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?
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.

$\begingroup$ One can certainly try to run the same argument on $S^2$. Things certainly break down in dimensions $n>2$, though, since in the evolution of heat equation, a pair of cancelling critical points of index $k$ and $k+1$ can appear whenver $0<k<k+1<n$. $\endgroup$ – John Pardon Jan 20 '15 at 3:58

1$\begingroup$ In dimension two, the only "bad" thing that can happen is that a "monkey saddle" could split into two index one critical points. But, as long as you start out with a function with nondegenerate critical points, this is not an issue: the maximum principle still implies that the number of critical points at any point of the heat evolution is no larger than the number of critical points of the initial condition. $\endgroup$ – John Pardon Jan 20 '15 at 4:01

$\begingroup$ So do you think with a similar argument we can deduce that the set of critical points of $u$ contains a closed curve on $S^2$? $\endgroup$ – User4966 Jan 20 '15 at 4:03

$\begingroup$ So as long as your $u$ on $S^2$ has nondegenerate critical points, you are reduced to analyzing the limiting behavior of the heat equation via spherical harmonics. This seems very tractable to me, though certainly not as easy as it is for $S^1$. $\endgroup$ – John Pardon Jan 20 '15 at 4:03

2$\begingroup$ Oh, I thought you wanted a lower bound on the number of critical points. Generically, the critical locus of a smooth function is finite; you won't be able to force a closed curve of critical points by imposing a finite set of conditions of the form $\int_{S^2}uf_j=0$ for functions $f_j:S^2\to\mathbb R$. $\endgroup$ – John Pardon Jan 20 '15 at 4:06
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.)

$\begingroup$ Sorry, I don't see a direct generalization to higherdimensional spheres. $\endgroup$ – Noam D. Elkies Jan 20 '15 at 4:23

1$\begingroup$ There is a generalization in $T^1$, though: if $u$ is orthogonal to $\cos n\theta$ and $\sin n\theta$ for each positive $n < d$ then $u$ has at least $2d$ critical points. (Your question is the case $d=2$; both John Pardon's proof and the one I gave readily generalize to arbitrary $d$.) $\endgroup$ – Noam D. Elkies Jan 20 '15 at 7:04