Knowing you can not use the minimum bound, there exists a function $f ( x , y )$ continuous in $\mathbb R ^ 2$ that has exactly two critical points which are (both) the minimum? Can you give me an example?
1 Answer
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Here's an example of a function with two maxima and no saddle points. I saw it on a mathematics professor's home page once—I think, someone who was even a prominent user here—and I simply cannot remember who they were. I hope someone will let me know in the comments so I can credit them.
Take a function with two maxima and a saddle point between them, say $$\begin{align} f(u,v)&=g(u,v-1)+g(u,v+1)\\ \text{where}\quad g(u,v)&=e^{-u^2-v^2}, \end{align}$$ and transform the domain to send the saddle point to infinity: $$\begin{align} u&=\operatorname{softmax}(x,-y^2)=\log(e^x+e^{-y^2}),\\ v&=y. \end{align}$$ Then $f$ as a function of $x$ and $y$ has two maxima and no saddle points.
Update: I just came across a note by Jerry Kazdan which gives a recipe for "a smooth function $f:\mathbb R^2\to\mathbb R$ having exactly $M$ local maxima, $s$ saddle points, and $m$ local minima, all non-degenerate, and no other critical points":
To construct an example, take a function $g$ with infinitely many non-degenerate maxima, saddles, and minima, for instance $g(x,y):=\sin\pi x\sin\pi y$. Pick the required number of maxima, saddles, and minima, Let $\Gamma$ be a smooth curve with no self-intersections that passes through these critical points and no others. Then let $\Omega$ be a tubular neighborhood of $\Gamma$ that contains no additional critical points. There is a diffeomorphism $\phi:\mathbb R^2\to\Omega$. The desired example is $f:=g\circ\phi$.