Let $u$ be a smooth function defined on the unit sphere $S^2$. Assume $u$ has two local maxima, two local minima, and two saddle points (a total of 6 critical points). Does there exist a plane $P$ passing through the origin such that $P\cap S^2$ contains at least three points $x_1,x_2,x_3$ with $\nabla u(x_i) \cdot n =0$, $i=1,2,3$, where $n$ is a vector normal to the plane $P$?

$\begingroup$ Do you know if what you want holds for all functions on $S^2$ with just two critical points? I.e., can you always find a geodesic orthogonal to three level sets of any function with just one local max and local min? $\endgroup$– Dmitri PanovNov 24, 2018 at 10:54

$\begingroup$ The result might be also true when there is only 2 critical points, but I am not sure. Having 6 critical points the result is most likely true. $\endgroup$– A random mathematicianNov 24, 2018 at 14:23

$\begingroup$ The plane $P$ cuts a great circle $C$ on the standard unit sphere $S^2$ (centered at the origin of $\mathbb{R}^3$). The condition $\nabla u (x)\cdot n=0$ means that the great circle $C$ is orthogonal at $x$ to the level curve of $u$ through $x$, or $x$ is one of the six critical points (We call this condition $Cond(C,x)$). We require that there is a great circle $C$ and three points $x_1,x_2,x_3$ on $C$ such that $Cond(C,x_i)$ holds for $i=1,2,3$. Do I get it correct? $\endgroup$– QfwfqNov 26, 2018 at 20:43

$\begingroup$ @Qfwfq Yes, you are right. $\endgroup$– A random mathematicianNov 26, 2018 at 22:28

1$\begingroup$ I thought the bounty is going to expire, so I wanted to give it to the best at that point. There seems to be no undo button. $\endgroup$– A random mathematicianDec 6, 2018 at 17:23
6 Answers
The goal of this updated answer is not to solve the question, but to write down some observations about it, mainly to illustrate some simple ideas that can be applied to the problem. I tried to make it an easy read. Note that so far the question stays completely open (as for 10/12/2018).
First, I want to slightly reformulate the question and state it in a formula free way.
Question 1. Let $u$ be any smooth function on the unit sphere $\mathbb S^2$. Is it true that there is a great circle on $\mathbb S^2$ tangent to the gradient of $u$ at three or more points? What if $u$ is a Morse function with exactly six critical points?
In other words, we are looking for a great circle orthogonal to the level sets of $u$ in at least three points. Since $\nabla u$ is zero at each critical point, in case a great circle passes through a critical point, it is tangent to $\nabla u$ at it.
An easier problem. When I first red the question, I confused it with the following one, which has a very simple solution, suggested by Pietro Major.
Exercise. Prove that for any smooth function $u$ on $\mathbb S^2$ there is a great circle tangent to three or more level sets of $u$.
Solution. We will assume the converse and deduce a contradiction with the Hairy ball theorem, https://en.wikipedia.org/wiki/Hairy_ball_theorem
Note that a circle $C$ is tangent to the level set $u=c$ at a point $p$ iff the restriction of $u$ to $C$ has a critical point at $p$. Now, by our assumptions, $u$ has exactly two critical points when restricted to any great circle $C$. Let us deduce that there is a nonvanishing vector field on $\mathbb S^2$. Indeed, take any point $x\in \mathbb S^2$. Let $C(x)$ be the great circle opposite to $x$. By our assumptions there exits a unique point $M(x)$ on $C(x)$ where $u_{C(x)}$ attains its maximum. Note, that $M(x)$ depends continuously on $x$. Now, let us take the segment $[xM(x)]$ of length $\frac{\pi}{2}$ an let $v(x)$ be the unit tangent vector to $[xM(x)]$ at the point $x$. We have constructed a continuous vector field $v(x)$ on $\mathbb S^2$ without zeros, which contradicts the Hairy ball theorem. QED.
Remark. I have not encountered Question 2 previously, and I am curious if this type of questions is well known. There seem to be plenty of ways to generalise it. For example, one can study round spheres $\mathbb S^n$ with $n>2$. Then for any smooth function $u$ on $\mathbb S^n$ there will be a hypersphere tangent two at least three level sets of $u$.
Back to Question 1. Now, I want make a few observations, some of which were made in other answers, concerning the original problem. I don't have an opinion whether the answer to Question 1 is positive or not, and think that it's worth to spend time looking for potential counterexamples, especially among function with exactly two critical points. In particular, there exist functions on $\mathbb S^2$ for which each great circle is orthogonal to at most $4$ levels. Also, I believe that there exist functions with arbitrary large number of critical points, such that each great circle is perpendicular to at most $10$ level sets (not certain if $10$ can be made $4$).
The next observation singles out a large class of functions for which one can find a circle orthogonal to at least $4$ level sets, this observation was used, in particular, by RBega2.
Observation. Let $u$ be a smooth function on $\mathbb S^2$. Suppose that there is a half sphere $D\subset \mathbb S^2$ with $k$ local minima or maxima and $l$ saddle points in it. Suppose that the great circle $\partial D$ doesn't pass through critical points of $u$. Then it is orthogonal to at least $2kl1$ level sets.
Proof. Consider the gradient vector field $\nabla u$ in $D$. Since $u$ has $k$ local minima or maxima and $l$ saddle points in $D$, the vector field $\nabla u$ has winding number $kl$ along $\partial D$. Since the unit tangent vector field to $\partial D$ has winding number $1$ we see that these two fields are perpendicular in at least $2kl1$ points on $\partial D$. QED.
One may wonder if this observation will give a positive answer to Question 1 for functions with a large enough number of critical points, just by studying the position of these points on $\mathbb S^2$. But this is not the case as the following lemma shows. In this lemma the set $X$ is interpreted as the set of local minima and maxima of a supposed function and $Y$ as the set of its saddle points, with $\#(X)\#(Y)=2$.
Lemma. For any $n\ge 2$ there is a set of $2n2$ points $$\{x_1,\ldots,\,x_n,\,y_1,\ldots,\,y_{n2}\}=X\cup Y\subset \mathbb S^2$$ such that for any halfsphere $D\subset \mathbb S^2$ one has $$\#(D\cap X)\#(D\cap Y)1\le 1.$$
Proof. Instead of constructing the set $X\cup Y$ in $\mathbb S^2$ we will construct it in $\mathbb R^2$, in which case the hypothesis should be checked for all halfplanes instead of halfspheres. We choose points $x_1,\ldots,x_n$ as the consecutive vertices of a regular $n$gone. Let $x_{n1,n}$ be the mid point of the segment $x_{n1}x_n$. Then for a small $\varepsilon$ and $i\in \{1,n2\}$ we set $y_i=(1\varepsilon)x_i+\varepsilon x_{n1,n}$. It is not hard to check that the constructed collection $X\cap Y$ satisfies necessary conditions. QED.
Next, I want to present a relatively explicit smooth function on $\mathbb S^2$ with two critical points for which all great circles are orthogonal to at most $4$ level sets.
Example. Instead of constructing $u$, we will construct its level sets. All the level sets will be constant curvature circles. Let us denote by $\gamma_{\varepsilon}(t)$ a length $\varepsilon$ arc of a fixed radius $\frac{\pi}{2}$ circle on $\mathbb S^2$, parametrized by its length. Using the notation $C(x,r)$ for a circle of radius $r$ centred at $x$, consider the following $1$parameter family of circles on $\mathbb S^2$:
$$C_t=C(\gamma_{\varepsilon}(t/\varepsilon), \pi/\varepsilon).$$
I claim that for the constructed family of circles (which are all disjoint) any great circle can be orthogonal at most four times to this family. Indeed, to be orthogonal to a circle, a great circle has to pass through its centre, but any great circle intersects $\gamma_{\varepsilon}$ in at most $2$ points, and $2\times 2\le 4$. QED.
What to do next? I would say, that in order to make some progress in the original problem, one should be able to solve the problems of the following type.
Question 3. Let $u$ be a smooth function on $\mathbb R^2$ with Morse point at $(0,0)$. Is it true, that for any $\varepsilon$ small enough there is a line $L$ in $\mathbb R^2$ on distance less than $\varepsilon$ from $(0,0)$ that is orthogonal to the level sets of $u$ in at least four points of $L$ in the ball $x^2+y^2\le \varepsilon^2$?
In other words, it might easily happen that the solution to Question 1 will come not from some global considerations, but from some local analysis.
I also think that the following question is interesting and should probably be solved if one wants to solve the original question.
Question 4. Does Question 1 has a positive solution for functions that at $C^{\infty}$ small perturbations of the coordinate function $x$ in $\mathbb R^3$ restricted to $\mathbb S^2$?

1$\begingroup$ Thank you Dmirti, I very much like your argument. However, my questions asks for three points at which the directional derivative is zero in the direction of the normal vector for a plan $P$. So we want the derivatives in the normal direction to a great circle to be zero, and not in the tangential direction. Clearly three critical points of $u$ restricted to a big circle does not imply the result. $\endgroup$ Nov 23, 2018 at 16:40

3$\begingroup$ Please keep the answer. In case I don't get a response for my question, I will revise the question to the one you answered and will accept the answer. Even if I get a response, we can form a different question and answer. The question you have answered is very interesting. $\endgroup$ Nov 23, 2018 at 19:10

$\begingroup$ Lemma is nice. Doesn't it hold just for any continuous function on $S^2$? (In the form: there is a great circle $C$ such that $fC$ has at least 2 local maxima). $\endgroup$ Nov 26, 2018 at 22:15

1$\begingroup$ Yes, it seems true. Proof: Assume by contradiction that $fC$ has a unique maximum point $P(C )$ for any great circle $C$. Then $P(C )$ depends continuously from $C$ by uniqueness+compactness. But after a halfrotation around any diameter of $C$, $C$ is mapped to $C$ again, which implies that $P(C )$ was on the diameter. But the diameter was arbitrary so $f$ is constant, a contradiction! $\endgroup$ Nov 26, 2018 at 22:28

4$\begingroup$ @PietroMajer You are right :) This is en.wikipedia.org/wiki/Hairy_ball_theorem :) . Indeed, for any point $x$ on $S^2$ choose the opposite circle and take on the circle the point $y$ where the function attains its max. Then choose the unit vector at $x$ that points to $y$. So one gets a continuos vector field on $S^2$, which contradicts the hairy ball theorem $\endgroup$ Nov 26, 2018 at 23:53
EDIT As DimitriPanov points out Claim 1' is false and this leads to an unfixable gap. I'll leave the answer in case someone finds it useful.
Ironically, it looks like I can salvage the proof only for any Morse functions $u$ with at most 4 critical points.
First some notation.
For any $\mathbf{v}\in \mathbb{S}^2$ let $D(\mathbf{v})=\{\mathbf{y}\in \mathbb{S}^2: \mathbf{y}\cdot \mathbf{v}\leq 0\}$ be the hemisphere on the side of the plane normal to $\mathbf{v}$ that does not contain $\mathbf{v}$. Let $C(\mathbf{v})=\partial D(\mathbf{v})$ be the great circle at the boundary of $D(\mathbf{v})$ whose (outward) normal is $\mathbf{v}$. Let $P=\{p: \nabla u (p)=0\}$ be the set of critical points of $u$. As $u$ is Morse, this is a finite set with an even number of points.
Assumption: Let us now assume that $\nabla u \cdot \mathbf{v}$ has at most two zeros on $C(\mathbf{v})$ for all $\mathbf{v}$.
I claim this assumption will lead to a contradiction and hence prove what you wanted, but for strictly fewer critical points than you requested.
Let $\Omega\subset \mathbb{S}^2$ be the set of $\mathbf{v}$ so that $\nabla u \cdot \mathbf{v}$ is never zero on $C(\mathbf{v})$ and let $\Omega'$ be the set of $\mathbf{v}$ so that $\nabla u \cdot \mathbf{v}$ is both positive and negative on $C(\mathbf{v})$. Notice both $\Omega$ and $\Omega'$ are open sets.
The assumption implies that, for $\mathbf{v}\in \Omega'$, $\nabla u\cdot \mathbf{v}$ vanishes at exactly two points on $C(\mathbf{v})$. It also ensures that if $\mathbf{v}\not\in \Omega\cup \Omega'$ then $\nabla u\cdot \mathbf{v}$ has a sign away from either one or two points.
Claim 1: (This has been corrected thanks to DmitriPanov's observation). Suppose $\mathbf{v}$ is such that $C(\mathbf{v})\cap P=\emptyset$. If $D(\mathbf{v})\cap P$ contains an even number of points, then $\mathbf{v}\in \Omega'$.
Proof: This follows from the PoincareHopf index theorem applied to the vector field $\nabla u$. Basically, the standing assumption ensures the boundary contributes $1$ (when $\mathbf{v}\not\in\Omega'$) and critical points contribute either $1$ (for maximum or minima) or $1$ (for saddle points). The claim follows by (for instance) working mod 2.
Claim 1': Suppose $\mathbf{v}$ is such that $C(\mathbf{v})\cap P=\emptyset$. If $\mathbf{v}\in \Omega'$, then either $D(\mathbf{v})\cap P$ contains an even number of points or both $D(\mathbf{v})\cap P$ and $D(\mathbf{v})\cap P$ contains at least three points of $P$.
Proof: PoincareHopf tells us that if $\nabla u$ contributes $0$ mod 2 along $C(\mathbf{v})$, then $D(\mathbf{v})\cap P$ contains an even number of points. Now suppose, $\nabla u$ contributes $1$ mod $2$ along the boundary. Let $\mathbf{T}:C(\mathbf{v})\to \mathbb{R}^3$ be the unit tangent to $C(\mathbf{v})$ compatible with the orientation. Write along $C(\mathbf{v})$ $$ \nabla u(p)=r(p) \cos \theta(p) \mathbf{T}(p)+r(p)\sin \theta(p) \mathbf{v} $$ where here $\theta:C(\mathbf{v})\to \mathbb{S}^1$ and $r>0$. The standing assumption ensures that $\sin \theta(p)=0$ only at two points $p_$ and $p_+$ and $\mathbf{v}\in \Omega'$ means that, up to relabelling $\sin(\theta(p))<0$ on $C_(\mathbf{v})$, the region between $p_$ to $p_+$, (relative to the orientation) and $\sin(\theta(p))>0$ on $C_+(\mathbf{v})$, the region between $p_+$ and $p_$. One verifies (e.g., by drawing a graph) that $\nabla u$ contributes $0$ to PoincareHopf if and only if $\cos(\theta(p_+))=\cos(\theta(p_))$ (e.g. $\theta(p_+)=0, \theta(p_)=\pi$ and) while $\nabla u$ contributes $1$ when $\cos(\theta(p_+))=\cos(\theta(p_))$ (e.g. $\theta(p_+)=\theta(p_)=0$).
In this latter case (which is the one we are interested in), the fact that $\nabla u$ is a gradient vector field, implies there are at least two points in both $C_(\mathbf{v})$ and $C_+(\mathbf{v})$ where $\sin \theta$ vanishes. In fact, one the two points will be a local maximum of $u_{C_\pm (\mathbf{v})}$ and the other will be a local minimum of $u_{C_\pm (\mathbf{v})}$. Let $q_+$ be the local maximum in $C_(\mathbf{v})$ and $q_$ be the local minima in $C_+(\mathbf{v})$. By analayzing level sets one obtains corresponding local maxima and minima of $u$, $q_'$ and $q_+'$ in $D(\mathbf{v})$ ** Edit: This reasoning only works if $q_$ and $q_+$ are actually absolute maximam on $C(\mathbf{v})$  i.e. if $q_$ only a local minima can't guarantee existence of $q_'$. **. That is, $D(\mathbf{v})$ contains at least two critical points of $u$ and so, by PoincareHopf, at least three points. This argument is symmetric in $\mathbf{v}$ and $\mathbf{v}$ so $D(\mathbf{v})$ also contains at least three critical points of $u$.
Returning to the main argument: Let $\Gamma=\bigcup_{p\in P} C(\mathbf{v}(p))$ where $\mathbf{v}(p)$ is chosen so $\mathbf{w}\in C(\mathbf{v}(p))$ if and only if $p\in C(\mathbf{w})$. Our standing assumption and the lower bound on the number of cirtical points ensures $C(\mathbf{v}(p))\neq C(\mathbf{v}(p'))$ for $p \neq p'$ as if $C(\mathbf{v}(p))=C(\mathbf{v}(p'))$ then $p$ and $p'$ are equal or antipodal. In the later case, one can find a $C(\mathbf{w})$ containing $p,p'$ and a third element $p''$ so $\nabla u(p'')\cdot \mathbf{w}=0$  contradicting the standing assumption (see User4966's argument below). Moreover, the standing assumption implies that $C(\mathbf{v}(p_1))\cap C(\mathbf{v}(p_2))\cap C(\mathbf{v}(p_3))=\emptyset$ for $p_1, p_2, p_3$ distinct. Indeed, if $\mathbf{w}\in C(\mathbf{v}(p_1))\cap C(\mathbf{v}(p_2))\cap C(\mathbf{v}(p_3))$ then $p_1,p_2, p_3\in C(\mathbf{w})$ so $\mathbf{w}$ violates the standing assumption.
Let $\Omega''=\mathbb{S}^2 \backslash \overline{\Omega'}$ so $\Omega''$ is open. Clearly, $\Omega\subset \Omega''$ (though they might not be equal).
Claim 2: If $P$ has at most $4$ points then, both $\Omega'$ and $\Omega''$ are nonempty and $\mathbb{S}^2\backslash \Gamma=\Omega'\cup \Omega''$ and each component of $\Omega'$ is contained in an open hemisphere.
Proof: Pick a $C(\mathbf{w})$ so $C(\mathbf{w})\cap P=\{p\}$. That is, so $\mathbf{w}\in C(\mathbf{v}(p))\subset \Gamma$ but $\mathbf{w}\not\in C(\mathbf{v}(p'))$ for any other $p'\in P$. By perturbing $\mathbf{w}$ slightly to $\mathbf{w}'$ one can ensure $C(\mathbf{w}')\cap P=\emptyset$ and $D(\mathbf{w}')\cap P=D(\mathbf{w})\cap P$ or $D(\mathbf{w}')\cap P=(D(\mathbf{w})\cap P)\backslash \{p\}$. Depending on which of the two cases occur $D(\mathbf{w}')\cap P$ contians either an even or odd number of points. Furthermore, as $P$ has at most $4$ elements $D( \mathbf{w}')$ and $D(\mathbf{w}')$ can't both contain at least three points of $P$. Hence, by Claim 1 and Claim 1', we can make $\mathbf{w}'$ lie in either $\Omega'$ or in $\Omega''$. That is $\mathbf{w}\in \partial \Omega'$ and also $\mathbf{w}\in \partial \Omega''$. It follows that $\Omega'$ and $\Omega''$ are nonempty and $\mathbf{w}\not\in \Omega'\cup \Omega''$. The second part of the claim follows as each component of $\Omega'$ lies in one of the components of $\mathbb{S}^2 \backslash C(\mathbf{v}(p))$ for some $p\in P$.
However we also have
Claim 3: There exists a connected component of $\Omega'$ that contains two antipodal points.
Proof: Let $F:\mathbb{S}^2 \to \mathbb{R}$ be defined by $$ F(\mathbf{v})=\int_{C(\mathbf{v})} \nabla u \cdot \mathbf{v}. $$ Clearly, $F$ is a continuous odd function. Our standing assumption ensures that $F^{1}(0)\subset \Omega'$. I claim that there is a connected component $Z$ of $F^{1}(0)$ that contains two antipodal points  see the second answer to this question for a proof (I didn't check this in detail but it seems intuitively clear). This implies that there is such a component for $\Omega'$ and proves the claim.
Since Claim 2 and Claim 3 are in contradiction the standing assumption is false, which gives what you want when $u$ has at most 4 critical points.

$\begingroup$ Thanks. Your approach sounds very reasonable. I will check the details. $\endgroup$ Dec 5, 2018 at 18:09

1$\begingroup$ Sorry RBega2 and User4966, I come back to the conclusion, that Claim 1 is not correct in the form it is stated. Namely, what RBega2 proves, is that $v\in \Omega'$ if $D(v)$ contains an even number of points. But the converse is not proven in the claim. It is not proven there that if $D(v)$ contains an odd number of points, then $v$ is not in $\Omega'$. Do you agree? If yes, could you please correct? I don't know if this affects what goes further. $\endgroup$ Dec 7, 2018 at 8:08

2$\begingroup$ I agree with the fact that the boundary contributes $1$ in this case. But I don't see how this will precludes $\nabla u\cdot v$ to vanish twice on $C(v)$. It is easy to construct a vector field along a great circle such that its index around the circle is equal to $1$, however it is perpendicular to the circle at two points. $\endgroup$ Dec 7, 2018 at 8:31

1$\begingroup$ @RBega2, thanks for the update. I have a question concerning Claim1'. Could you point me to the place in its proof, where you use the global assumption of the problem  that EACH great circle in $S^2$ is perpendicular to at most two level sets of the function. If one doesn't use this assumption, the claim can not work. So far I only see that in its proof you are working with one circle  $C(v)$. I can construct a function on $S^2$ with just two critical values, such that there is a great circle that splits two critical points and is perpendicular to exactly two level sets. $\endgroup$ Dec 8, 2018 at 9:42

1$\begingroup$ @DmitriPanov After taking a walk I realized how to construct an example like you describe and realize it illustrates a (seemingly unfixable) gap. $\endgroup$– RBega2Dec 8, 2018 at 15:00
This isn't a complete answer: it only shows how far a simple topological argument can get, and proves a weaker result.
Given a great circle the special points you are looking for are the ones where the hemisphere to which the gradient points to changes. Hence their number is generically even. A sufficient condition for having at least four special points is that the degree of the gradient map along the great circle is either at least 3 or less than 3. That degree is the sum of indices of critical points within a hemisphere (1 for an extremum, 1 for a saddle). Unfortunately there are configurations of critical points for which there exist no such hemisphere... However the result is true if one looks at arbitrary circles rather than great circles, as one can see by perturbing a circle going through three critical points.

$\begingroup$ There seems to be a few gaps in the argument. The number of special points could be odd. Also " If there are only two such points on a great circle, this means the great circle is split in only two regions, hence by index theory each hemisphere has index zero, which is impossible" does not seem to be true. One can easily construct a function which has exactly two such points on a great circle. $\endgroup$ Nov 28, 2018 at 5:43

$\begingroup$ Not seeing how special points could be odd, but I agree with your second concern. I'm pretty sure that type of argument is the right one, although it needs to be fixed $\endgroup$– alesiaNov 28, 2018 at 18:58

1$\begingroup$ ok the argument doesn't suffice. I edited the answer. $\endgroup$– alesiaNov 28, 2018 at 20:22
I think you can find $x_1,x_2,x_3$. I'm not sure what follows is a(n almost) proof or something totally useless, also because here it's quite late at the moment.
I'll occasionally argue informally, and I'll also say "line" to mean "great circle in $\mathbb{S}^2$".
1) Let $S$ be one of the two saddle points, and set $x_1:=S$. Let $C_\infty$ be the "$\infty$"shaped critical locus to which $S$ belongs [Edit: okay, you can have other possibilities for the topology of the critical locus, but I don't think it matters so much: we could consider a maximum point instead of the point $S'$ below. Further edit: thinking more carefully, I believe the only possibility for the critical locus is being two disjoint $8$shaped curves with the saddles being the singular points, and containing the maxima and minima inside the "petals" of the two $8$shaped curves, exactly one for petal]. Consider the two tangent lines $r_1,r_2$ to the critical locus $C_\infty$ passing through $S$. Consider one of the two (closed) angle sectors determined by these two tangents that contains a topological circle subset of $C_\infty$. Considering lines $r_1'$ (resp. $r_2'$) through $S$ and close to $r_1$ (resp. $r_2$) in the interior of said angle sector and considering the angles formed with the curve $C_\infty$ (which is a smooth curve away from $S$) we see those angles have opposite sign, so there must be a point $Q\in C_\infty \setminus S$ at which the line $r$ (through $S,Q$) is orthogonal to $C_\infty$ at $Q$.
2) Consider a regular level curve $E$ very close to $C_\infty$ and such that its "Jordan disk" (on the side of $S$) contains $C_\infty$ in its interior. Consider the (most away from $S$, in the sector of $Q$) point $P$ of $E\cap r$. If a critical point different from $S$ belongs to $r$, we're done ($x_2:=Q$ and $x_3:=$ that point). Otherwise, call the hemisphere $D_{}$ determined by $r$ and $S'$ (the other saddle point) "internal" and the other one "external". Consider the tangent $\ell$ to $E$ at $P$, and the angle it forms with the orthogonal to $r$ at $P$. If the "outside part" of $\ell$ is leaning towards $S$ then set $P'':=P$, $r':=r$, and skip step 3).
3) If the "outside" of $\ell$ is not leaning towards $S$, then there is an "internal" (i.e. belonging to the hemisphere determined by $r'$ and $S'$) point $P'$, lying on $E$ and close to $P$, such that the line $r'$ through $SP'$ is orthogonal to $E$ at $P'$. Set $x_2:=P'$.
A regular level curve $E'$ very close to $C_\infty$ and outside its Jordan disk will meet $r'$ in a point $P''$ that "leans towards" $S$ (meaning the outside part $\ell''$ of the tangent to $E'$ at $P''$ is leaning towards $S$).
4) Let $E''$ be a regular level curve whose Jordan disk (not containing $S$) contains $S'$, and all of its critical locus $C_\infty'$, in its interior. Now we have a $1$parameter family of regular level curves $\{C_t\}_{t\in[a,b]}$ with $C_a=E''$ and $C_b=E'$. Let $D_t$ be the Jordan disk of $C_t$ containing $S'$. Let
$$\bar{t}:=\mathrm{sup}\{t\in[a,b]\mid D_t\subseteq D'_{}\}$$
(here D_{}' is to $r'$ what $D_{}$ was to $r$).
At points of $r'\cap C_{\bar{t}}$, $C_{\bar{t}}$ is tangent to $r'$.
5) Let $T$ be the tangent point in $r'\cap C_{\bar{t}}$ nearest to $P''$ [Edit: actually, not just the nearest, but the nearest starting from $P''$ and traveling along $r'$ in the direction opposite to $E'$]. For $0<\epsilon<<1$, consider $C_{\bar{t}+\epsilon}$. Near $T$, $C_{\bar{t}+\epsilon}$ intersects $r'$ in two points $T_{+},T_{}$. Let $T_{+}$ be the one closest to $P''$. The oriented angles
 that $E'$ forms with the orthogonal to $r'$ at $P''$, and
 that $C_{\bar{t}+\epsilon}$ forms with the orthogonal to $r'$ at $T_{+}$
have opposite signs. So there is a $t\in[\bar{t}+\epsilon,b]$ such that $C_t$ is orthogonal to $r'$. Set $x_3:=C_t\cap r'$.
Edit1:: Actually, I have missed the case in which the other critical locus $C_\infty'$ is not contained in the "interior" i.e. in the hemisphere determined by $r'$ and $S'$. But I think one could directly consider a nonsaddle critical point $M$ instead of $S'$ and reason more easily.
Edit2: I realized it could happen that the point constructed last, namely $x_3$, could be coincident with $x_2$. If this happens, instead of $S$ (or of a nonsaddle critical point $M$ as per Edit1), make the same construction with even another different nonsaddle critical point: the new point $x_3$ thus constructed will not be equal to $x_2$ because it cannot lie on a level curve which is part of the family $\{C_t\}_{t\in[a,b]}$ encircling $M$.
Edit3: I realize now that edit2 might in fact not quite work: the critical loci of the saddles could be $8$shaped curves and all the maxima and minima could be each one inside a "petal" of one of the two $8$shaped curves (I've written this in the edit in Step 1). So there could be no other maxima/minima, outside the disks of $C_a$ and $C_b$, to use. But the $x_3=x_2$ incident looks quite like a "non generic" condition: $x_2$ and $x_3$ are found by rather different constructions so maybe moving $r'$ a little bit and taking a different $E$ (among the same family of level sets near $C_\infty$) could make the condition $x_2=x_3$ disappear.

$\begingroup$ How do we guarantee that the angles have opposite signs in step 5? It seems to me that there could be no sign change near $r' \cap C_{\bar{t}}$. $\endgroup$ Nov 28, 2018 at 5:39

$\begingroup$ Up to a diffeo of a small neighbourhood of $T$, we can pretend $T$ is $(0,0)$, $r'$ is the $x$axis, the outwards normal to $r'$ is the $y$axis, and $C_{\bar{t}}$ looks like $y=x^{2n}$ for some $n\geq 1$. Then $C_{\bar{t}+\epsilon}$ will look like $y=x^{2n}+\varepsilon$ with $\varepsilon<<1$. Now $T_{\pm}$, for small $\epsilon$, look like the points $(\pm \varepsilon^{1/2n},0)$ (or $(\mp \varepsilon^{1/2n},0)$). The tangents to $T_+$ has positive angular coefficient. (...) $\endgroup$– QfwfqNov 28, 2018 at 18:32

$\begingroup$ (...) Now, following $r'$ keeping the same orientation, we have a small neighbourhood of $P''$ in which we read $P''$ as $(0,0)$, the negative $x$axis corresponds to $r'\cap D_{E'}$ (where $D_{E'}$ is the open Jordan disk of $E'$), and the positive $x$axis corresponds to $r'\cap(S^2\setminus D_{E'})$, and the outwardpointing normal to $r'$ at $P''$ becomes the $y$axis. By construction, the tangent to $E'$ at $P''$ (which is leaning toward the set $D_{E'}$ where $S$ belongs) corresponds to a line $y=mx$ with negative $m$ in the $(x,y)$ coordinates. Does it make sense? $\endgroup$– QfwfqNov 28, 2018 at 18:33

$\begingroup$ Okay, maybe instead of taking $T$ to be just the closest to $P''$, we should take $T$ to be the first we encounter starting from $P''$ and following $r'$ along the direction that corresponds to the positive $x$axis (in the orientation above). $\endgroup$– QfwfqNov 28, 2018 at 18:40

1$\begingroup$ I've thought a little bit more, and so far I haven't found serious flaws except for the $x_2=x_3$ thing. I think that one last problem (x_2=x_3) may make my proof not work, and I can't make serious sense of the "make a small generic perturbation to the construction" heuristics I invoked in Edit3  Worse, I'm not even positive that such heuristics is true. $\endgroup$– QfwfqDec 2, 2018 at 20:36
Edit: there seems to be a problem with this (see comments on maps from the Klein bottle to the projective plane), but I'm not exactly sure where.
Let $S'$ be the sphere minus the critical points. Consider the map $p:S'\to P^2$ that sends each point to the direction (in projective space) of its gradient rotated clockwise by $90$ degrees. The preimages of a direction must lie in the great circle orthogonal to that direction, and in fact they are exactly the special points (the ones we look for) in that great circle.
So we want to show that the preimages of $p$ cannot all have cardinality $0$, $1$ or $2$. Preimages with cardinality $1$ form a curve $C$. Indeed, special points on a given great circle are those where the gradients shifts from one hemisphere to the other, so their number is generically even.
Define now $S''$ to be $S'$ with $C$ removed. Let's argue by contradiction that the preimages of $p:S''\to P^2$ cannot all have cardinality $0$ or $2$. Because all preimages of directions in $p(S'')$ have cardinality $2$, we get that $p:S''\to p(S'')$ is a 2fold covering space. Hence it must be a Galois cover: there exists a homeomorphism $\phi$ of $S''$ that permutes each fiber.
Now consider the action of $\phi$ on the "boundary" of $S''$. There are two parts in this boundary, one coming from the critical points, and one coming from $C$. Viewed in the blowup, the part corresponding to a critical point is a union of arcs of circles in the tangent plane (and the full unit circle if $C$ doesn't go through that critical point). Under $p$, each such circle arc maps to an arc of great circle orthogonal to that critical point. The image great circle arcs coming from different critical points belong to distinct great circles, meaning that an interior point in the blowup at a critical point cannot map to an interior point in the blowup at another critical point under $\phi$. Indeed, by continuity, this would imply the existence of open circle arcs mapping to each other despite their images by $p$ belonging to distinct great circles.
Consider the pairs of points $(a,b)$ with $a$ lying on a level set that's very close to the maximum of the function, and same for $b$ but with the minimum. The pair realizing the smallest distance is necessarily a fiber as the great circle it generates is orthogonal to the level sets. Generically, and in the limit of infinitesimally small level sets, both points in that pair are in the interior of the blowup at their critical point. The conclusion of the above paragraph means that $\phi$ cannot permute this particular fiber, which is a contradiction.

2$\begingroup$ This time the problem is in the phrase "Because all preimages of directions $p(S')$ have cardinality 2.... Hence it must be a Galois cover". Consider a projection from $S^2$ to a plane, this is not a Galois cover, since some points have 2 perimages, some have $0$. In fact, for any oriented surface you can construct a map to any different surface, so that each point has either $0$ or $2$ perimages. $\endgroup$ Nov 30, 2018 at 9:42

1$\begingroup$ Ok, I have a problem now with the following bit: "In the limit, the neighborhood of a critical point is mapped by p to the set of directions orthogonal to the critical point. Hence the images of these neighborhoods are distinct, meaning that a neighborhood cannot map to another neighborhood under $\phi$". It seems that you assume that the neighbourhood of a critical point is mapped to a neighbourhood of some other point. But why is this the case? The involution that you study is discontinuous in a neighbourhood of a critical point so its image might be a neighbourhood of a whole line on $P$ $\endgroup$ Dec 2, 2018 at 18:06

1$\begingroup$ Ok. But you seem to claim that a small neighbourhood of a critical point should be mapped to something small. Do you claim this? If yes, then why? $\endgroup$ Dec 2, 2018 at 19:11

1$\begingroup$ How do you know that other other boundary components are mapped to different sets by $p$? $\endgroup$ Dec 2, 2018 at 19:26

2$\begingroup$ Yes, you can. I would still advise you to think of the Klein bottle K, if you don't mind. Then two exceptional circles (that correspond to two blow ups) are two circles that are disjoint in $K$ and have odd selfintersection. Klein bottle K has an involution $\sigma$, such that it's quotient is a Mobius band. One can arrange $\sigma$ so that these two circles go to two odd circles in the Mobius band that have just one intersection point. Now you embed the Mobius band in the projective plane so that two circles go to two different lines. Voila... $\endgroup$ Dec 3, 2018 at 0:25
Your question asks for 6 points, at least 3 of them in an equatorial plane, while at least one further such point is at the according pole.
So just think about an inscribed octahedron. It has even 4 points at the equatorial plane and both other points are on the according poles.
Now think about the spherical coordinates, i.e. $\vartheta$ for the angular height above/below the equator and $\varphi$ for the circle coordinate along the equator.
Next think about the function $f(\vartheta=0,\varphi)=1+\cos^2(\varphi)$. That one takes its extrema along the equatorial circle alternatingly at the 4 vertices of the inscribed square.
Finally those extrema ought be localized with the height direction as well. That is the maxima along the circle within the plane ought become maxima wrt. the height direction too and the minima along the circle within the plane ought become minima wrt. the height direction too. For that purpose we take advantage of the same function with orthogonal orientation, i.e. of $1+\sin^2(\varphi)$ and superimpose that increasingly at tropical levels, whereas the former one should decrease there.
At the poles this second function only survives and provides there obviously the coincidence of a minimum and a maximum wrt. orthogonal directions, i.e. your required saddle points.
Thus you might want to set up that function something like $$f(\vartheta,\varphi)=1+\cos(\vartheta)\cos^2(\varphi)+(1\cos(\vartheta))\sin^2(\varphi)$$
 rk

$\begingroup$ I might have misunderstood something, but isn't the function $u$ given in advance? $\endgroup$– QfwfqNov 26, 2018 at 21:57

1$\begingroup$ Yes, $u$ is given in advance and the goal is to prove there exists three points with the stated properties. $\endgroup$ Nov 26, 2018 at 22:30