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Let there be a function $f(x,y)$ defined and differentiable on the open set $(-\infty..\infty,-\infty..\infty)$ which has exactly one point $(a,b)$ such that $\nabla f(a,b)$ is the zero vector, and which has a positive Hessian determinant at $(a,b)$. Then $(a,b)$ is an absolute extrema of $f$.

wllmkphrt
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2 Answers2

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You can find counterexamples to your assumption.

Note $$ \Psi(x)=\begin{cases} e^{-\frac{1}{1-x^2}} & \text{ for } \vert x \vert < 1\\ 0 & \text{ otherwise} \end{cases}$$ the bump function.

Now take the function $$f(x,y)=x +10\Psi(x)\Psi(y).$$

You can prove that $f$ has no absolute extrema, as $\lim\limits_{x \to +\infty} f(x,0)=+\infty$ and $\lim\limits_{x \to -\infty} f(x,0)=-\infty$.

However $f$ has a local maximum at $(0,0)$.

mathcounterexamples.net
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  • This is wrong in two ways: There are two critical points, and $(0,0)$ is not one of them. – Robert Israel Nov 04 '15 at 21:28
  • @Robert. You're right. I have to modify the example. However, my understanding of the OP is that there is exactly one point $(a,b)$ with gradient equal to $0$ and Hessian definite (positive or negative), vs. exactly one critical point. Such example seems possible to create or do you have even doubt with this? – mathcounterexamples.net Nov 04 '15 at 21:35
  • I'm getting four critical points. $(2.37854,0)$, $(1,0)$, $(.898237,0)$, and $(.331236,0)$ – wllmkphrt Nov 04 '15 at 22:07
  • Nevermind. It is a piecewise so only the last two would apply. – wllmkphrt Nov 04 '15 at 23:10
  • There are examples with only one critical point, that critical point being a local max, but no global max. – Robert Israel Nov 04 '15 at 23:18
  • I believe I can visualize this idea. Do you know of any function which would fit the criteria? – wllmkphrt Nov 04 '15 at 23:29
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    See e.g. http://math.stackexchange.com/questions/121326/unique-critical-point-does-not-imply-global-maximum-global-minimum – Robert Israel Nov 04 '15 at 23:41
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Let $f(x,y)=x^2 + y^2 (1 - x)^3$ $$f_x=2 x - 3 (1 - x)^2 y^2==0$$ $$f_y=2 (1 - x)^3 y$$ the only point where the gradient is vanishes is at $(0,0)$

$$f_{xx}f_{yy}-f_{xy}^2=6 x^2 y + 2 (1 - x^3) (2 - 6 x y^2)\underset{x=0=y}{=}4>0$$ We have minimum there since $$f_{xx}=2 - 6 x y^2\underset{x=0=y}{=}2>0$$ However, $f(0,0)=0$ and $f(5,1)=-39$ therefore the minimum is local.

Michael Medvinsky
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