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I am no expert on PDE and analysis but I am looking for certain technique from PDE. Let $D_2$ be the Laplace operator and $f$ is an eigenfunction, i.e., $D_2 f=\lambda f$ for some $\lambda>1$. (or even weakly $\lambda\gg 1$)

Let $D_3$ be a degree 3 partial differential operator, which may be explicitly written down. We assume that $D_3 f=\lambda_3 f$. Is there any technique which may prove $$|\lambda_3|\leq c\lambda ^{3/2}$$ for some $c>0$.

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2 Answers 2

Under reasonable boundary conditions, you will have $$\|D_3f\|_{L^2}\le C\|f\|_{H^3}\le C\|f\|_{H^4}^{1/2}\|f\|_{H^2}^{1/2}\le C\|\Delta^2f\|_{L^2}^{1/2}\|\Delta f\|_{L^2}^{1/2}\le C\lambda^{3/2}\|f\|_{L^2}.$$

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Hello, thanks for answer. but what is H^4 and H^3? are they sobolev norm? or the norm on Hardy space? –  Jeep Wrangler Aug 27 '13 at 15:25
    
@Jeep Wrangler : it's the Sobolev norms –  Jung Wen Chen Oct 13 '13 at 15:52

If you are on a compact domain $\Omega$in $\mathbb{R}^n$ you need to impose some elliptic boundary conditions or it is difficult to make predictions. Assume for simlicity that $f$ satisfies the Dirichlet boundary conditions, i.e., it vanishes on the boundary of $\Omega$. Then

$$\lambda_3\Vert f\Vert_{L^\infty}=\sup_{x\in\Omega} |D_3f(x)| \leq C\lambda^{\frac{3}{2}} \Vert f\Vert_{L^2}\leq C'\lambda^{\frac{3}{2}} \Vert f\Vert_{L^\infty}$$

This gives you

$$\lambda_3\leq C' \lambda^{\frac{3}{2}}. $$

Edit. In my earlier post I misquoted the result I used. I have fixed the errors. For details see my source X. Bin: Derivatives of the spectral function and Sobolev norms of eigenfunctions on a closed Riemann manifold, Annal. Glob. Annalysis, vol. 26(2004), 231-252, Theorem 1.2.

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