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Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is a strictly quasiconcave and continuous function and $f(x) < f(y)$ for $x, y \in \mathbb{R}^n$ and $f$ is increasing such that $x_1 << x_2 $ ($x_2$ is elementwise greater than $x_1$) implies $f(x_1) < f(x_2) $. Consider a convex combination of $x$ and $y$, $\lambda x + (1- \lambda) y$ for some $\lambda \in (0,1)$ and the corresponding value of $f$ as $\overline{f}_\lambda = f(\lambda x +(1-\lambda)y)$.

For a given $\lambda$, is it possible to find $\lambda' \neq \lambda $ such that $\overline{f}_\lambda = \overline{f}_{\lambda'}$? By the continuity of $f$ it is possible to find a value of $\lambda$ that satisfies $\overline{f}_\lambda = \overline{f}$ for some $\overline{f} \in (f(x), f(y))$, but I wonder if such decomposition is unique if $f$ is strictly quasiconcave. Note that for $f : \mathbb{R} \to \mathbb{R}$ uniqueness can be relatively easily shown by a similar argument as in Convex function has unique zero.

(edit: added an additional assumption that $f$ is increasing).

Max
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  • What is the definition of $x<<y$?? – daw Sep 25 '20 at 11:01
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    I meant $x_i < y_i$ for all $i = 1,...,n$ where $x_i, y_i$ are $i$th elements of $x,y \in \mathbb{R}^n$ – Max Sep 25 '20 at 11:01
  • It can be shown that a strictly quasiconcave function has a unique maximizer $\mathbf{x}^$, and also that the function is strictly increasing on the line segment from any point $\mathbf{y}\neq\mathbf{x}^$ to $\mathbf{x}^*$. – iarbel84 Sep 25 '20 at 13:49

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No. Take $f(x)=-x^2$, $x=-1$, $y=2$. Then $f$ is not injective on $[x,y]$.

daw
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  • Thanks. I wonder what happens if $f$ is increasing? I modified the question accordingly. – Max Sep 25 '20 at 10:21
  • Sorry, can you elaborate? I don't see it trivial especially given that the domain is multidimensional. – Max Sep 25 '20 at 10:39
  • I thought about $f:R\to R$... – daw Sep 25 '20 at 11:10
  • Even $f: R \to R$ case should be helpful. Thanks! – Max Sep 25 '20 at 11:15
  • I guess for $f: R \to R$ case the method is quite similar as in https://math.stackexchange.com/questions/2616257/convex-function-has-unique-zero. – Max Sep 25 '20 at 11:19