Number of critical points of smooth functions on $S^1$ Let $u$ be a smooth function on the unit circle $S^1$ such that $\int_{S^1}ux_j=0$, for $j=1,2$. Is the number of critical points of $u$ strictly bigger than 2?   
 A: An elementary argument: write $u$ as a function of
 $\theta \in T^1 := {\bf Z} / 2 \pi {\bf Z}$,
where $(x_1,x_2) = (\cos \theta, \sin \theta)$.  Then each of
$$
\int_{T^1} u'(\theta) \, d\theta,
\quad
\int_{T^1} u'(\theta) \cos \theta \, d\theta,
\quad
\int_{T^1} u'(\theta) \sin \theta \, d\theta
$$
vanishes (the first is clear, and the other two are obtained from
$$
\int_{T^1} u(\theta) \sin \theta \, d\theta
 = \int_{T^1} u(\theta) \cos \theta \, d\theta
 = 0
$$
by integration by parts).
Thus if $u'$ has only finitely many zeros in $T^1$
then it has at least four sign changes, by an argument
that I already gave for
this
Math.SE question.
Assume $u'$ had only two sign changes in each period,
say at $\theta_1$ and $\theta_2$.  Then we could find reals $A,B,C$
such that $t(\theta) = A + B \cos \theta + C \sin \theta$ has sign changes
at the same $\theta_1$ and $\theta_2$ and nowhere else; but then
$u'(\theta) \, t(\theta)$ is either everywhere $\geq 0$ or everywhere $\leq 0$,
but is not everywhere zero, which contradicts
$\int_{T^1} u'(\theta) \, t(\theta) \, dx = 0$, and we're done.
(To find $A,B,C$ we can go back to the $S^1$ picture:
 consider the points
$(\cos \theta_1, \sin \theta_1)$ and $(\cos \theta_2, \sin \theta_2)$
on the circle $x_1^2 + x_2^2 = 1$, and join them by a line
$A+Bx_1+Cx_2 = 0$, which meets the circle at those two points
and thus nowhere else.)
A: Yes, in fact the number of critical points of $u$ is at least four.
Expand $u:S^1\to\mathbb R$ as a Fourier series:
$$u(\theta)=a_0+\sum_{i=1}^\infty a_j\cos(j\theta)+b_j\sin(j\theta)$$
Then your condition may be stated as $a_1=b_1=0$ (I assume by $x_1$ and $x_2$, you mean to be identifying $S^1$ with $\{x_1^2+x_2^2=1\}$, so $x_1=\cos\theta$ and $x_2=\sin\theta$ in the notation above).
Now let $u(t,\theta)$ be the solution to the heat equation $u_t=u_{\theta\theta}$ with initial value $u(0,\theta)=u(\theta)$.  By the maximum principle, we have:
$$\#\operatorname{crit}(u(\cdot))\geq\#\operatorname{crit}(u(t,\cdot))$$
for all $t\geq 0$.  On the other hand, we have the explicit solution:
$$u(t,\theta)=a_0+\sum_{i=1}^\infty a_je^{-j^2t}\cos(j\theta)+b_je^{-j^2t}\sin(j\theta)$$
Let $j_0=\min\{j\geq 1:(a_j,b_j)\ne(0,0)\}$.  If $j_0=\infty$ (i.e. only $a_0$ is nonzero), then $u$ is constant and there is nothing to prove, so therefore we have $1<j_0<\infty$.
Now the behavior of the function $u(t,\cdot)$ for large $t$ is evidently dominated by the leading terms:
$$u(t,\theta)=a_0+e^{-j_0^2t}\left[a_{j_0}\cos(j_0\theta)+b_{j_0}\sin(j_0\theta)\right]+\cdots$$
Writing $a_{j_0}\cos(j_0\theta)+b_{j_0}\sin(j_0\theta)=\Re\left[(a_{j_0}-ib_{j_0})e^{j_0\theta}\right]$, it follows that for sufficiently large $t$, we have:
$$\#\operatorname{crit}(u(t,\cdot))=2j_0$$
Since $j_0\geq 2$, this gives the desired result.
