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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?

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  • $\begingroup$ @StinkingBishop check the 3rd point about the unique minimum. $\endgroup$
    – IamKnull
    Sep 10 '19 at 18:27
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    $\begingroup$ If $f(0,0)=0$, the function is not 'positive'. $\endgroup$
    – paw88789
    Sep 10 '19 at 18:28
  • $\begingroup$ 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 $f(x,y)\ge 0, \ \forall x,y\in\mathbb{R}$ $\endgroup$ Sep 10 '19 at 18:31
  • $\begingroup$ What does $\approx$ mean in this context? The same as $\sim$ in web.mit.edu/broder/Public/asymptotics-cheatsheet.pdf ? $\endgroup$ Sep 10 '19 at 18:39
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    $\begingroup$ @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. $\endgroup$ Sep 10 '19 at 19:09
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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 $0<|\epsilon|,|\delta|<C$. 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.

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