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Let $f(\theta)$ be a fixed positive $2\pi-$periodic $C^1$ function on $\mathbb{R}$ with $$\int_0^{2\pi}f(\theta)\cos\theta d\theta=\int_0^{2\pi}f(\theta)\sin\theta d\theta=0,$$ Does for any $2\pi-$periodic $C^1$ function $\phi$ satisfy $\int_0^{2\pi}f(\theta)\phi(\theta)d\theta=0$, we have the following inequality?$$\int_0^{2\pi}\phi(\theta)^2d\theta\leq\int_0^{2\pi}(\phi'(\theta))^2d\theta$$

Remark: If $f(\theta)\equiv1$, then its the Classical Poincare Inequality.

Moreover, I have the following question:

Let $(M^n,g)$ be a compact Riemannian manifold with $\partial M=\emptyset$, suppose $\lambda_1$ is the first eigenvalue of the Laplace operator on $(M^n,g)$ and $\{\phi_1,\phi_2,\cdots,\phi_k\}$ are the first eigenfunctions. Let $f:M\rightarrow\mathbb{R}^+$ be a fixed smooth function with $\int_M f\phi_i d\mu=0,i=1,2,\cdots,k$. Does for any smooth function $\phi:M\rightarrow\mathbb{R}$ satisfy $\int_M f\phi d\mu=0$, we have the following inequality? $$\lambda_1\int_M\phi^2d\mu\leq\int_M|\nabla\phi|^2d\mu.$$

Does anyone have ever know about this kind of problems? Thank you very much if you can give any useful guides.

Many thanks if you can give any reference!

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  • $\begingroup$ The answer is no in both cases. (The first case is a special case of the 2nd question.) In the first case take $f(\theta)=\cos(2\theta)$ and $\phi(\theta)=1$. $\endgroup$ – Liviu Nicolaescu Mar 31 '15 at 15:54
  • $\begingroup$ @LiviuNicolaescu : I also fell into this trap, but here, $f$ is actually assumed to have a nonvanishing mean (having a constant sign), so a modified Poincaré inequality holds, albeit with a different constant. $\endgroup$ – Hachino Mar 31 '15 at 15:59
  • $\begingroup$ @Hacino Missed that. $\endgroup$ – Liviu Nicolaescu Mar 31 '15 at 16:05
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Disclaimer : I initially misunderstood the question and rewrote the answer accordingly.


Regarding question 1 : the positivity condition implies in particular that the first Fourier coefficient of $f$ is nonzero, call it $f_0$. The orthogonality condition on $\phi$ reads :

$$\phi_0 f_0 + \langle \phi - \phi_0, f - f_0\rangle = 0, \ \ (*)$$

i.e.

$$\phi_0 = - \frac{1}{f_0^2} \langle \phi - \phi_0, f - f_0\rangle $$

where $\langle.,.\rangle$ denotes the standard $L^2$ inner product.

Call $R(\phi) = \frac{\int |\nabla \phi|^2 dx}{\int |\phi|^2 dx}$. We have, for all $\phi$ satisfying $(*)$ :

$$R(\phi)^{-1} = 1 + \frac{1}{f_0^2} \frac{|\langle \phi - \phi_0, f - f_0\rangle|^2}{\|(\phi - \phi_0)'\|_{L^2}^2}. $$

Using Cauchy-Schwarz inequality and the usual Poincaré inequality on $\phi - \phi_0$, we get :

$$R(\phi)^{-1} \leq 1 + \frac{\|f-f_0\|_{L^2}^2}{f_0^2} \ \ (**) $$

and thus a Poincaré inequality on $\phi$. What is actually necessary and sufficient here is that $f$ should have a nonvanishing mean (otherwise said, $f$ should not be orthogonal in $L^2$ to the constant function).

Regarding the second question, you can't possibly expect $\lambda_1$ as a constant in any case, because this constant highly depends on the mutual importance of the constant coefficient of $f$ and the the other ones, as $(**)$ shows. Whether you can recover a uniform constant instead of $\lambda_1$ or not depends on a claim like "a positive periodic function has a uniform fraction of its energy concentrated in the constant mode". The latter is false in general, peaked functions giving counterexamples. Modifying them so as to ensure orthogonality to some eigenfunctions of the Laplacian is another task.

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    $\begingroup$ Thanks very much, but I need to remind you that I assume $f$ is a positive function here. $\endgroup$ – Alfred Chern Mar 31 '15 at 14:33
  • $\begingroup$ @AlfredChern : The answer has been updated. $\endgroup$ – Hachino Mar 31 '15 at 15:43
  • $\begingroup$ Thanks very much for your help! But I need to claim that $f$ is a positive function here, not only $f_0\neq0$, and I also know about the $f_0\neq0$-case. Exactly, if $f$ is a positive function, then I can prove that the conditional functional $R(\phi)$ have a lower bound $\frac{1}{4}$ in the special case. Can you give any reference? Anyway, thanks very much! $\endgroup$ – Alfred Chern Apr 6 '15 at 11:20
  • $\begingroup$ If $f$ is positive, in particular, its mean is nonvanishing. Perhaps positivity implies that your function may be expanded only in terms of even powers of $\sin$ and $\cos$, which in turn implies that the minimal frequency appearing in Fourier series is $2$, hence the lower bound of $\frac 14$. But that's only a guess. $\endgroup$ – Hachino Apr 7 '15 at 7:36
  • $\begingroup$ Thanks very much! I prove that by analysis the zero points of $\phi(\theta)$ and using the Wirtinger inequality. The lower bound $\frac{1}{4}$ is optimal and can't attains. $\endgroup$ – Alfred Chern Apr 8 '15 at 0:24

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