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I have a question about a simple proposition, I suppose that this is something well-known or a special case of something well-known:

Let $D\subset\mathbb{R}^{2}$ be the closed unit disk in the plane and $f:D\rightarrow\mathbb{R}$ be a smooth function. Suppose that the restriction $f|_{\mathbb{S}^{1}}$ has exactly $n$ local extrema, but none of them is a local extremum of $f$. Then the function $f$ has at least $\frac{n}{2}+1$ local extrema inside $D$.

Note that it follows from the condition that the level line of $f$ at a local extremum of $f|_{\mathbb{S}^{1}}$ touches the boundary from inside. Some examples are given below as level lines diagrams.

drawing

Here closed level lines correspond to local extrema of $f$ and crossing points of level lines correspond to saddles (possibly degenerated). As the condition is simple and clear and the proof is quite elementary*, I intend to propose this at a student competition. However, if this is well-known or folklore, it might not be suitable to propose it, so any references are welcome.

*/ In fact I have an elementary proof of a weaker form of the statement, claiming the existence of at least $\frac{n}{2}+1$ critical points of $f$ inside $D$.

Remarks. 1) This proposition may be completed by adding a second part: Let $\varphi:\mathbb{S}^{1}\rightarrow\mathbb{R}$ be a smooth function with $n$ local extrema, then it may be extended to a smooth $f:D\rightarrow \mathbb{R}$ satisfying the above conditions with exactly $\frac{n}{2}+1$ local extrema (and thus lying in the interior of $D$). Presumably this is true, but I don't have a proof.

2) For a given even $n$, we may ask about the number of (topologically) different diagrams up to symmetry (rotation, reflection) with exactly $\frac{n}{2}+1$ local extrema and without other critical points; but it seems to be a very hard task.

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  • $\begingroup$ Those pictures do not really look like a proof to me. Besides, to make the statement closer to the truth, one should count the extrema with some carefully defined multiplicities; otherwise, there may be just a single extremum unside. The multiplicity is, probably, the index of the gradient vector field, in which case the statement becomes just the Poincaré–Hopf theorem with boundary. $\endgroup$ Feb 6, 2015 at 10:15
  • $\begingroup$ @ Alex Degtyarev: I don't think that "there may be just a single extremum unside" - if $P$ and $Q$ are an absolute minimum and an absolute maximum of the restriction, then the conditions guarantee at least 2 absolute extrema of $f$ inside $D$ - a minimum and a maximum. $\endgroup$ Feb 6, 2015 at 10:29
  • $\begingroup$ If you allow degenerate minima, and (w.l.o.g.) $\phi>0$ we can take $f(x)=x^2\phi(x/|x|)$. $\endgroup$ Feb 6, 2015 at 10:39
  • $\begingroup$ @PietroMajer maybe it's the same example :) $f=r^2+r^3\sin(n\theta)$ in the polar coordinates. If my math is correct, it has no critical points in $0<r<2/3$. Rescale it to the unit disk. $\endgroup$ Feb 6, 2015 at 10:46
  • $\begingroup$ Ah, sorry, I see that you inhibit extrema, not just critical points on the boundary. Then, indeed, there must be two. $\endgroup$ Feb 6, 2015 at 10:52

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