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Let $f:{\bf R}^n\to {\bf R}$ ($n\geq 2$) be a $C^1$ function. Is it true that $$\sup_{x\in {\bf R}^n}f(x)=\sup_{x\in {\bf R}^n}f(x+\nabla f(x))\hskip 3pt ?$$

Biagio Ricceri
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    Notice that this cannot be reduced to the almost trivial 1-dimensional case by looking at the line through $x$ and $x+\nabla f(x)$. – Yaakov Baruch Nov 29 '19 at 08:18
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    Note that $f\left(x+\nabla f(x)\right) \leq \sup_{x\in \mathbb{R}^n}f(x)$ and hence $\mathrm{sup}{x\in \mathbb{R}^n}f\left(x+\nabla f(x)\right) \leq \mathrm{sup}{x\in \mathbb{R}^n}f(x)$. Now assume that there exists $x_0$ such that $f(x_0) = \mathrm{sup}{x\in \mathbb{R}^n}f(x)$, then $\nabla f(x_0) = 0$ and hence $f(x_0+\nabla f(x_0)) = f(x_0) = \mathrm{sup}{x\in \mathbb{R}^n}f(x)$. So you are interested in the case when $f$ does not attain its global maximum. Am I right or I am lost? – Slup Nov 29 '19 at 15:58
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    @Slup: This is an interesting observation. You prove that the statement holds if the sup is a max, but as you pointed out yourself, there are other scenarios. For example, your argument also works for $f(x-\nabla f)$, but in this case, the claim is false ($f=|x|^2/2$). – Christian Remling Nov 29 '19 at 16:36
  • I haven't checked the details, and I doubt it works as is, but a counterexample could perhaps be found along the lines of $z=(1+\sin\big(\sqrt{x^2+y^2}-2\arctan{(\frac{y}{x}})\big)(e^{x^2+y^2}-1)$? Here is a graph (up to stretches): https://tinyurl.com/sq4c4wc . – Yaakov Baruch Dec 01 '19 at 13:38
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    A trivial remark: no such function can be bounded from below; moreover, $\liminf (f(x) / |x|^2) \leqslant -\tfrac{1}{2}$. Indeed: otherwise, for every $p \in \mathbb{R}^n$, $f(x) + (x - p)^2/2$ would have a global minimum at some $q \in \mathbb{R}^n$, and so $0 = \nabla f(q) + (q - p)$, that is, $p = q + \nabla f(q)$. In other words, the range of $x + \nabla f(x)$ is all of $\mathbb{R}^n$. – Mateusz Kwaśnicki Dec 02 '19 at 07:26
  • @MateuszKwaśnicki : It seems to me that you proved that the equality in question ** does** hold if $\liminf(f(x)/|x|^2)>-1/2$. Perhaps, instead of "no such function can be bounded from below", you meant "there is no counterexample with a function bounded from below". – Iosif Pinelis Dec 03 '19 at 05:05
  • @IosifPinelis: Right, I think I meant "no counterexample" . (Alternatively, one could think this is a responce to Yaakov Baruch's comment, but this was not the case.) – Mateusz Kwaśnicki Dec 03 '19 at 05:45
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    In any case my suggestion does not need to be bounded below. It can be replaced with $f(x,y)=\sin\big(\sqrt{x^2+y^2}-2\arctan(\frac{y}{x})\big)(e^{x^2+y^2}-1)$. https://tinyurl.com/qq3o8kb . I really don't expect this to work, but the (naive) idea is that any point on the crest or close to it be thrown off the tangent of the crest and down the very steep slope. (And one has to check where all the other points go too, but at least the region where $f$ is below some large enough finite bound is connected, as it would need to be.) – Yaakov Baruch Dec 03 '19 at 10:44
  • @MateuszKwaśnicki Doesn't your argument also show that any counterexample should also satisfy $\limsup(f(x)/|x|^2)\geq -\frac12 $ ? – username Apr 20 '21 at 15:06
  • I think this only gives $\liminf$, but, honestly, after a year and a half, I hardly even remember the question, let alone my comment. :-) – Mateusz Kwaśnicki Apr 20 '21 at 18:14

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