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Does there exist a convex function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that for some $x_1 \in \mathbb{R}$ and $x_2 \in \mathbb{R}$ we have \begin{equation} \lim_{y \rightarrow + \infty} f(x_1,y) = - \infty, \\ \lim_{y \rightarrow + \infty} f(x_2,y) > - \infty ? \end{equation} Geometric intuition suggests to me that such a behavior at infinity is not possible, but I could not prove that such a function cannot exist.

NOTE 1. Note that if we fix $x \in \mathbb{R}$, the function $g(y)=f(x,y)$ is convex, so that $g$ is nonincreasing over $\mathbb{R}$ or it is strictly increasing over some half-line $[\bar{y},\infty)$. In any case the limit \begin{equation} \lim_{y \rightarrow \infty} g(y) = \lim_{y \rightarrow \infty} f(x,y) \end{equation} exists (finite or infinite).

NOTE 2. Please note that $f$ is required to be defined on all $\mathbb{R}^2$, otherwise the function $f(x,y)=- \sqrt{xy}$ defined on $\mathbb{R}^2_{+}= \{ (x,y) \in \mathbb{R}^2 : x \geq 0, y \geq 0 \}$ would be a simple example of such a behavior. Let us also note that there is no convex function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that for some $x_1 \in \mathbb{R}$ and $x_2 \in \mathbb{R}$ we have \begin{equation} \lim_{y \rightarrow + \infty} f(x_1,y) = \infty, \\ \lim_{y \rightarrow + \infty} f(x_2,y) < \infty . \end{equation} This follows from Rockafellar, Convex Analysis, Theorem (8.6).

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Let $x_3 = 2x_2-x_1$, so that $x_2 = (x_1+x_3)/2$. Since $$ f(x_2, y)\le \frac12 (f(x_1, y) + f(x_3, y)) $$ it follows that $f(x_3, y) \to \infty$ as $y\to+\infty$. This contradicts the result you quoted at the end.

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  • $\begingroup$ What a fantastic proof! I could not see by myself that the statement I was trying to prove was implied by the statement I already knew! Thank you very very very .... much for your invaluable help. $\endgroup$ – Maurizio Barbato Jun 2 '18 at 15:42

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