# Proof global minimum in two dimensions WITHOUT hessian but based on limit behavior

Consider a function $$\Phi \colon D \to \mathbb{R}$$, $$D = \{(x,y) \in (0,\infty)^2 \mid x < y\}$$, $$\Phi$$ differentiable on $$D$$. It is known, that $$(x^\ast,y^\ast)$$ is the only critical point in $$D$$ and we know about the limits that $$\lim_{\substack{x\to 0\\ y \to 0}} \Phi(x,y)=\lim_{\substack{x\to 0\\ y \to \infty}} \Phi(x,y)=\lim_{\substack{x\to \infty\\ y \to \infty}} \Phi(x,y)=\infty$$ and $$\lim_{x \uparrow y} \Phi(x,y) = \infty \;\; \forall\; y \in (0,\infty) \qquad \text{and} \qquad \lim_{y \downarrow x} \Phi(x,y) = \infty \;\; \forall\; x \in (0,\infty),$$ i.e., $$\Phi$$ goes to infinity on the boundary of $$D$$.

Is that sufficient to claim that $$\Phi$$ attains a global minimum in $$(x^\ast,y^\ast)$$? I think that it is due to a combination of Fermat's theorem on critical points and the extreme value theorem, but I would appreciate a second opinion. The function $$\Phi$$ is really messy and calculating the hessian is not an option.

• Since $\Phi$ is differentiable and defined in an open set, the only possible global minimizer will be $(x^*, y^*)$. On the other hand, he conditions you provide, together with the continuity of $\Phi$, should guarantee the existence of a global minimum (which must be attained in $(x^*,y^*)$). – PierreCarre Nov 12 '20 at 15:10
• @PierreCarre Thank you for your answer! Would you mind to give some reference for the existence of the global minimum and the fact that $(x^\ast, y^\ast)$ is the only candidate? – spitzen Nov 12 '20 at 15:57
• Just look at any calculus textbook and you'll find the proof that if $\Phi$ is differentiable at an interior point $X$ and attains a min/max at that point, $X$ must be a critical point. In the situation you describe: a differentiable function in an open set, any min/max must be a critical point. – PierreCarre Nov 12 '20 at 16:26
• The existence of the global minimum is trickier. Given the conditions, for any $M>0$ you can think of the set $X_M =\{(x,y) \in D: \Phi(x,y) \leq M\}$. If $X_M=D$, $M$ is a lower bound for $Phi$ in $D$. If $X_M \ne D$, $X_M$ is a compact set and $Phi$ will have a global minimum on $X_M$. – PierreCarre Nov 12 '20 at 16:36