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Is $\displaystyle f(x_1,x_2) = x_1 - \frac{1}{x_2}$ a convex function? What if we restrict the values of this function to the positive reals?

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user90593
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1 Answers1

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It is convex for $x_2 < 0$, not for $x_2 > 0$ (there it is concave).

Robert Israel
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