1
$\begingroup$

In $\mathbb{R}$, if there is only one critical point, and it is a local minimum, it is also the global minimum (because on every path to a different global minimum, there would have to be another critical point).

In $\mathbb{R}^n$ with $n > 1$, this does not hold, e.g., for $f(x,y) = x^2 + y^2(1-x)^3$, $(0,0)$ is the only critical point and a local minimum, but obviously not the global minimum.

Now let's consider any path from this local minimum to the global minimum (or any point $(x,y)$ s.t. $f(x,y) < f(0,0)$ for that matter).

My question is: Is it correct that on such a path, every partial derivative has to be zero at least once (but not necessarily at the same point, therefore no further critical point)?

$\endgroup$
3
  • 1
    $\begingroup$ There is no global minimum! Suppose $g(0) = (0,0)$ and $g(t_0) = (x_0,y_0)$, with $f(g(t_0)) < 0$. How would you deduce from the fact that $f\circ g$ has a critical point in $0<t<t_0$ that $\partial f/\partial x$ must vanish on the path $g(t)$, $0<t<t_0$? $\endgroup$ Commented yesterday
  • $\begingroup$ Well that's basically the question, I'm not saying it's right, that's just the intuition I had. In 1D thats how it works, I tried to find a way to extend that intuition to $n$ dimensions. Of course on the path there is not going to be a critical point, but I thought there might have to be one if we only look at the "slice" wrt. to each input, just not at the same point. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ Yes, there is a critical point of the restriction of $f$ to any such path, but that just tells you (by the chain rule) that $\nabla f$ is orthogonal to the tangent vector of the curve at some point. $\endgroup$ Commented 3 hours ago

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.