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? 



Number of critical points of smooth functions on $S^1$
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.


