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Apr 27, 2011 at 22:29 comment added J.C. Ottem How does that contradict what I wrote?
Apr 27, 2011 at 21:40 comment added James Rohal I agree with some of what you said, however there are choices of $f$ where each connected component of $f \neq 0$ contain one and only one critical point of $g$ (all of which are local maxima). For example, $f = 4-10x^2$. The following figure has three connected components where $f$ and $g$ are the red and blue curves, respectively: i.imgur.com/xRhLV.png
Apr 27, 2011 at 20:41 comment added J.C. Ottem This might be better suited for math.stackexchange.com by the way..
Apr 27, 2011 at 20:40 comment added J.C. Ottem Yes, $g$ is bounded on the whole real line and eventually decreasing as $x\to\pm \infty$, hence $g$ has a local minimum on the components with large $\|x\|$. Now note that any local maximum point of of $g$ must lie in the interior of $\{x:f\neq\0\}$ since $g\ge 0$ everywhere. Conversely, every connected component of $\{x:f\neq\0\}$ has a local maximum, since $f$ cannot be identically 0 on the entire component.
Apr 27, 2011 at 19:55 history asked James Rohal CC BY-SA 3.0