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Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?

derivatives differential-geometry orthogonality submanifold tangent-spaces

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asked Jun 21 at 17:32

playdisplaydis

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  Please make the title more informative.
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  – YuiTo Cheng
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3

$begingroup$

Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?

derivatives differential-geometry orthogonality submanifold tangent-spaces

share|cite|improve this question

asked Jun 21 at 17:32

playdisplaydis

4444 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Please make the title more informative.
  $endgroup$
  – YuiTo Cheng
  Jun 22 at 4:07
  

  
  
  
  

add a comment | 

$begingroup$

Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?

derivatives differential-geometry orthogonality submanifold tangent-spaces

share|cite|improve this question

asked Jun 21 at 17:32

playdisplaydis

4444 bronze badges

$endgroup$

share|cite|improve this question

asked Jun 21 at 17:32

playdisplaydis

4444 bronze badges

share|cite|improve this question

asked Jun 21 at 17:32

playdisplaydis

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asked Jun 21 at 17:32

playdisplaydis

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asked Jun 21 at 17:32

playdisplaydis

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2 Answers
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Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).

The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.

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answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

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$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.

Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$

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answered Jun 21 at 17:37

MaxMax

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5

$begingroup$

Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).

The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.

share|cite|improve this answer

answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

196k2424 gold badges153153 silver badges272272 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).

The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.

share|cite|improve this answer

answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

196k2424 gold badges153153 silver badges272272 bronze badges

$endgroup$

add a comment | 

$begingroup$

Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).

The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.

share|cite|improve this answer

answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

196k2424 gold badges153153 silver badges272272 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

196k2424 gold badges153153 silver badges272272 bronze badges

answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

196k2424 gold badges153153 silver badges272272 bronze badges

answered Jun 21 at 17:36

José Carlos SantosJosé Carlos Santos

196k2424 gold badges153153 silver badges272272 bronze badges

4

$begingroup$

$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.

Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$

share|cite|improve this answer

answered Jun 21 at 17:37

MaxMax

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add a comment | 

4

$begingroup$

$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.

Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$

share|cite|improve this answer

answered Jun 21 at 17:37

MaxMax

19.9k11 gold badge1212 silver badges4646 bronze badges

$endgroup$

add a comment | 

$begingroup$

$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.

Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$

share|cite|improve this answer

answered Jun 21 at 17:37

MaxMax

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$endgroup$

share|cite|improve this answer

answered Jun 21 at 17:37

MaxMax

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answered Jun 21 at 17:37

MaxMax

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answered Jun 21 at 17:37

MaxMax

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