# Continuous function satisfying three properties

I'm searching for a function of two variables $$f(x,y)$$ satisfying

• $$f$$ is continuous and positive,
• $$\lim_{(\epsilon,\delta)\rightarrow 0}\frac{f(\epsilon,\delta)}{\epsilon\delta} =0$$
• f(0,0)=0 is the unique minimum of $$f$$ (and thus unique zero).

Can you provide an analytical expression for such a function?

• @StinkingBishop check the 3rd point about the unique minimum. Sep 10 '19 at 18:27
• If $f(0,0)=0$, the function is not 'positive'. Sep 10 '19 at 18:28
• I'd say that forbidding the function to have zeros and be positive would be requiring it to be strictly positive. Anyhow what i meant is $f(x,y)\ge 0, \ \forall x,y\in\mathbb{R}$ Sep 10 '19 at 18:31
• What does $\approx$ mean in this context? The same as $\sim$ in web.mit.edu/broder/Public/asymptotics-cheatsheet.pdf ? Sep 10 '19 at 18:39
• @Tanatofobico: here is what Stinking Bishop means. The definition of the limit of $f(x,y)$ as $(x,y)\to(a,b)$ requires that $f(x,y)$ be defined in some deleted neighborhood around $(a,b)$. In other words, we assume there is an $r$ so that $f$ is defined for all $(x,y)$ satisfying $\color{red}{0<}|(x,y)-(a,b)|<r$ (which is a "deleted" nbhd, because the $\color{red}{0<}$ forces us to delete the point $(a,b)$ from the nbhd). Your expression $\frac{f(x,y)}{xy}$ is not defined on any such deleted neighborhood of $(0,0)$, though. That's because any such nbhd contains points where $x,y$ are zero. Sep 10 '19 at 19:09

There isn't such function. Suppose the second condition is valid: this would mean that, for some $$C>0$$, $$|f(\epsilon,\delta)|<|\epsilon\delta|$$ for all $$\epsilon$$, $$\delta$$ satisfying $$0<|\epsilon|,|\delta|. Now fix one such $$\epsilon\ne 0$$, and let $$\delta\to 0$$: from continuity of $$f$$ we have that $$|f(\epsilon, 0)|\le 0$$, which would imply that $$f(\epsilon, 0)=0$$ - contradictory with $$(0,0)$$ being a unique minimum.