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Hello. Can somebody help me with the following question that I have thought over for quite some time, to no avail?

Let $f$ be a smooth function (class $\mathrm{C}^{\infty}$), $f:\mathbb{R}^n \longrightarrow \mathbb{R}$ and suppose that $f$ is a positive convex function and we define $$ \varphi: \mathbb{R}^n \longrightarrow \mathbb{R}^n, \varphi(X) = \frac{\operatorname{grad}(f)(X)}{f(X)} $$

My question is this:

Is it true that the image of $\varphi$ is a convex set?

This is not my research area so I will appreciate any help or comment.

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Where does this question come from? –  Igor Rivin Oct 27 '11 at 8:12
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Now that a counterexample has been posted: might it still be true if $f$ is logarithmically convex? Equivalently (with $g=\log f\phantom.$): if $g: {\bf R}^n \rightarrow {\bf R}$ is convex, is the image of ${\rm grad}(g)$ a convex subset of ${\bf R}^n$? –  Noam D. Elkies Oct 27 '11 at 16:11
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2 Answers

No. Consider $f(x,y)=e^x+y^2$, then $\varphi(x,y)=(e^x,2y)/(e^x+y^2)$. The image of $\varphi$ has only one point $(1,0)$ on the axis $y=0$. The points $a:=\varphi(0,1)=(\frac12,1)$ and $b:=\varphi(0,-1)=(\frac12,-1)$ belong to the image of $\varphi$ but their midpoint $\frac{a+b}2 = (\frac12,0)$ does not.

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Greetings everyone.

I have a similar problem. In my case, the function $f$ is $$ f:\mathbb{R}^n \longrightarrow \mathbb{R}, f(X)=\sum_{i=1}^{m}a_i^2 \operatorname{e}^{2\langle X, \alpha_i \rangle} $$ where $a_1, \ldots, a_m$ are not null, $\alpha_1, \dots, \alpha_m$ are any vectors in $\mathbb{R}^n$ and $\langle \cdot , \cdot \rangle$ is the usual inner product of $\mathbb{R}^n$.

I know that my problem is related to the Atiyah-Guillemin-Sternberg Convexity theorem, but I don't know how to prove it in my case by using elementary methods. I would really appreciate any comment.

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Looks like this should be a new question, not an answer! Is the image of the logarithmic gradient here just the convex hull of $\lbrace 2 \alpha_i \rbrace$? If so then it can probably be proved using something like the Brouwer fixed-point theorem; I don't know how elementary that would be for your purposes. –  Noam D. Elkies Oct 28 '11 at 16:30
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