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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 ...


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 ...


the three-dimensional Fourier transform $F(\vec{p})$, of the radial function $f(r)$ has Fourier transform $$F(\vec{p})=\int_0^\infty dr\int_0^\pi d\theta \int_0^{2\pi}d\phi\;e^{ ipr\cos\theta}f(r) r^2\sin\theta$$ $$\qquad=\frac{4\pi}{p}\int_0^\infty rf(r)\sin(pr)\,dr,\;\;{\rm with}\;\;p=|\vec{p}|.$$ so you're asking when the Fourier-sine-transform $S(p)$ ...


The magic words are "Bochner's theorem" (which says that the Fourier transform is positive for positive definite functions). For more on radial functions see Theorem 2.4.1 in this book. (Greg Fasshauer)


the $a_i$'s are complex coefficients in the asymptotic expansion $a(h)=\sum_{i=0}^\infty a_i h^i$; a more explicit expression is given on page 275-276 of Guillemin and Sternberg:


No. The inequality $||\hat{f}||_{L^p} \lesssim || \hat{|f|} ||_{L^{p}}$ does not hold for $p \neq 2$. This is, perhaps, easier to see in the case of Fourier series on the circle. A sketch of a contstruction is given in my answer to another mathoverflow quesion here. This example can be modified to work in Euclidean space as well.


Not sure if this is true, but me may write that \begin{eqnarray} \frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+\varepsilon|\xi|^2+i} &=& \frac{\tau+|\xi|^4+i}{\tau+|\xi|^4+\varepsilon|\xi|^2+i} \frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+i} \\ &:=& m_{\varepsilon}(\tau, \xi) \hat{F}(\tau, \xi) \end{eqnarray} and from the case $\varepsilon = 0$, you ...


see On positivity of Fourier transforms; one sufficient condition is that $f''(x)>0$ for all $x>0$. there are other sufficient conditions, see for example On the positivity of Fourier transforms a necessary and sufficient condition is not known.

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