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Apr 28, 2017 at 18:02 comment added Pietro Majer @Neil: yes, as soon $n\ge2$. It's quite easy to construct a smooth function with any prescribed finite set $S$ of critical points, together with prescribed levels and Morse indices, plus, possibly, a set $S'$ of finitely many other critical points. To kill these additional critical points include them into an unbounded simple curve $\Gamma\sim [0,\infty)$, disjoint from $S$; consider a diffeo $\phi:\mathbb{R}^n\to\mathbb{R}^n\setminus\Gamma$ which is the identity outside a suitable nbd of $\Gamma$. Then $f\circ\phi$ has exactly $S$ as a critical set, and coincides with $f$ on a nbd of $S$.
Apr 28, 2017 at 16:47 comment added Neil Strickland Some examples include $xy$ (with a single saddle) and $x\,\sin(y)$ (with infinitely many isolated saddles) and $(e^x\cos(y)-1)^2+(e^x\sin(y))^2$ (with infinitely many isolated local minima) and $x^2$ (with a curve of weak local minima extending to infinity). I'm not sure if it is possible to have both an isolated local minimum and an isolated local maximum. If you have a closed curve of weak local minima then inside there must be a local maximum (by compactness) but the value must be zero at both the minima and maxima so it is zero everywhere inside the curve.
Apr 28, 2017 at 16:44 history edited Danilo Socovan CC BY-SA 3.0
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Apr 28, 2017 at 16:35 comment added Piero D'Ancona Like $(g(x)-c)^3$? I'm not sure I understand your last remark in view of your question
Apr 28, 2017 at 16:33 comment added Piero D'Ancona @jean sure contains :)
Apr 28, 2017 at 16:32 review Reopen votes
Apr 29, 2017 at 4:16
Apr 28, 2017 at 16:28 history edited Danilo Socovan CC BY-SA 3.0
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Apr 28, 2017 at 15:38 history edited Danilo Socovan
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Apr 28, 2017 at 15:37 review Reopen votes
Apr 28, 2017 at 16:06
Apr 28, 2017 at 15:18 history edited Danilo Socovan CC BY-SA 3.0
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Apr 28, 2017 at 15:15 comment added Danilo Socovan @PieroD'Ancona thank you very much! The one you said, $f(x) = (g(x)-c)^2$ is useful but only let me represent functions $f:\mathbb{R}^n \to \mathbb{R}^+$. I would need an alternative which allows me to also represent $f:\mathbb{R}^n \to \mathbb{R}$
Apr 28, 2017 at 15:13 history closed Stefan Kohl
Michael Renardy
Denis Serre
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Apr 28, 2017 at 14:47 comment added Jean Duchon @PieroD'Ancona the zero set of $f$ contains the zero set of $\nabla f$
Apr 28, 2017 at 14:45 comment added Danilo Socovan @CarloBeenakker , I mean that I want to find a function $f(\boldsymbol{x})$ which is 0 for each $\boldsymbol{x}$ where $\nabla f(x)$ = 0. i.e: if $f(\boldsymbol{x})=\boldsymbol{a}^T\boldsymbol{x}$ then $\nabla f(\boldsymbol{x})=\boldsymbol{a}$. As a consequence, $f(\boldsymbol{x})=\boldsymbol{a}^T\boldsymbol{x} = \nabla f(\boldsymbol{x})^T\boldsymbol{x} = 0$ if $\nabla f(\boldsymbol{x})=0$.
Apr 28, 2017 at 14:38 comment added Piero D'Ancona I think he means that the zero set of $f$ is contained in the zero set of $\nabla f$
Apr 28, 2017 at 14:35 comment added Carlo Beenakker apologies, I don't understand the question:$ \nabla f=0$ means $f={\rm constant}$, what else is there to say?
Apr 28, 2017 at 14:35 comment added Piero D'Ancona Like $f(x)=(g(x)-c)^2$, with $\nabla g\neq0$ everywhere and $c$ being in the range of $g$?
Apr 28, 2017 at 14:26 comment added Danilo Socovan Sorry for the notation. I tried to fix it in the question. I mean $\boldsymbol{x}$ is an $\mathbb{R}^n$ input vector, and $\nabla f(x)$ the gradient of $f(\boldsymbol{x})$ w.r.t vector $\boldsymbol{x}$. If it is not clear yet, just let me know.
Apr 28, 2017 at 14:23 history edited Danilo Socovan CC BY-SA 3.0
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Apr 28, 2017 at 13:21 review Close votes
Apr 28, 2017 at 15:08
Apr 28, 2017 at 13:13 comment added Carlo Beenakker what do you mean by $\nabla_w$ ?
Apr 28, 2017 at 12:56 review First posts
Apr 28, 2017 at 13:02
Apr 28, 2017 at 12:48 history asked Danilo Socovan CC BY-SA 3.0