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Let D be a domain in R^n, and let f be an eigenfunction of the Laplacian with Dirichlet boundary condition with eigenvalue $\lambda$. Assume that f has L^2 norm 1. I want to know if I can say anything about the Sobolev s-norm of f (interms of s \lambda and D) ?

In particular, I want to know if it is true that |f|_s is like \lambda^{\frac{s}{2}}.

Same question for the Neumann and dbar-Neumann boundary conditions.

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  • $\begingroup$ When you say Sobolev s-norm, are you working in the Hilbert space $H^{s}_{0}$ (At least for the Dirichlet BC)? Any conditions on $s$? $\endgroup$
    – MLevi
    Commented Nov 5, 2009 at 16:30

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There is an estimate of the form

|f|_s < C(D, \lambda, s)|f|_0

where |f|_0 = L^2 norm of f and |f|_s = Sobolev s-norm. There are different ways to get this estimate, depending on how sharp an estimate you need and the regularity of the boundary of D.

In particular, you can apply any a priori estimate for the Sobolev norm of the solution f to

Laplacian(f) = h

in terms of h and bootstrap to get whatever you want.

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  • $\begingroup$ Yes. What I am trying to figure out is the precise nature of the constant C(D,\lambda,s). e.g. is it true that C is like \lambda^{\frac{s}{2}} $\endgroup$
    – Debraj Chakrabarti
    Commented Nov 6, 2009 at 8:03
  • $\begingroup$ This sounds right to me, and, if it weren't for the boundary terms, easily proved by integration by parts. I am sure the boundary integrals can be handled, if the boundary of D is smooth, but someone else will need to explain in more detail how. $\endgroup$
    – Deane Yang
    Commented Nov 6, 2009 at 18:15
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Here are some of my thoughts on the question. Fix $s\in(0,\frac{1}{2})$. Then $C:=\sup_{r\geq 0}\frac{(1+r^{s})^{2}}{1+r}$. Notice then that $\int(1+|\xi|^{s})^{2}||\widehat{f}(\xi)|^{2}d\xi\leq C\int(1+|\xi|)^{2}|\widehat{f}(\xi)|^{2}d\xi\int |\widehat{f}(\xi)|^{2}d\xi$. Now, as mentioned by Deane, there may be some issues with the boundary $\partial D$. Suppose that $f\in H^{1}_{0}(D)\cap H^{2}(D)$ and $-\triangle f = \lambda f$ (We maybe be able to drop the second order regularity of $f$ if more regularity is assumed on the boundary for example). After integrating by parts and using perhaps using some sort of Poincaré inequality (need some sort of boundedness for the domain), one can see by integration by parts that $||f||_{H^{1}_{0}(D)}\sim\lambda$. I THINK that $||f||_{H^{s}(D)} \sim\big(\int(1+|\xi|^{s})^{2}|\widehat{f}(\xi)|^{2}d\xi\big)^{\frac{1}{2}}$, but I'm not sure. In fact this might be another question... I'm not very familiar with fractional Sobolev spaces - much less on subsets of $\mathbb{R}^{n}$. If it were true (it should be true for $s$ an integer - See Evans page 282), then your result would be that $||f||_{H^{s}(D)}\lesssim_{D}C_{n}\lambda$ (modulo $D$ because of the Poincaré inequality - which would require some sort of boundedness of one coordinate). This was my first idea. I'm sure there are better ideas/results. I hope this helps.

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  • $\begingroup$ Sorry for the $C_{n}$. I meant $C$ $\endgroup$
    – MLevi
    Commented Nov 7, 2009 at 7:25
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    $\begingroup$ Is it possible to just prove the desired inequality for s equal to a positive integer (avoiding Fourier transform) and then infer the result for all s by interpolation? It seems to me that the desired estimates can be inferred from results or techniques presented in, say, Gilbarg-Trudinger or one of Stein's books on singular integral operators. But I'm not sure and don't have the time to check. I confess to having never learned any of this properly. $\endgroup$
    – Deane Yang
    Commented Nov 7, 2009 at 18:33
  • $\begingroup$ @Deane You might be right. One may need to use some C-Z theory. $\endgroup$
    – MLevi
    Commented Nov 7, 2009 at 20:39
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Let me make the question more precise -- f is normalized to have L^2 norm 1.

Then f is in $H^1_0(D)$ and the Sobolev 1-norm of f is \lambda.

We want the norm of f in H^s(D).

I'll be glad if anybody can help.

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    $\begingroup$ @Debraj: you should edit your original question, instead of providing clarification in an answer. It looks like you didn't register your account, so if you're having trouble logging in to edit, please read mathoverflow.net/questions/24/theres-a-bug-with-logging-in $\endgroup$
    – Kim Morrison
    Commented Nov 5, 2009 at 21:21
  • $\begingroup$ @Debraj: Morrison has a good point, but now that your question has been made more clear, what do you impose on $s>0$? Have you thought of using the Fourier transform and trying some interpolation? $\endgroup$
    – MLevi
    Commented Nov 5, 2009 at 21:30
  • $\begingroup$ Debraj: I would be surprised if you could obtain an explicit formula for the norm. More likely I would expect you can obtain a bound in $H^s$ for $0 < s < 1$ by an interpolation inequality as MLevi has suggested. You will probably have something like: $||u||_{H^s} \leq C ||u||_{L^2}||u||_{H^1}$. $\endgroup$
    – Dorian
    Commented Oct 14, 2010 at 21:08

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