(gamma (0)) = int_0^1 frac{operatorname{d}}{operatorname{d}t} f(gamma(t)) operatorname{d}t = int_0^1 Df_{gamma(t)} (dot{gamma}(t)) operatorname{d}t = 0,
end{align*}
where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.
answered Mar 6 '14 at 10:08
Dragoslav
23816
add a comment |
up vote
2
down vote
Although for the function of several variables there is no Mean Value Theorem, there is another similar result.
Suppose $f$ maps a convex open set $E subset mathbb{R}^d$ into
$mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real
number $M$ s.t. $left| {f'left( x right)} right| leqslant M$
for every $x in E$. Then $left| {fleft( a right) - fleft( bright)} right|leqslant Mleft| {a - b} right|$ for all $a,b in E$.
In this case, we can put $M=0$ since $Df(a) = 0$.
answered Mar 6 '14 at 9:52
Junefi
1103
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
add a comment |
up vote
1
down vote
Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Here is little more precise answer: Let $a, b in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $gamma :[0, 1] to U$ from $a$ to $b$. Then
begin{align*}
f(b) -f(a) &= f(gamma(1))- f(gamma (0)) = int_0^1 frac{operatorname{d}}{operatorname{d}t} f(gamma(t)) operatorname{d}t = int_0^1 Df_{gamma(t)} (dot{gamma}(t)) operatorname{d}t = 0,
end{align*}
where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.
answered Mar 6 '14 at 10:08
Dragoslav
23816
add a comment |
up vote
3
down vote
Here is little more precise answer: Let $a, b in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $gamma :[0, 1] to U$ from $a$ to $b$. Then
begin{align*}
f(b) -f(a) &= f(gamma(1))- f(gamma (0)) = int_0^1 frac{operatorname{d}}{operatorname{d}t} f(gamma(t)) operatorname{d}t = int_0^1 Df_{gamma(t)} (dot{gamma}(t)) operatorname{d}t = 0,
end{align*}
where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.
answered Mar 6 '14 at 10:08
Dragoslav
23816
add a comment |
up vote
3
down vote
up vote
3
down vote
Here is little more precise answer: Let $a, b in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $gamma :[0, 1] to U$ from $a$ to $b$. Then
begin{align*}
f(b) -f(a) &= f(gamma(1))- f(gamma (0)) = int_0^1 frac{operatorname{d}}{operatorname{d}t} f(gamma(t)) operatorname{d}t = int_0^1 Df_{gamma(t)} (dot{gamma}(t)) operatorname{d}t = 0,
end{align*}
where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.
answered Mar 6 '14 at 10:08
Dragoslav
23816
Here is little more precise answer: Let $a, b in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $gamma :[0, 1] to U$ from $a$ to $b$. Then
begin{align*}
f(b) -f(a) &= f(gamma(1))- f(gamma (0)) = int_0^1 frac{operatorname{d}}{operatorname{d}t} f(gamma(t)) operatorname{d}t = int_0^1 Df_{gamma(t)} (dot{gamma}(t)) operatorname{d}t = 0,
end{align*}
where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.
answered Mar 6 '14 at 10:08
Dragoslav
23816
answered Mar 6 '14 at 10:08
Dragoslav
23816
answered Mar 6 '14 at 10:08
Dragoslav
23816
answered Mar 6 '14 at 10:08
Dragoslav
23816
23816
add a comment |
add a comment |
up vote
2
down vote
Although for the function of several variables there is no Mean Value Theorem, there is another similar result.
Suppose $f$ maps a convex open set $E subset mathbb{R}^d$ into
$mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real
number $M$ s.t. $left| {f'left( x right)} right| leqslant M$
for every $x in E$. Then $left| {fleft( a right) - fleft( bright)} right|leqslant Mleft| {a - b} right|$ for all $a,b in E$.
In this case, we can put $M=0$ since $Df(a) = 0$.
answered Mar 6 '14 at 9:52
Junefi
1103
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
add a comment |
up vote
2
down vote
Although for the function of several variables there is no Mean Value Theorem, there is another similar result.
Suppose $f$ maps a convex open set $E subset mathbb{R}^d$ into
$mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real
number $M$ s.t. $left| {f'left( x right)} right| leqslant M$
for every $x in E$. Then $left| {fleft( a right) - fleft( bright)} right|leqslant Mleft| {a - b} right|$ for all $a,b in E$.
In this case, we can put $M=0$ since $Df(a) = 0$.
answered Mar 6 '14 at 9:52
Junefi
1103
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
add a comment |
up vote
2
down vote
up vote
2
down vote
Although for the function of several variables there is no Mean Value Theorem, there is another similar result.
Suppose $f$ maps a convex open set $E subset mathbb{R}^d$ into
$mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real
number $M$ s.t. $left| {f'left( x right)} right| leqslant M$
for every $x in E$. Then $left| {fleft( a right) - fleft( bright)} right|leqslant Mleft| {a - b} right|$ for all $a,b in E$.
In this case, we can put $M=0$ since $Df(a) = 0$.
answered Mar 6 '14 at 9:52
Junefi
1103
Although for the function of several variables there is no Mean Value Theorem, there is another similar result.
Suppose $f$ maps a convex open set $E subset mathbb{R}^d$ into
$mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real
number $M$ s.t. $left| {f'left( x right)} right| leqslant M$
for every $x in E$. Then $left| {fleft( a right) - fleft( bright)} right|leqslant Mleft| {a - b} right|$ for all $a,b in E$.
In this case, we can put $M=0$ since $Df(a) = 0$.
answered Mar 6 '14 at 9:52
Junefi
1103
answered Mar 6 '14 at 9:52
Junefi
1103
answered Mar 6 '14 at 9:52
Junefi
1103
answered Mar 6 '14 at 9:52
Junefi
1103
1103
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
add a comment |
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
But here the domain need not be convex so you can not apply mean value theorem.
– XYZABC
Dec 2 '17 at 3:32
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
@XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant.
– Andrés E. Caicedo
Nov 27 at 12:38
add a comment |
up vote
1
down vote
Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
add a comment |
up vote
1
down vote
Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
answered Mar 3 '14 at 10:50
Siminore
30.2k23368
30.2k23368
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
add a comment |
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
Can you explain a little bit more your answer?
– ILoveMath
Mar 6 '14 at 9:26
add a comment |
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I think you don't need $U$ to be open
– dani_s
Mar 6 '14 at 10:13
@dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:mathbb R^2tomathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake.
– user127096
Mar 8 '14 at 4:35