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][1]


  [1]: https://i.sstatic.net/zAEBo.png

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.

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.