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Prove that $$f(x,y)=\exp(-x^2-y^2)$$ is a strictly quasi-concave function, that is,$\forall (x_i,y_i)\in \Re^2, (x_i,y_i)\neq (x_j,y_j), \lambda\in(0,1)$, $$f(\lambda x_1+(1-\lambda) x_2,\lambda y_1+(1-\lambda) y_2)>\min[f(x_1,y_1),f(x_2,y_2)].$$

I'm trying to use triangle inequality but so far without success. Perhaps some insight will be useful.

bluemaster
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2 Answers2

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The logarithm is a strictly increasing monotonic function, so if $a>b$ then $\log a>\log b$ and vice-versa. Take logarithms then $$ \log (e^{-x^2-y^2})=-2(\log x+\log y) $$ Using just one dimension, suppose that the maximum is at $x_1$ then $$ \log \exp -(\lambda x_1+(1-\lambda)x_2)^2= -2(\lambda x_1+(1-\lambda)x_2)\\ >-2x_1 $$ by the usual inequalities.

Suzu Hirose
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Another approach is to observe a function $f$ is (strictly) quasi-concave if and only if all the upper contours set $\{x: f(x)\ge \alpha\}$ are (strictly) convex.

wxydx00
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