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Let $\exp : \mathbb{R}^2 \to \mathbb{R}^2$ be the function given by $\exp(x_1,x_2) := (e^{x_1}, e^{x_2})$. Suppose that $f : \mathbb{R}^2 \to \mathbb{R}$ is a smooth (i.e. $\mathcal{C}^2$) convex function.

Is $f\circ \exp : \mathbb{R}^2 \to \mathbb{R}$ convex, i.e. for $0\le \lambda \le 1$, do we have $$\lambda f(e^{x_1}, e^{x_2}) + (1 - \lambda) f(e^{y_1}, e^{y_2}) \ge f(e^{\lambda x_1 + (1 - \lambda)y_1}, e^{\lambda x_2 + (1 - \lambda)y_2})$$ for all $(x_1,x_2), (y_1, y_2) \in \mathbb{R}^2$?

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First, we know that $e^x$ is a convex, non-decreasing function. \begin{align} (f\circ\exp)(\lambda x_1 + (1 - \lambda)y_1,\lambda x_2 + (1 - \lambda)y_2)=f(\exp(\lambda x_1 + (1 - \lambda)y_1,\lambda x_2 + (1 - \lambda)y_2))\\ \le f(\lambda\exp(x_1)+(1-\lambda)\exp(y_1),\lambda\exp(x_2)+(1-\lambda)\exp(y_2))\\ \le \lambda f(\exp(x_1),\exp(x_2))+(1-\lambda)f(\exp(y_1),\exp(y_2)) \end{align}

The second line is only true if the function $f$ is non-decreasing, otherwise the inequality does not hold.

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  • $\begingroup$ Okay. How about the case where $f$ is not necessarily non-decreasing? $\endgroup$ – user17982 Jul 15 '16 at 7:48

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