To prove that the vector $\nabla{f}(x_0)$ is orthogonal to the tangent vector to "an arbitrary smooth curve" passing through $x_0$ on the level set determined by $f(x)=f(x_0)$ the following proof is outlined in Chong and Zak :

It takes a special curve $\gamma$ lying in the level set and passing through $x_0$ and parameterized by a continuously differentiable function $g:\Bbb R \to \Bbb R^n$ such that $g(t_0) =x_0$ and $Dg(t_0)=v \neq 0$, so that $v$ is a tangent vector to $\gamma$ at $x_0$. Then it proves that $\nabla{f}(x_0) ^T v =0$. This proof is not correct. It does not consider any arbitrary smooth curve. The condition that $v$ is tangent to $\gamma$ at $x_0$ may not be true always. Am I correct ?

| cite | improve this question | | | | |

The curve must satisfy $\gamma(0) = x_0$, $f(\gamma(t)) = f(x_0)$ for $t \in [0,\delta)$ and $\gamma'(0) = v$. Then $\langle \nabla f(x_0), v \rangle = 0$ from the chain rule.

Unless $\nabla f(x_0) = 0$, it cannot be true for arbitrary $v$.

If $v \bot \nabla f(x_0)$, and $\nabla f(x_0) \neq 0$, then we can use the implicit function theorem to construct a suitable curve.

If $v = 0$, it is trivial, just let $\gamma(t) = x_0$.

If $v \neq 0$, consider the function $\phi(\alpha,\beta) = f(\alpha v + \beta \nabla f(x_0)+x_0) - f(x_0)$. Note that $\frac{\partial \phi(0,0)}{\partial \beta} = \| \nabla f(x_0)\|^2 \neq 0$, and use the implicit function theorem with $\phi(\alpha,\beta) = 0$ to find a differentiable $\xi$ such that $\phi(\alpha, \xi(\alpha)) = 0$, $\xi(0) = 0$, and $\frac{\partial \xi(0)}{\partial \alpha}= - \frac{\frac{\partial \phi(0,0)}{\partial \alpha}}{\frac{\partial \phi(0,0)}{\partial \beta}} = 0$.

Then let $\gamma(t) = t v + \xi(t) \nabla f(x_0)+ x_0$. Note that $\gamma(0) = x_0$, $f(\gamma(t)) = f(x_0)$, and $\gamma'(0) = v$.

| cite | improve this answer | | | | |
  • $\begingroup$ The inputs $f$ takes is quite confusing, isnt $v$ a vector?? $\endgroup$ – Vishesh Oct 14 '13 at 6:22
  • $\begingroup$ I meant while defining $\phi(\alpha,\beta)$. $\endgroup$ – Vishesh Oct 14 '13 at 6:24
  • $\begingroup$ Yes, $v$ is a vector, $\alpha, \beta$ are scalars. $\endgroup$ – copper.hat Oct 14 '13 at 6:37
  • $\begingroup$ But $x_0$ is a point, right??I am still confused.Sorry about this. So $f$ takes both vector and points?? $\endgroup$ – Vishesh Oct 14 '13 at 7:21
  • $\begingroup$ Draw a picture. I am just using the notation you used in the question. $f$ is defined on $\mathbb{R}^n$. we have $v,x_0, \nabla f(x_0) \in \mathbb{R}^n$, $\alpha,\beta \in \mathbb{R}$, so $\alpha v + \beta \nabla f(x_0)+x_0 \in \mathbb{R}^n$. $\endgroup$ – copper.hat Oct 14 '13 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.