Barycentric cuts through a convex body
Zuzana Patáková Martin Tancer Uli Wagner
math.CO cs.CG math.PR


[
    Franck Pereira Dos Santos
    December 30, 2023
=============================




Let K be a convex body in ^n (i.e., a compact convex set with nonempty
  interior). Given a point p in the interior of K, a hyperplane h passing
  through p is called barycentric if p is the barycenter of K ∩
  h. In 1961, Grünbaum raised the question whether, for every K, there
  exists an interior point p through which there are at least n+1 distinct
  barycentric hyperplanes. 


  Two years later, this was seemingly resolved
  affirmatively by showing that this is the case if p=p_0 is
  the point of maximal depth in K. However, while working on a related
  question, we noticed that one of the auxiliary claims in 

 the proof
  is incorrect. Here, we provide a counterexample; this re-opens Grünbaum's
  question.

It follows from known results that for n ≥ 2, there are always at least
  three distinct barycentric cuts through the point p_0 ∈ K of maximal
  depth. Using 
  tools related to  Morse theory

  we are able to improve this bound: four distinct barycentric cuts through p_0 are guaranteed if n ≥ 3. 





§ INTRODUCTION



    Grünbaum's questions.
 Let K be a convex body in ^n (i.e., compact convex set with nonempty interior). Given an interior point p∈ K, a hyperplane h passing through p is called barycentric if p is the barycenter (also known as the centroid) of the intersection K∩ h. In 1961, Grünbaum <cit.> raised the following questions (see also <cit.>):


Does there always exist an interior point p∈ K through which there are at least n+1 distinct barycentric hyperplanes?  



In particular, is this true if p is the barycenter of K?

Seemingly, Question <ref> was  answered affirmatively by Grünbaum himself <cit.> two years later, by using a variant of Helly's theorem to show that there are at least n+1 barycentric cuts through the point of K of maximal depth (we will recall the definition below).
The assertion that Question <ref> is resolved has also been reiterated in other geometric literature <cit.>. However, when working on Question <ref>, which remains open, we identified a concrete problem in Grünbaum's argument for the affirmative answer for the point of the maximal depth. 
The first aim of this paper is to point out this problem, which re-opens Question <ref>. 



    Depth, depth-realizing hyperplanes, and the point of maximum
depth.
 In order to describe the problem with Grünbaum's argument, we need a
few definitions. Let p be a point in K. For a unit vector v in the unit
sphere S^n-1⊆^n,  let h_v = h_v^p:={x∈^n ⟨
v,x-p⟩ =0} be the hyperplane orthogonal to v and passing through p, and let H_v = H_v^p:= {x∈^n ⟨ v,x-p⟩≥ 0} be the half-space bounded by h_v in the direction of v. Given p, we define the depth function δ^p
S^n-1→ [0,1] via δ^p(v) = λ(H_v ∩
K)/λ(K), where λ is the Lebesgue measure (n-dimensional volume) in ^n.
The depth of a point p in K is defined as 
(p, K) := inf_v ∈ S^n-1δ^p(v). It is easy to
see[Given v, v' ∈ S^n-1, λ(H_v ∩
K) and λ(H_v'∩ K) differ by at most λ((H_v Δ H_v')
∩ K) where Δ is the symmetric difference. For ε > 0
and v and v' sufficiently close, λ((H_v Δ H_v')
∩ K) < ελ(K) as K is bounded.]
that
δ^p is a continuous function, therefore the infimum in the definition is
attained at some v ∈ S^n-1. Any hyperplane h_v through p such that (p, K) =
δ^p(v) is said to realize the depth of p. Finally, a point
of maximal depth in K is a point p_0 in the interior of K such that
(p_0, K) := max(p, K) where the maximum is taken over all points in
the interior of K.[We remark that our depth function slightly differs from the
function f(H,p) used by Grünbaum <cit.>. However, the
point of maximal depth coincides with the `critical point' in <cit.> and 
hyperplanes realizing the depth for p_0 coincide with the `hyperplanes through
the critical point dividing the volume of K in the ratio F_2(K)'.]
The point of maximal depth always exists (by compactness of S^n-1) and
it is unique (two such points would yield a point of larger depth on the segment between them).



    Many depth-realizing hyperplanes?
 Grünbaum's argument has two ingredients. The first is the following result, known as Dupin's theorem <cit.>, which dates back to 1822:


If a hyperplane h through p realizes the depth of p then it is barycentric with respect to p.

Grünbaum refers to Blaschke <cit.> for a proof; for a more recent reference, see <cit.>.[The idea of the proof is simple: For contradiction assume that h realizes the depth of
p but that the barycenter b of K ∩ h differs from
p. Let v ∈ S^n-1 be such that h = h_v and (p, K) = δ^p(v).
Consider the affine (d-2)-space ρ in h passing through p and
perpendicular to the segment bp. Then by a small rotation of h along ρ
we can get h_v' such that δ^p(v') < δ^p(v) which contradicts
that h realizes the depth of p.
Of course, it remains to check the
details.]
A stronger statement will be the content of
Proposition <ref> below.

The second ingredient in Grünbaum's argument is the following assertion (which in <cit.> is deduced using a variant of Helly's theorem, without providing the details).


If p_0 is the point of K of maxiumal depth, then there are at least n+1 distinct hyperplanes through p_0 that realize the depth.

If correct, Postulate <ref>, in combination with Dupin's theorem, would immediately imply an affirmative answer to Question <ref>. 
However, it turns out that this step is problematic. Indeed, there is a counterexample to Postulate <ref>:



Let K = T × I ⊆^3 where T is an 
equilateral triangle and I is a line segment (interval) orthogonal to T, and let p_0 ∈ K be the point of maximal depth  (which in this case coincides with the barycenter of K). Then there are only 3 hyperplanes realizing the depth of p_0. 
 

 We believe that Proposition <ref> can be
  generalized to higher dimensions in the sense that, for every n, there are
  only n depth-realizing hyperplanes through the point of maximal depth in
  Δ× I ⊆^n, where Δ is a regular
  (n-1)-simplex. However, we did not attempt to work out the details
  carefully, because Kynčl and Valtr <cit.> informed us about
  stronger counterexamples: For every n, there exists a convex body K ∈^n such that there are only 3 depth-realizing hyperplanes through the point of maximal depth in K. Therefore, we prefer to keep the proof of Proposition <ref> as
simple as possible and focus on dimension 3.




 We emphasize that Proposition <ref> does not preclude an affirmative answer to Grünbaum's Question <ref> (nor to Question <ref>), since T × I contains infinitely many distinct barycentric hyperplanes through p_0.
Thus Grünbaum's questions remain open. 


We also remark that a weakening of Postulate <ref> is known to be true (see the `Inverse Ray Basis Theorem <cit.>, using the proof from <cit.>):[We remark that the second condition in the statement of the result in <cit.> is equivalent to the statement that 0 ∈ U, in our notation.]^,[Sketch of the inverse ray basis theorem:
if there is a closed hemisphere C ⊆ S^n-1 which does not contain
a point of U, let v be the centre of C. Then a small shift of p_0 in
the direction of v yields a point of larger depth, a contradiction.]



Let U ⊆
S^n-1 be the set of vectors u such that δ^p_0(u) = (p_0, K). Then 0 ∈ U.


In the special case that U is in general position, the cardinality of U is
at least n+1 (otherwise U < n and U would not contain
the origin, by general position), which proves Postulate <ref> in
this special case. However, U need not be always in general position. For
example, in the case K = T × I in ^3 = ^2 × of
Proposition <ref>, the set U contains three vectors in the
plane through the origin parallel with T. This is also the way we arrived at
the counterexample from Proposition <ref>.








Inverse Ray Basis Theorem immediately implies that three barycentric
hyperplanes are guaranteed in dimension at least 2.


  
Let K be a convex body in ^n where n ≥ 2 and p_0 be the point of
  maximal depth of K. Then there at least three distinct barycentric
  hyperplanes through p_0.









  Let U be the set from Proposition <ref>.  Then, 0 ∈ U and U ⊆ S^n-1 imply
together |U| ≥ 2. However, if |U| = 2, then U = {u,-u} for some u ∈
S^n-1. This necessarily means (p_0, K) = δ^p_0(u) =
δ^p_0(-u) = 1/2 as δ^p_0(u) + δ^p_0(-u) = 1. Then for any other v ∈ S^n-1 we get
min{δ^p_0(v), δ^p_0(-v)}≥ 1/2 which implies δ^p_0(v) =
δ^p_0(-v) = 1/2 as well. Therefore v ∈ U contradicting |U| = 2.)




    Four barycentric cuts via critical points of C^1
functions.
 Using tools related to 
 Morse theory, we are able to obtain one more
barycentric hyperplane, provided that n ≥ 3.



Let K be a convex body in ^n where n ≥ 3 and p_0 be the point of
  maximal depth of K. Then there are at least four distinct hyperplanes h such that p_0 is the barycenter of K ∩ h. 


Here we should also mention related work of Blagojević and
Karasev <cit.> and
<cit.>. They show that there are at least
μ(n) barycentric hyperplanes passing through some interior point of
K (not necessarily the point of maximal depth), where μ(n):=min_f
max_p∈ S^n |f^-1(p)| is the minimum multiplicity of any
continuous map f^n→ S^n (here, P^n is the n-dimensional
real projective space).  By calculations with Stiefel–Whitney classes, they
obtain lower bounds for μ(n) that depend in a subtle (and non-monotone) way
on n (see <cit.>). For example, μ(n)≥n/2+1 if n=2^ℓ -2, but for values of n of the form n=2^ℓ-1
(e.g., for n=3) their methods only give a lower bound of μ(n)≥ 2.


Our argument in the proof of Theorem <ref> is, in certain sense,
tight. For completeness we discuss this in Section <ref>.


In what follows, 
we view 
S^n-1 as a smooth manifold with its 
standard differential structure. A key tool in the proof of Theorem <ref> is the following close connection between barycentric hyperplanes and the critical points of the depth function:



  Let K ⊆^n be a convex body and p be a point in the interior of
  K. Then the corresponding depth function δ^p  S^n-1→
  is a C^1 function. In addition, v ∈ S^n-1 is a critical point of
  δ^p (that is, Dδ^p(v) = 0, where Df(v) denotes the total derivative of a function f
  at v) if and only if
  h_v is barycentric. 


As mentioned earlier, Proposition <ref> generalizes Dupin's
theorem.


Indeed, if h = h_v
realizes the depth, then v is a global minimum of δ^p, hence h is barycentric by Proposition <ref>. 

In the proof, we closely follow computations by Hassairi and
Regaieg <cit.> who stated an extension of Dupin's
theorem to absolutely continuous probability measures.


As explained in <cit.> (see Proposition 29, Example 7, and
the surrounding text in <cit.>), the extension of Dupin's
theorem


does not hold in the full generality stated
in <cit.>, and it requires some additional assumptions.
However, a careful check of the computations of Hassairi and
Regiaeg <cit.> in the special case of uniform probability
measures on convex bodies reveals not only Dupin's theorem but all items of
Proposition <ref>.






















Regarding the proof of Theorem <ref>, the 

Inverse Ray Basis Theorem (Proposition <ref>) and Corollary <ref>

imply that δ^p_0 has at least three global minima. This
gives three barycentric hyperplanes via Proposition <ref>.
Furthermore, we also get three maxima of δ, as a maximum
appears at v, if and only if a minimum appears at -v (note that h_v =
h_-v). However, it
should not happen for a C^1 function on S^n-1 that it has only such critical
points. We will show that there is at least one more critical point, which
yields another barycentric hyperplane via Proposition <ref>. Namely, we
show the following proposition.







 
Let n ≥ 2 and let f  S^n → be a C^1 function. Let m_1,
  …, m_k be (not necessarily strict) local minima or maxima of f, where k ≥ 3. Then there exists
  u ∈ S^n, different from m_1, …, m_k, such that Df(u) = 0. 
















This finishes the proof of Theorem <ref> modulo
Propositions <ref> and <ref>.
(Proposition <ref> is applied with k = 6.)

The main idea beyond the proof of Proposition <ref> is
that if we have at least three local minima or maxima, then we should also
expect a saddle point (unless there are infinitely many local extremes).

This would be
an easy exercise for Morse functions (which are in particular C^2) via Morse
theory (actually, the Morse inequalities would provide even more critical points).
Working with C^1 functions adds a few difficulties, but all of
them can be overcome.




    Relation to probability and statistics.
 

The depth function, as we define it above is a special case of the (Tukey) depth of a probability
measure in ^d, a well-known notion in statistics <cit.>. More precisely, given a probability measure  on ^d
and p ∈^d, we can define (p, ) := inf_v ∈ S^n-1(H_v).
Then (p, K) is a special case of the uniform probablity measure on a convex body K, i.e., 

(A) := λ(A)/λ(K) for A Lebesgue-measurable. 



We refer
to <cit.> for an extensive recent survey making
many connections between the depth function in statistics and geometric
questions.


There is a vast amount of literature, both in computational geometry and statistics, devoted
to computing the depth function in various settings (which is not easy in
general). We refer, for example,
to <cit.> and the references
therein. From this point of view, understanding the minimal possible number of
critical points of the depth function is a quite fundamental property of the
depth function. Via Proposition <ref>, this is essentially equivalent to Grünbaum's questions.














    Organization.
 Proposition <ref> is proved in
Section <ref>; Proposition <ref> is proved in
Section <ref>; and Proposition <ref> is proved in
Section <ref>.




§ FEW HYPERPLANES REALIZING THE DEPTH



In this section we prove Proposition <ref>, assuming Proposition <ref>.




    Preliminaries.
 Let us recall that given a bounded measurable set Y ⊆^n of
positive measure, the
barycenter of Y is defined as 

    Y = ∫_^n x χ_Y(x) dx/∫_^nχ_Y(x) dx = 1/λ(Y)∫_Y x dx

where χ_Y is the characteristic function and the integral is
considered as a vector in ^n. 


If Y splits as a disjoint union Y
= Y_1 ⊔⋯⊔ Y_ℓ of sets of positive measure then 

    Y = 1/λ(Y)(∑_i=1^ℓλ(Y_i) 
      Y_i)

 which easily follows from (<ref>).

If h is a hyperplane, and Y ⊆ h has positive (n-1)-dimensional
Lebesgue measure inside h, then the formula for the barycenter is analogous
to (<ref>):

    Y = ∫_h x χ_Y(x) dλ_n-1(x)/∫_h
      χ_Y(x) dλ_n-1(x) = 1/λ_n-1(Y)∫_Y x
      dλ_n-1(x)

where, for purpose of this formula, λ_n-1 denotes the
(n-1)-dimensional Lebesgue measure on h. 

If h ⊆^n is a hyperplane whose 
orthogonal projection π(h) onto ^n-1×{0} 
(the first n-1 coordinates) 
equals ^n-1×{0}, 
then π(Y) = π( Y).


Let 
T ⊆^2 be 
an equilateral triangle with 
  (T) = 0 and I = [-1,1]. Then (K) = 0. In addition, because the
  point of maximal depth p_0 is unique and invariant under isometries of K,
  we get p_0  = 0.

We will use the following notation: a, b, c are the vertices of T and
  α, β, and γ are lines perpendicular to T passing through
  a, b, and c respectively.

Now let h be a hyperplane passing through 0. We want to find out whether
  h realizes the depth. We will consider three cases: 
  
  
   
  * h is perpendicular to T; 
   
  * h is not perpendicular to T and all
  intersection points of h with α, β, and γ belong to K; 
  
  * h is not perpendicular to T and at least one
  of the  intersection points of h with α, β, and γ does 
  not belong to K.
  

  In case (<ref>), we will find three candidates for hyperplanes
  realizing the depth. Then we show that there is no hyperplane realizing the
  depth in cases (<ref>) and (<ref>), which shows that only the three
  candidates from case (<ref>) may realize the depth. They
  realize the depth because we have at least three hyperplanes realizing the
  depth by the discussion in the introduction above
  Theorem <ref>.
  














  Let us focus on case (<ref>). 
This is the same as considering the lines realizing the depth in an
 equilateral triangle.
  
  
  It is easy to check and well known (see
  e.g. <cit.>) that the depth of the equilateral
  triangle 
  
  is 4/9 and it is realized by lines parallel with the sides of the
  triangle. It follows that we can reach depth 4/9 in K by hyperplanes perpendicular to
  T and parallel with the three sides of T, and all other hyperplanes from
  case (i) bound a portion of K strictly larger than 4/9 on each of their sides.







 Case (<ref>) is very easy: It is easy to compute that each hyperplane of type 
  (<ref>) splits K into two parts of equal volume 1/2.
  Therefore, no such hyperplane 
  realizes the depth.

 Finally, we investigate case (<ref>). Here we show that no hyperplane h
  of case (<ref>) is barycentric. Therefore, by Theorem <ref>,

  it cannot realize the depth either.

  We aim to show that 0 is not the barycenter of h ∩ K. Let U be the
  orthogonal projection of h ∩ K to the triangle T. Equivalently, we want to show
  that 0 is not the barycenter of U. We also realize that U = T ∩ S,
  where S is an infinite strip obtained as the orthogonal projection of h
  ∩ (^2 × I) to ^2×{0}
  ; see Figure <ref>.


Let s be the center line of S. This is the line where h meets the plane
of T. 
  We remark that 0 belongs to s and in addition U is a proper subset of T (otherwise we would
  be in case (<ref>)). We again distinguish three cases: 
  
  
   
  * none of the vertices a, b, c belongs to U,
   
  * one of the vertices a, b, c belongs to U,
   
  * two of the vertices a, b, c belong to U.
  

  In all the cases we will show U ≠ T.

  



  In case (a), s splits one of the vertices of T from the other two.
  Without loss of generality, a is on one side of s and b and c are on
  the other side. The center line s also splits U into two parts. Let W'
  be the (closed) part on the side of a, W” be the mirror image of W' along S
  and W := W' ∪ W”. Note that W is a proper subset of U; indeed, since T = 0 and T is equilateral, the line s splits the segment ab closer to b and the
  segment ac closer to c. By the symmetry of W, the barycenter W
  belongs to the line s. However, this means that the barycenter of U is
  not on s; it is on the bc side of s. Formally, this follows
  from (<ref>) for the decomposition U = W ⊔ (U ∖ W). 

In case (b), without loss of generality, U contains c. Then T ∖
U is the union of two triangles T_a and T_b. Let κ be the line
parallel with ab passing through 0. Without loss of generality, up to
rotating T, κ is the x-axis. From (<ref>), we get
0 =  T = 1/λ(T)(λ(U)  U + λ(T_a)  T_a +
λ(T_b)  T_b). The barycenters T_a and T_b are below
the line κ or on it. At least one of these barycenters is strictly
below (T_a is on κ if and only if c belongs to the closure of
T_a, and similarly with T_b). 







Therefore, U must be strictly above
κ if the above equality is supposed to hold. 

In case (c), it is even more obvious that U ≠ T. Without loss of
generality U contains b and c. Then T ∖ U is a triangle T_a.
Since both T and T_a are convex and T_a does not contain T, we have T_a ≠ T. Therefore T
≠ U follows from (<ref>) for the decomposition T = U
⊔ T_a.




§ CRITICAL POINTS OF THE DEPTH FUNCTION


Here we prove Proposition <ref>. We follow <cit.>
with a slightly adjusted notation and adding a few more details here and there.


Without loss of generality, we can assume that the point p coincides
with the origin and we suppress it from the notation. That is, we write
δ for the depth function instead of δ^p.



Let e_1, …, e_n be the canonical basis of ^n and let

    S_j+^n-1 = {u = ∑_i=1^n u_ie_i ∈ S^n-1; u_j >
    0}5mm  and 5mm S_j-^n-1 = {u = ∑_i=1^n u_ie_i ∈ S^n-1; u_j <
    0}

be the relatively open hemispheres  of S^n-1 with poles at e_j and
-e_j, for j ∈ [n]. 
These sets
form an atlas on S^n-1.

Let us consider j ∈ [n].
Given x ∈^n and i ∈ [n], x_i denotes the ith coordinate of
x, that is x= ∑_i=1^n x_ie_i.  With a slight abuse of the notation, we
identify ^n-1 with the subspace of ^n spanned by e_1, …,
e_j-1, e_j+1, …, e_n. Let x̂ := ∑_i=1, i ≠ j^n x_ie_i
∈^n-1.








 Following <cit.> we consider the diffeomorphisms u ↦β(u) = -û/u_j between S_j+^n-1 and ^n-1 or
  between S_j-^n-1 and ^n-1. 



We will check the required properties of δ locally at each
of the 2n hemispheres S_j+^n-1 or S_j-^n-1 (with respect to the aforementioned
diffeomorphisms). Given that all cases are symmetric, it is sufficient to focus
only on the S_n+^n-1 case. 
  That is, from now on, we assume that j = n and ^n-1 is spanned
  by the first (n-1) coordinates in the convention above. Given a point x
  ∈^n, we also write it as x = ( x̂; x_n).









 Now, for y ∈^n-1 we consider the hyperplane h'_y in ^n containing the origin and defined by 
 
    h'_y = {(x̂; x_n) ∈^n  x_n = ⟨ y,x̂⟩}.






Note that if u ∈ S^n-1_j+, then h'_β(u) = {x ∈^n⟨ x, u ⟩ = 0}.


In particular, since p is the origin, h'_β(u) coincides with h_u used in
the introduction for definition of the depth function. This also means
that the map
y ↦ h'_y provides a parametrization of a family of those
hyperplanes containing the origin which do not contain e_n. We also
set H'_y to be the positive halfspace bounded by h'_y:

    H'_y = {(x̂; x_n) ∈^n  x_n ≥⟨ y,x̂⟩}.

Again, if u ∈ S^n-1_j+, then H'_β(u) coincides with H_u from
the introduction (here we use u_n > 0).









Now, we consider the map f ^n-1→ defined by 

    f(y) = λ (H'_y ∩ K) = ∫_^n-1∫_⟨ y,
     x̂⟩^∞χ_K(x̂; x_n) dx_n dx̂,

where χ_K is the characteristic function of K. 
When y = β(u) for some u ∈ S^n-1_j+, then f(β(u)) =
δ(u). Therefore, given that the map u →β(u) is a diffeomorphism,
it is sufficient to prove that f is a C^1 function and that
β(v) ∈^n-1 is a critical point of f if and only if
h'_β(v) = h_v is barycentric.











The aim now is to differentiate f(y) with respect to y.
We will show that the total derivative equals

    Df(y) = - ∫_^n-1x̂·χ_K(x̂; ⟨ y, x̂⟩) d x̂

considering the integral on the right-hand side as a vector.
Deducing (<ref>) is a quite routine computation skipped
in <cit.>.[When compared with formula (3.1)
in <cit.>, we obtain a different sign in front of the
integral. This is caused by integration over the opposite halfspace.] 
However, this is the step in the proof of Theorem 3.1 
in <cit.> which reveals that some extra assumptions
in <cit.> are necessary. Thus we carefully
deduce (<ref>) at the end of this proof for completeness.








We will also see that all partial derivatives of f are continuous
which
means that f is a C^1 function
which is one of our required conditions. Now we want to show that
Df(β(v)) = 0 if and only if h_v is barycentric.

First, assume that Df(β(v)) = 0.
This gives

    0 = ∫_^n-1x̂·χ_K(x̂; ⟨β(v), x̂⟩) d x̂/∫_^n-1χ_K(x̂; ⟨β(v), x̂⟩) d x̂

which means that 0 is the barycenter of K ∩ h'_β(v) from the
definition of h'_β(v). 

On the other hand, if 0 is the barycenter of K ∩ h'_β(v),
then we deduce (<ref>) which implies Df(β(v)) = 0.



It remains to show (<ref>). For this purpose, we compute partial
derivatives ∂/∂ y_k f(y), 1 ≤ k ≤ n-1. 
In the following computations, recall that e_k stands for the standard basis vector
for the kth coordinate and let ∫_a^b := -∫_b^a if a > b. We get

    ∂/∂ y_kf(y) 
         =  lim_t → 01/t∫_^n-1(∫_⟨ y + t e_k, x̂⟩^∞χ_K(x̂; x_n) d x_n - ∫_⟨ y, x̂⟩^∞χ_K(x̂; x_n) d x_n) dx̂
       =  lim_t →
        0∫_^n-11/t∫_⟨ y, x̂⟩ + t x_k^⟨ y, x̂⟩χ_K(x̂; x_n)
          d x_n dx̂.


Let y, x̂∈^d-1 be such that ( x̂; ⟨ y, x̂⟩)
∉∂ K. Then we get

    lim_t→ 01/t∫_⟨ y, x̂⟩ + t x_k^⟨ y, x̂⟩χ_K(x̂; x_n)
    	      d x_n = -x_kχ_K(x̂; ⟨ y, x̂⟩),

 because ( x̂; ⟨ y, x̂⟩)
 ∉∂ K implies that the function χ_K(x̂; x_n) as a
 function of x_n is constant on the interval (⟨ y, x̂⟩ - |t x_k|, ⟨ y, x̂⟩ + |t x_k|) for small enough
 |t|. Therefore, by the dominated convergence theorem, 

    ∂/∂ y_kf(y) = ∫_^n-1-x_k χ_K(x̂; ⟨
     y, x̂⟩) dx̂.

 
 For fixed y, the condition ( x̂; ⟨ y, x̂⟩)
  ∉∂ K holds for almost every x̂ because ( x̂;
  ⟨ y, x̂⟩) ∈ h_y and h_y passes through the interior of
  K (through the origin). 




  By another application of dominated convergence
  theorem, we realize that the right hand side of (<ref>) is
  continuous in y (this time, we consider a sequence y^i → y and we
  observe that χ_K(x̂; ⟨ y^i, x̂⟩) →χ_K(x̂; ⟨ y, x̂⟩) for almost every x̂). Therefore the total
  derivative of f at any y exists and (<ref>) gives the
  formula (<ref>).




  In the last paragraph of the proof above we crucially use the convexity of K.
  Without convexity, there is a compact nonconvex polygon K' ⊆^2,
  with 0 in the interior, such that there is y with the property that the
  set of those x̂ for which ( x̂; ⟨ y, x̂⟩)
   ∈∂ K' has positive measure;
   
 
  see Figure <ref>.
  In fact, even (<ref>) does not hold for K'. Here we took K' to be
  the polygon from Example 7 of <cit.>, and we refer the reader to that paper for more
  details.














































































































































§ REEB GRAPHS AND C^1 FUNCTIONS, OBSOLETE



In this section, we prove Proposition <ref>. We start with some
preliminaries.




Let M be a manifold. Given a continuous function f M → and s ∈ 
we define the level set L_s := {w ∈ M  f(w) = s}. Moreover, we define an equivalence relation ∼
on M, where u ∼ v if f(u) = f(v) and u and v belong to the
same connected component of the level set L_f(u)=L_f(v). The Reeb graph of f is defined as
the quotient G = G_f := M_/∼. By q = q_f we denote the quotient
map q  M → G. Furthermore, by the definition of the Reeb graph, the map f induces 
a well-defined and continuous map Φ G → given by
Φ([u]) = f(u), where [u] is the equivalence class of u ∈ M with
respect to ∼. 

We remark that, in the most general case, a Reeb graph as defined above need not be a graph, nor even a Hausdorff topological space. 
However, if M is a smooth manifold, and f is a C^1 function, then by the following proposition the local structure of the Reeb graph is well-behaved in the neighborhood of points u for which the total derivative Df(u) is nonzero. 































Let M be a smooth n-manifold, n ≥ 1, f be a C^1 map and u ∈ M
  be such that Df(u) ≠ 0. Then
there is an open neighborhood N(u) of u such that Φ is a homeomorphism
when restricted to q(N(u)).



The first sentence below was half-modified by Uli but the second half
remained in the original setting. Now the sentence is a bit odd. I do not step
in because I do not know what was the intention of the modification. Uli, are
you willing to fix it?

This proposition might be folklore, but the usual
setting in which Reeb graphs are considered is when the function f is assumed to Morse, or at
least C^2 functions; see e.g. <cit.>. We were not able to find a suitable reference for C^1
functions. Therefore, we provide a proof via the implicit function theorem and
standard ε-δ-arguments (which appear in the proof in the guise of 
repeatedly choosing sufficiently small neighborhoods).





  The key in the proof the proposition will be the following
  claim:


  There is a neighborhood N'(u) of u such that Φ is injective when
  restricted to q(N'(u)). Equivalently, for each v,w ∈ N'(u), if f(v) = f(w), then v and w belong to the same component of the level set L_f(v).

  Updates: This claim is still used in the new proof; however,
  Proposition <ref> is not used. Thus, after checking the proof,
  this claim should be turned into a proposition. It is used only in ^n
  which should simplify the proof. Also, it is sufficient to state in the form
  after the word `Equivalently'. On the other hand, it should be stated that v
  and w can be connected by a path in the level set. But this immediately
  follows from the proof. 
  
  In the Claim above, Uli changed `Equivalently' to `Thus'. Perhaps
  based on the suggestion of the referee. This we have to discuss! Both
  implications are used! (The intended meaning is that the level set is
  considered in the whole manifold.) I have also changed the numbering so that
  it is more apparent that the claim is embedded in the proof of the theorem.
  Reverted `Thus' into `Equivalently'. Do you agree?


  First we show how the proposition follows from the claim, then we prove the
  claim.
  
Consider a closed ball B(u) around u which is fully contained in N'(u). We
  get that Φ(q(B(u))) is a connected compact subset of  as a
  continuous image of a connected set is connected and a continuous image of a
  compact set is compact, that
  is, Φ(q(B(u))) is a
bounded closed interval (or a point but the point can be easily excluded as f
  cannot be constant on N'(u)). Moreover, because q(B(u)) is compact and
  Φ is injective, Φ must be a
  homeomorphism between q(B(u)) and Φ(q(B(u))). Now, it is
sufficient to set N(u) to be the interior of B(u).

Now we want to prove Claim <ref>. First we prove the claim
  under the additional assumption that M = ^n. Then we deduce the claim in
  general.





Let us assume that M = ^n. Without loss of generality assume
that ∂ f/∂ x_n(u) > 0, otherwise we permute the
coordinates and/or swap x_n and -x_n. Consistently with the previous
  section, given x ∈^n, we write x =
(x̂, x_n) where x̂∈^n-1 and x_n ∈. Now we
consider the C^1 function F ^n-1××→ defined as
F(x̂, t, x_n) := f(x̂, x_n) - t. Note that
∂ F/∂ x_n = ∂ f/∂ x_n. We also
observe that F(û, f(u), u_n) = 0. Therefore, by the implicit
function theorem, there is an open neighborhood N” of (û,
f(u))
in ^n-1× such that there is a C^1 function g  N”→ with g(û, f(u)) = u_n and that F(v̂, t,
g(v̂, t)) = 0 for any (v̂, t) ∈
N”. From the definition of F this gives

    f(v̂, g(v̂, t)) = t.


By possibly restricting the neighborhood to a smaller set, we can assume
that N” is the Cartesian product of a neighborhood N”(û) of
û in ^n-1 and N”(f(u)) of f(u) in , and that both
N”(û) and N”(f(u)) are open balls. 
Moreover, 

we can assume
that ∂ F/∂ x_n(v̂, t, v_n) > 0 for any
(v̂, t, v_n) ∈ N”× N”'(u_n) where N”'(u_n) is some
neighborhood of u_n in , again a ball. Now we possibly further restrict
  N”(û) and N”(f(u)) so that g(v̂, t) belongs to N”'(u_n) for any
(v̂, t) ∈ N”.



The condition on the partial derivative of F
implies that for every (v̂, t) ∈ N” the equation F(v̂,
t, x_n) =
0 has at most one solution x_n ∈ N”'(u_n). Therefore it has a unique
solution x_n = g(v̂,
t). In other words we get:

    If  f(v̂, x_n) = t,  then  x_n = g(v̂,t).



Now, we define N'(u) := Ψ^-1(N”) where Ψ^n-1×→^n-1× is defined as Ψ(v) = (v̂, f(v)) for any v
∈^n-1×. 
In particular (v̂, f(v)) belongs to N” for any v ∈ N'(u).














































  Now we prove Claim <ref> in the form of the second sentence in
  the claim.
  Let t := f(v) = f(w). From (<ref>) we get v_n = g(v̂,
  t) and w_n = g(ŵ, t). Now, let us consider an arbitrary path
  P in  N”(û) connecting v̂ and ŵ. Let
  us map P to a path P_t := {(x̂, g(x̂, t))x̂∈
  P}. This is a path connecting v and w in ^n-1×. We will
  be done once we show P_t ⊆ L_t. This means that we are
  supposed to show that f(x̂, g(x̂, t)) = t
 
  for
  every x̂∈ P which follows from (<ref>). 

  This finishes the proof of
  Claim <ref> under the additional condition that M = ^n.

In the paragraph above, I have reverted the changes. You have changed
  t to f(v) in the sentence `This means...'. I have previously did the change L_f(v)→ L_t (maybe it
  wasn't noticed). Then, I think that it is better to have t in the next
  sentence. And I also think that it is beeter to use L_t than L_f(v),
  because t is used everywhere above and it does not really distinguish
  f(v) and f(w). If you insist on your changes, I kept them in the
  comments.
  
  













Is the last paragraph understandable. I was quite struggling with this.


  Now, let us drop the assumption that M = ^n. Let
  Ω(u) be an open ball containing u and ψ be a
  diffeomorphism from an open ball B ⊆^d to Ω(u). 
  Let Φ_ψ G → be given as Φ_ψ([x]) = f ∘ψ(x) where the equivalence class is taken with respect to level sets
      in B.

  The condition
  Df(u) ≠ 0 is equivalent with D(f∘ψ)(ψ^-1(u)) ≠ 0.

Therefore, by the claim for M = ^n, we get that there is a neighborhood
  N'_ψ⊆ B of ψ^-1(u) such that if v' and w' in
  N'_ψ satisfy f ∘ψ (v') = f ∘ψ(w'), then they belong to
  the same component of their level set in B. Let us set N'(u) = ψ^-1(N_ψ). The
  condition above says that if v = ψ(v') and w = ψ(w') satisfy f(v)
  = f(w), then they belong to the same component in their level set in Ω(u) (because
  ψ is, in particular, a homeomorphism), therefore in
  M as well. This proves Claim <ref> and finishes the proof of
  the proposition.

































Now, we are ready for the proof of Proposition <ref>.


 
 First, we can assume that all local extrema m_1,
  …, m_k are strict. Indeed, if some of them is not strict, say m_1,
  then we can find u ≠ m_1, …, m_k with Df(u) = 0 in a neighborhood
  of m_1.
 
  Not yet done: The set U in the proof of Proposition 4
  (currently <ref>) is not quite the
  U you obtain
  from Proposition 6 (currently <ref>). This needs more explaining. Also, having a picture of
  manifold, a function on the manifold and its associated Reeb graph and
  explaining f vs. Phi vs. q (maybe illustrating the proof of Proposition 4)
  would make Section 4 more readable.
  
  Let U = S^n∖{m_1,…, m_k}. The set U is connected
  (even path connected) because
  n ≥ 2. For contradiction, let us assume that Df(u) ≠ 0 for every u
  ∈ U. From Proposition <ref> we get that Φ is a
  homeomorphism between q(U) and the image Φ(q(U)). As Φ(q(U)) is

  connected, it must be an interval, and because Φ(G) is bounded, we
  get that Φ(q(U)) is a bounded interval. Let a < b be the endpoints of
  this interval.
 
  We also observe that for every i ∈ [k] the equivalence classes [m_i] consist of a single
  point because m_i are local extrema. In other words, each m_i is also an
  equivalence class in the Reeb graph. Because k > 3, there are either two
  points among [m_1], …, [m_k] mapped to a by Φ or two such points mapped to b
  or one such point mapped into the interval (a,b). We will exclude each case.



First, let us assume that two of the points are mapped to a or b.
    Without loss of generality, Φ([m_1]) = Φ([m_2]) = a. For i = 1,2,
    let B(m_i) ⊆ S^n be a closed ball around m_i such that m_i
    is a strict minimum on B(m_i); let s_i := min{f(u) u ∈∂
    B(m_i)}; and let t be a real number satisfying a < t < s_1,s_2, and
    t ≠ f(m_j) for j ∈ [k]. In particular, the level set L_t is
    nonempty and connected because Φ is an isomorphism on U.
    Because f(m_1) = f(m_2) = a and f(u) > t for u ∈∂ B(m_1) or
    u ∈∂ B(m_2), we get that L_t intersects both the interior of
    B(m_1) and the interior of B(m_2); however these two intersections are
    separated by ∂ B(m_1) and ∂ B(m_2) which contradicts
    connectedness of L_t.






  It remains to exclude the case f(m_i) ∈ (a,b) for some i. The
  argument is similar. Without loss of generality f(m_1) ∈ (a,b) and m_1
  is a local minimum. Define B(m_1) and s_1 as above and consider t satisfying f(m_1)
  < t < s_1, and t ≠ f(m_j) for j ∈ [k]. The level set L_t
  intersects the interior of B(m_1) as before. However, it also intersects the (open)
  complement S^n ∖ B(m_1) because f(u) > t on ∂ B(m_1)
  whereas the global minimum of f on S^n does not belong to B(m_1).
  Therefore, ∂ B(m_1) again separates L_t, a contradiction. 































































§ ONE MORE CRITICAL POINT








In this section, we prove Proposition <ref>. 





Given a manifold M and a continuous function f M → and s ∈ we define the level set L_s := {w ∈ M  f(w) = s}. 
In the proof of Proposition <ref> we will need that the
level sets are well-behaved in the neighborhoods of points u for which the
total derivative Df(u) is nonzero.
















Let n≥ 1, f^n → be a C^1 function and u ∈^n be
  such that Df(u) ≠ 0. Then there is a neighborhood N(u) of u such that 


  for every v,w ∈ N(u) if f(v) = f(w), then v and w can be connected
  with a path within the level set L_f(v). (It is allowed that this path
  leaves N(u) provided that it stays in L_f(v).)





Without loss of generality assume
that ∂ f/∂ x_n(u) > 0, otherwise we permute the
coordinates and/or swap x_n and -x_n. Consistently with the previous
  section, given x ∈^n, we write x =
(x̂, x_n) where x̂∈^n-1 and x_n ∈. Now we
consider the C^1 function F ^n-1××→ defined as
F(x̂, t, x_n) := f(x̂, x_n) - t. Note that
∂ F/∂ x_n = ∂ f/∂ x_n. We also
observe that F(û, f(u), u_n) = 0. Therefore, by the implicit
function theorem, there is an open neighborhood N' of (û,
f(u))
in ^n-1× such that there is a C^1 function g  N' → with g(û, f(u)) = u_n and that F(v̂, t,
g(v̂, t)) = 0 for any (v̂, t) ∈
N'. From the definition of F this gives

    f(v̂, g(v̂, t)) = t.


By possibly restricting the neighborhood to a smaller set, we can assume
that N' is the Cartesian product of a neighborhood N'(û) of
û in ^n-1 and N'(f(u)) of f(u) in , and that both
N'(û) and N'(f(u)) are open balls. 
Moreover, 

we can assume
that ∂ F/∂ x_n(v̂, t, v_n) > 0 for any
(v̂, t, v_n) ∈ N' × N”(u_n) where N”(u_n) is some
neighborhood of u_n in , again a ball. Now we possibly further restrict
  N'(û) and N'(f(u)) so that g(v̂, t) belongs to N”(u_n) for any
(v̂, t) ∈ N'.



The condition on the partial derivative of F
implies that for every (v̂, t) ∈ N' the equation F(v̂,
t, x_n) =
0 has at most one solution x_n ∈ N”(u_n). Therefore it has a unique
solution x_n = g(v̂,
t). In other words we get:

    If  f(v̂, x_n) = t,  then  x_n = g(v̂,t).



Now, we define N(u) := Ψ^-1(N') where Ψ^n-1×→^n-1× is defined as Ψ(v) = (v̂, f(v)) for any v
∈^n-1×. 
In particular (v̂, f(v)) belongs to N' for any v ∈ N(u).

  Let t := f(v) = f(w). From (<ref>) we get v_n = g(v̂,
  t) and w_n = g(ŵ, t). Let us consider an arbitrary path
  P [0,1] → N'(û) connecting v̂ and ŵ. Let
  us `lift' P to a path P_t  [0,1] →^n-1× given by 
  P_t(s)  := (P(s), g(P(s), t)). This is a path connecting v and w. We will
  be done once we show P_t([0,1]) ⊆ L_t. This means that we are
  supposed to show that f(P(s), g(P(s), t)) = t
 
  for
  every s ∈ [0,1] which follows from (<ref>). 





  
  


  
*Total derivatives and gradients.
Let f ^n → be a C^1 function. Then for any u ∈^n, the
total derivative Df(u) is represented by a row vector
( ∂/∂ x_1f(u), …, ∂/∂
  x_nf(u)) (if Df(u) exists). By Df(u) we mean the Euclidean norm
  of this vector. The gradient ∇ f(u) is the same vector transposed

    ∇ f(u) := ( ∂/∂ x_1f(u), …,
      ∂/∂
        x_nf(u))^T.

Then ∇ f(u)  = Df(u), and in addition Df(u)(∇ f(u)) =
Df(u)^2. 


Let x ∈^n and ρ > 0, by B(x,ρ)
⊆^n we
denote the 
compact ball of radius ρ centered in x with respect to the
standard Euclidean metric.

















  
  Let f ^n → be a C^1 function, let x ∈^n and let
  ζ, ρ >
  0. Assume that Df(u)≥ζ for every u ∈ B(x, ρ). Then
  there is v ∈ B(x, ρ) such that f(v) ≥ f(x) + ζρ/2.





Let 
  
    K := {y ∈ B(x, ρ) f(y) ≥ f(x) + ζ/2y -
      x}.

This is a closed therefore compact set, it is also nonempty because x ∈ K.



  Let M := max{f(y) y ∈ K}, for contradiction M <
  f(x) + ζρ/2. Let w ∈ K be such that f(w) = M. Note that
  for every v ∈∂ B(x, ρ) ∩ K we get f(v) ≥ f(x) +
  ζρ/2 because v-x = ρ in this case. Thus, in
  particular, w ∉∂ B(x, ρ). See Figure <ref>.

  Consider the derivative at w in the direction of the gradient ∇
  f(w).
  From properties of the total derivative, we get
  
    lim_t → 0|f(w + t ∇ f(w)) - f(w) -
      Df(w)(t∇ f(w))|/t∇ f(w) = 0.

 Therefore, for small enough t > 0 we get 
  
    |f(w + t ∇ f(w)) - f(w) -  tDf(w)^2| ≤ζ/2t
      Df(w).
 
  Consequently,
  
    f(w) + tDf(w)^2 - f(w + t ∇ f(w)) ≤ζ/2t
        Df(w).

  Let u = w + t ∇ f(w). If t is small enough then u ∈ B(x, ρ).
  Using Df(w)≥ζ this gives

    f(u) ≥ f(w) + tDf(w)^2 - ζ/2t
          Df(w)≥ f(w) + ζ/2tDf(w) = f(w) +
          ζ/2t∇ f(w).



  Because w ∈ K and t∇ f(w) = u-w, we further get

    f(u) ≥ f(x) + ζ/2(w - x + u - w) ≥ f(x) +
      ζ/2(u-x).

  Equation (<ref>) gives that u ∈ K while (<ref>)
  gives f(u) > M. This is a contradiction
with the choice of M.



 First, we can assume that all local extrema m_1,
  …, m_k are strict. Indeed, if some of them is not strict, say m_1,
  then we can find u ≠ m_1, …, m_k with Df(u) = 0 in a neighborhood
  of m_1.

Next, because k ≥ 3, there are at least two local maxima or two local
  minima among m_1, …, m_k. Without loss of generality, m_1 and m_2
  are local maxima. 
 
  Now, let us
  consider a path γ [0,1] → S^n such that γ(0) = m_1 and
  γ(1) = m_2. Let min_f(γ):=
  min{f(γ(t))   t ∈ [0,1]} (the minimum exists by compactness)
  and let s := sup(min_f(γ))
  where the supremum is taken over all γ as above. 

 Before we proceed with the formal proof, let us sketch the main idea of the
  proof; see also Figure <ref>. For contradiction assume that Df(u) ≠ 0 for every u ∈
  S^n∖{m_1, …, m_k}. Consider γ such that min_f(γ) is very close to s. We
  will be able to argue that we can assume that such γ is not close to
  any of the other extremes m_3, …, m_k. This guarantees that
  Df(γ(t)) is bounded from 0 for every t ∈ [0,1] except the
  cases when γ(t) is close to m_1 or m_2. Using
  Lemma <ref>, we will be able to modify γ to γ' with 
 min_f(γ') > s obtaining a contradiction with the definition of s.

  


 In further consideration, we consider the standard metric on S^n obtained by
  the standard embedding of S^n into ^n+1 and restricting the Euclidean
  metric on ^n+1 to a metric on S^n. For every i ∈ [k], we pick 
  two closed metric[By a metric ball we mean a ball with a given
  center and radius. This way, we distinguish a metric ball from a general
  topological ball.] balls B_i and B'_i centered in m_i. Namely, B_i
  is chosen so that m_i is a global extreme on B_i. We also assume that the
  balls B_i are pairwise disjoint. Next, we distinguish whether m_i is a
  local maximum or minimum. If m_i is a local maximum, let us define a_i :=
  max{f(x) x  ∈∂ B_i}. Note that f(m_i) > a_i as m_i is
  a global maximum on B_i. Then we pick a closed ball B'_i centered in m_i
  inside B_i so that f(x) > a_i for every x ∈ B'_i. 
  If m_i is a local
  minimum, we proceed analogously. We set a_i :=
    min{f(x) x  ∈∂ B_i} and we pick B'_i so that f(x)
    < a_i for every x ∈ B'_i. 
    For later use, we also define a'_i := min{f(x) x ∈ B'_i} for
    i ∈{1,2}. Note that a'_i > a_i. 
    
    Given a path γ connecting
    m_1 and m_2, we say that γ is avoiding if it does not
    pass through the interior of any of the balls B'_3, …, B'_k. 
    

  

    Let γ be a path connecting m_1 and m_2. Then there is an avoiding
    path γ̅ connecting m_1 and m_2 such that min_f(γ̅) ≥min_f(γ).
  

  
  Assume that γ enters a ball B'_i for i ∈{3, …, k}. Let us
    distinguish whether m_i is a local maximum or minimum.

    First assume that m_i is a local maximum. Then min_f(γ) ≤ a_i
    because γ has to pass through ∂ B_i. By a homotopy,
    fixed outside the interior of B'_i
    we can assume that γ avoids m_i (here we use
    n ≥ 2); see, e.g., the proof of Proposition 1.14 in <cit.>
    how to perform this step.[We point out that the current online
    version of <cit.> contains a different proof of Proposition 1.14.
    Therefore, here we refer to the printed version of the book.] In addition, by further homotopy fixed outside the
    interior of B'_i we can modify γ so that it avoids the interior of
    B'_i (the second homotopy pushes γ in direction away from m_i).
    This does not affect min_f(γ) because f(x) > a_i for every x ∈
    B'_i. 

    Next let us assume that m_i is a local minimum. Then min_f(γ) < a_i
    because γ has to pass through ∂ B'_i (this is not a
    symmetric argument when compared with the previous case). Modify γ
    by analogous homotopies as above; however, this time with respect to B_i
    (so that γ completely avoids the interior of B_i). Because
    min_f(γ) < a_i and f(x) ≥ a_i for x ∈∂ B_i, the
    minimum of γ cannot decrease by these modifications.
    By performing these modifications for all B'_i when necessary, we get the
    required γ̅.
  






  
  Now, let us consider a diffeomorphism ψ S^n ∖{m_k}→^n given by the stereographic projection (in particular, it maps closed
  balls avoiding m_k to closed balls). Let g ^n
  → be defined as g := f∘ψ^-1. Let n_i := ψ(m_i) for i
  ∈ [k-1].  Once we find v ∈^n, v ≠ n_1, …, n_k-1 such
  that Dg(v) = 0, then u := ψ^-1(v) is the required point with
  Df(u)=0.  Note that n_1, n_2 are still local maxima of g and n_3,
  …, n_k-1 are local maxima or minima. We also set D_i := ψ(B_i)
  and D'_i := ψ(B'_i) for i ∈ [k-1] and C_k := ψ(B_k ∖{m_k}), C'_k := ψ(B'_k ∖{m_k}). The sets D_i and D'_i
  are closed (metric) balls centered in n_i whereas C_k and C'_k are
  complements of open (metric) balls in ^n. Let K be the compact set
  obtained from ^n by removing the interiors of D'_1, …, D'_k-1,
  C'_k. Let us fix small enough η > 0 such that the closed
  η-neighborhoood K_η of K avoids n_1, …, n_k-1. We will
  also use the notation K_η/3 for the closed η/3-neighborhood
  of K. See Figure <ref>.



  Assume, for contradiction, that K_η does not contain v with Dg(v) =
  0. Because K_η is compact and g is C^1, there is ζ > 0 such
  that Dg(w)≥ζ for every w ∈ K_η. 

  For every w ∈
K_η/3 let N(w) be the neighborhood given by Proposition <ref> (the neighborhood is
considered in the whole ^n not only in K_η/3). 




By possibly restricting N(w) to smaller sets,
we can assume that each N(w) is open and fits into a ball of radius 2/3η. (In particular, if w ∈ K_η/3, then N(w) ⊆ K_η.)



  There is ε > 0 such that for every x ∈ K_η/3 the metric ball
  B(x,ε) ⊆^n centered in x of radius ε fits into N(w) for
  some w ∈ K_η/3.



This is just a modification of the Lebesgue number lemma.


  Let us consider the
  open cover Ø of K_η consisting of all sets N(w) together with the
  relative interiors of the sets B'_1 ∩ (K_η∖ K_η/3),
  …, B'_k-1∩ (K_η∖ K_η/3),
  C'_k ∩ (K_η∖ K_η/3) (all sets are relatively open in K_η). Note that the newly added sets are
  disjoint from K_η/3. Let ε > 0 be the standard Lebesgue number with
  respect to the cover Ø, that is, for every x ∈ K_η, the ball
  B(x,ε) fits into one of the sets of Ø; see <cit.>. Then the required claim
  holds with this ε because if x ∈ K_η/3, then x does not belong to
  any of the newly added sets of Ø.


Let ε be the value obtained from Claim <ref>. Because
some ball B(x, ε) fits into some N(w) which fits into a ball of
radius 2/3η, we get ε≤2/3η.




Let us consider a path γ in S^n such that

  
  * s - min_f(γ) < a'_1 - a_1;
  
  * s - min_f(γ) < a'_2 - a_2; and
  
  * s - min_f(γ) < ζε/4.

By Claim <ref>, we can assume that
γ is avoiding.
We will start modifying γ to γ' with
min_f(γ') > s, which will be the required contradiction. 
Let α :=
ψ∘γ; see the diagram at Figure <ref>. Then α connects n_1 and n_2, and α avoids
the interiors of D'_3, …, D'_k-1 and C'_k; see Figure <ref>. 


Because, α is a continuous function on the compact interval [0,1], we
get, by the Heine-Cantor theorem, that α is uniformly continuous. In particular, there is δ > 0
such that if t_1, t_2 ∈ [0,1] with |t_1 - t_2| ≤δ, then
α(t_1) - α(t_2)≤ε/3. Let us consider a
positive integer ℓ > 1/δ. We will be modifying α in two
steps. First, we get α” such that α”(t) > s if t =
j/ℓ for some j ∈{0, …, ℓ}. Then we modify α”
individually on the intervals (j/ℓ, j+1/ℓ) for j ∈{0,
…, ℓ-1} obtaining α' with min_g(α') > s. (Given a path β [0,1] →^n connecting n_1
and n_2, we define min_g(β) := min{g(β(t)) t ∈ [0,1]} =
min_f(ψ^-1∘β).)
The required γ' will be obtained as ψ^-1∘α'. 

For the first step, let us first say that an interval I_j = [j/ℓ,
j+1/ℓ] where j ∈{0, …, ℓ-1} requires a
modification if g(α(t)) ≤ s for some t ∈ I_j. This in
particular
means that α(t) ∈ K for this t: Indeed, this follows from (s1) and
(s2). We already know that α avoids the interiors of D'_3, …,
D'_k-1 and C'_k. It remains to check that α(t) does not belong to
the interiors of D'_1 and D'_2 as well. Because α has to meet
∂ D_1 and ∂ D_2, we get that min_f(γ) = min_g(α)
≤ a_1, a_2 from the definition of a_1 and a_2. By (s1) and (s2), we get
s < a'_1, a'_2. Therefore, from the definition of a'_1 and a'_2, we get
that α(t) cannot belong neither to D'_1 nor to D'_2 as required.

By the uniform continuity, the fact that g(α(t)) ≤ s for some t ∈
I_j implies that α(I_j) belongs to the closed
ε/3-neighborhood of K. In particular, α(I_j) belongs
to K_η/3 as ε≤2/3η < η.

Now, for each I_j which requires a modification, consider the open
ε-ball U_j ⊆^n centered in α(2j+1/2ℓ).
(Note that, 2j+1/2ℓ is the midpoint of I_j.) From the previous
considerations, the centre of each U_j belongs to K_η/3 and the whole
U_j is a subset of K_η.

Now we perform the first step. Consider t = j/ℓ for some j ∈{0,…, ℓ}. If g(α(t)) > s, then we do nothing. Note that this
includes the cases j = 0 or j=ℓ. If g(α(t)) ≤
s, then both intervals I_j-1 and I_j require a modification. 


By the uniform continuity, the open ball V_j
⊆^n
centered in α(t) of radius 2ε/3 is a subset of both
U_j-1 and U_j; see Figure <ref>. We observe that V_j is a subset of K_η as V_j
⊆ U_j.



In particular, by the
definition of ζ, we get that Dg(w)≥ζ for every w ∈
V_j. By Lemma <ref>, used on a closed ball of a slightly smaller
radius ε/2, there is a point v in V_j such that 

    g(v) ≥
    g(α(t)) + ζε/4≥min_g(α) +
    ζε/4 = min_f(γ) +
    ζε/4.

Using (s3), we get g(v) > s. Now, by a homotopy, we modify α to
α” so that
it stays fixed outside the interval (t - 1/4ℓ, t+1/4ℓ), the
modification of α occurs only in V_j and α”(t) = v; see
Figure <ref>.
We perform
these modifications simultaneously for every t = j/ℓ with
g(α(t)) ≤ 
s. This is possible as the intervals [t - 1/4ℓ, t+1/4ℓ] are
pairwise disjoint. This way, we obtain the required α”.








Finally, we perform the second step of the modification. Let I_j = [j/ℓ, j+1/ℓ] be an
interval requiring a modification. We already know that
g(α”(j/ℓ)) > s and g(α”(j+1/ℓ)) > s. In
addition, we know that both α”(j/ℓ) and
α”(j+1/ℓ) belong to U_j as they belong to V_j or
V_j+1. We set α'(j/ℓ) := α”(j/ℓ) and
α'(j+1/ℓ) := α”(j+1/ℓ). Next, we aim to
define α' on (j/ℓ, j+1/ℓ), which is the interior
of I_j, so that min(g(α'(I_j))) > s. By Claim <ref>, U_j
fits into some N(w) for some w ∈ K_η/3. (Here we use that the center
of U_j belongs to K_η/3.) Now, Proposition <ref> implies that
α'(j/ℓ) and α'(j+1/ℓ) may be connected by
a path P[0,1] →^n such that g(P(t)) > s for every t ∈ [0,1]:
Indeed, let us assume that, without loss of generality,
g(α'(j/ℓ)) ≥ g(α'(j+1/ℓ)) > s. First, draw
P as a straight line from α'(j/ℓ) towards α'(j+1/ℓ) until
we reach a (first) point x ∈ U_j ⊆ N(w) with g(x) =
g(α'(j+1/ℓ)); of course, it may happen that x =
α'(j+1/ℓ). Then by Proposition <ref>, x and
α'(j/ℓ) can be connected within the level set L_g(x);
see Figure <ref>.
(This may mean that P leaves N(w), or even K_η, but this is not problem for the
argument.)
Altogether, we set α' on I_j so that it follows the path P, and this
we do independently on each interval requiring a modification. Other intervals
remain unmodified. 


From the construction, we get min_g(α') > s; therefore the path γ'
:= ψ^-1∘α' satisfies min_f(γ') = min_g(α') > s which
contradicts the definition of s.





§ DEPTH-LIKE FUNCTIONS WITH FEW CRITICAL POINTS






    Bipyramid over a triangle.
 In ^3, we have a candidate example
of a convex body, namely the regular bipyramid B over an equilateral
triangle T, such that there are exactly four barycentric hyperplanes (with
respect to the barycenter of B, which coincides with the point of maximal
depth in this case).
On the one hand, this is not surprising, because this is n + 1
hyperplanes, where n = 3 is the dimension of the ambient space. On the other
hand, if this is true, then it answers negatively, in dimension 3, a question 
from <cit.>, whether 2^n - 1 barycentric hyperplanes
always exist.

More concretely, we conjecture that the only barycentric hyperplanes are the
following: three planes perpendicular to T which meet T in lines realizing
the depth of T (these would be the hyperplanes realizing the depth), and the
plane of T (this is the one extra plane).  Unfortunately, in this case, it is
not so easy to analyze the depth function as in the case of T × I.





    A function with four critical points and many properties of the depth.

Let us recall that the depth function δ S^n-1→ [0,1] 
on a convex body satisfies the following properties: 




 
  *  δ(v) = 1 - δ(-v);
 
  *  0 ∈ U where U ⊆ S^n-1 is the set of
   the points where δ attains the minimum (by Proposition <ref>);
 
  *  |U| ≥ 3, (by Corollary <ref>);
 
  *  δ is C^1 (by Proposition <ref>);
 
  *  if U is finite, then δ has at least one more pair of opposite critical points (by Proposition <ref>
and by (<ref>)).


We will show that our argument in the proof of Theorem <ref> is tight in
the sense that for n ≥ 3 there exists a function δ'  S^n-1→ [0,1]
satisfying (<ref>)–(<ref>) with equalities in (<ref>) and (<ref>). 










In order to define δ', it will be much more convenient to reparametrize
δ'. Thus, we will exhibit δ” S^n-1→ [-1, 1] which
satisfies (<ref>)–(<ref>) with equalities in (<ref>) and (<ref>) but δ”(v) =
-δ”(-v) instead of (<ref>). Then the required δ' is obtained as
1/2δ” + 1/2.

This time we decompose ^n as ^n-2×^2 and for a point x ∈^n we write x = (y; z) where y =
(y_1, …, y_n-2) ∈^n-2 and z = (z_1, z_2) ∈^2. The idea is to define δ”
separately on the sphere S^n-3×{0} so that there is only one pair
of opposite critical points here (this will be the extra pair from (<ref>)),
separately on the sphere {0}× S^1 so that there are three pairs of
critical points (these will be three global minima and three global maxima from
(<ref>)), and then merge the two constructions so that the resulting function is
smooth and no new critical points arise. Unfortunately, the details are
somewhat tedious.

We actually define δ” on ^n∖{0} considering S^n-1 as a subset of ^n∖{0}. 

Now, we set

    δ”(y,z) = 1/10(2y - y^3)y_1 + 1/2z(z_1^3 - z_1z_2^2 - 2z_1z_2^2).

We remark that the expression (z_1^3 - z_1z_2^2 - 2z_1z_2^2) is nothing else
then the real part (z^3), where z = (z_1, z_2) is identified with the
complex number z_1 + iz_2. From (<ref>) we easily see that
δ” is smooth on  ^n∖{0}, therefore its restriction to S^n-1 is smooth as well as
the inclusion S^n-1⊆^n∖{0} is a smooth embedding. 
We also easily check that δ”(y,z) = - δ”(-y, -z).

From now on, let us assume that (y,z) ∈ S^n-1, that is y^2 + z^2 = 1.
If y ≠ 0, we get

    (2y - y^3)y_1 = (1 - (1 - y^2)^2)y_1/y =
      (1-z^4)y_1/y,

and if z ≠ 0, we get (in complex numbers)

    z(z_1^3 - z_1z_2^2 - 2z_1z_2^2) = z^4  (z^3)/z^3 =
      z^4 ((z/z)^3).

Altogether (<ref>), (<ref>) and (<ref>)
give

    δ”(y,z) = 
      
        (1-z^4)1/10y_1/y + z^4 1/2((z/z)^3)
      if  y,z ≠ 0,
    1/2(z^3) 
      if  y = 0,
    1/10y_1
     if  z = 0.


In particular, (<ref>) implies that for y,z ≠ 0, δ”(y,z) is a convex combination of 1/10y_1/y and 1/2((z/z)^3), which attain values in
[-1/10, 1/10] and [-1/2,1/2] respectively.  Therefore
δ”(S^n-1) ⊆ [-1/2,1/2].

Now we check that δ” attains exactly three global minima on S^n-1.
We observe that δ”(0,e^i(2k+1)/3)π) = -1/2 for k = 0,1,2
by (<ref>). Therefore δ” attains the
minimum at these three points. On the other hand, we realize that these are the
only three points where δ”(y,z) = -1/2. Indeed, if y = 0, then
δ”(y,z) = -1/2 only if (z^3) = -1, which occurs only if
z = e^i(2k+1)π/3 for k = 0,1,2. If z = 0, then δ”(y,z) ≥
-1/10 by (<ref>). Finally, if y,z ≠
0, then the convex combination from (<ref>) 
has the strictly
positive coefficient (1- z^4) at 1/10y_1/y, which
implies δ”(y,z) > 1/2((z/z)^3) ≥ -1/2. This
characterization of global minima also gives property (<ref>).

It remains to check that there is exactly one extra pair of opposite critical
points of δ”. This could be done via Lagrange multipliers but the
computations seem to be slightly tedious, thus we provide a different argument.
In advance, we announce that these extra critical points
will be (e_1, 0) and (-e_1,0), where e_1 ∈ S^n-3⊆^n-2 is the first coordinate
vector e_1 = (1,0,…, 0). We will rule out all other options, thus
these points have to be indeed critical by Proposition <ref>.

Let (y,z) ∈ S^n-1 be a critical point. If y = 0, then δ”(0, z) =
1/2(z^3) by (<ref>) when restricted to {0}×
S^1 ⊆ S^n-1 (where 0 ∈^n-2 in this case). Therefore (0,
z) has to be critical point of the restriction as well. It is easy to analyse
that the only critical points of 1/2(z^3) are of the form z =
e^ikπ/3 where k ∈{0,…, 5}, which are the minima and the maxima
opposite to the minima. If z = 0, then δ”(y,0) = 1/10y_1
when restricted to S^n-3×{0}⊆ S^n-1 (where 0 ∈^2 in this case). Again, it is easy to analyze that e_1 and -e_1 are the
only critical points. (Here they are the maximum and minimum in the restriction
respectively, but they are not even local extremes on whole S^n-1.)
Finally, we consider the case y, z ≠ 0. First, we fix y and let z vary
subject to y^2 + z^2 = 1, which implies that z is fixed as well. Then δ”(y, z) =  a_y + b_y((z/z)^3) by (<ref>) where
a_y and b_y are constants depending on y. This implies that if (y,z) is
critical, then z/z = e^ikπ/3 where k ∈{0,…, 5}. Next, we fix
z and let y vary. By a similar idea as above, we deduce that if (y,z) is
critical, then y = (± y_1, 0,…,0). Finally, let us fix both y and z
and consider the
2-plane ρ(y,z) given by all vectors (t_y y, t_z z) for t_y, t_z ∈. This 2-plane meets S^n-1 in a circle. For (t_y y, t_z z) in
ρ(y,z) ∩ S^n-1 the equation (<ref>) gives

    δ”(t_y, t_z)    = (1 - t_z^4 z^4)1/10y_1/y +
      t_z^4z1/2((z/z)^3)
       = 1/10y_1/y + t_z^4
      z^4(1/2((z/z)^3) -
      1/10y_1/y).

Therefore (t_y y, t_z z), for t_y, t_z ≠ 0, may be the critical point of
δ” only if 

    1/2((z/z)^3) = 1/10y_1/y

 which is independent of the values t_y and t_z. If we recall the previous
 two conditions on the critical point (y,z), we get ((z/z)^3) = ±
 1 and y_1/y = ±1, therefore (<ref>) may not hold
 simultaneously. This finishes the analysis of the critical points of
 δ”.




§ ACKNOWLEDGEMENTS
 We thank Stanislav Nagy for introducing us to Grünbaum's questions,
for useful discussions on the topic, for providing us with many references, and
for comments on a preliminary version of this paper. We thank Jan Kynčl and
Pavel Valtr for letting us know about a more general counterexample they found.
We thank Roman Karasev for providing us with
references <cit.> and for comments on a
preliminary version of this paper. Finally, we thank an anonymous referee for
many comments on a preliminary version of the paper which, in particular,
yielded an important correction in Section <ref>.















