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Consider a bounded, connected and open subset $\Omega\subset \mathbb{R}^d$ and the Dirichlet Laplacian $-\Delta$ acting in $L^2(\Omega)$.

Then we know that the eigenvalues of $-\Delta$ form an increasing sequence of eigenvalues $0<\lambda_1<\lambda_2\leq \lambda_3\leq....$ and we have a corresponding orthonormal basis of eigenfunctions in $L^2(\Omega)$. Furthermore the eigenfunctions are smooth functions on $\Omega$.

If now $\Omega$ is sufficiently regular (e.g. smooth), then the eigenfunctions will vanish pointwise for all $x\in\partial \Omega$. I'm wondering if this is still true when no regularits assumptions are made? Do one has pointwise $\varphi(x)=0$, $\forall x\in\partial \Omega$ and all eigenfunctions $\varphi$?

Best regards,

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    $\begingroup$ No. The standard example is a domain with a point removed. The eigenfunctions are the same as if the point were not removed. $\endgroup$
    – Michael Renardy
    Commented Sep 6, 2014 at 13:16
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    $\begingroup$ Another problem is that it's unclear how to even define $\varphi(x)$ for $x\in\partial\Omega$ for a general domain. $\endgroup$
    – Christian Remling
    Commented Sep 6, 2014 at 19:21
  • $\begingroup$ Thank you very much! @Christian: Why should there be a problem? Do you have an exceptional situation in mind? $\endgroup$
    – supersnail
    Commented Sep 7, 2014 at 5:54
  • $\begingroup$ @supersnail: The eigenfunctions are originally defined as elements of $L^2$ (or perhaps as distributions, depending on the approach you take), so extra work is needed to assign meaning to pointwise values. $\endgroup$
    – Christian Remling
    Commented Sep 7, 2014 at 20:42
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    $\begingroup$ The similar problem appears already in the case of the Laplace equation; so not only for the eigenfunctions. The Wiener criterion characterizes the points where the solution attains continuous boundary values. The so called regular points (i.e. the bdry points where the function behaves nicely) can also be classified into different categories. One interesting example (apart from e.g. the punctured ball) is the Lebesgue spine. $\endgroup$
    – Juhana Siljander
    Commented Sep 24, 2014 at 20:01

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In case this helps, the answer is yes for a Lipschitz domain.

From McLean's book Strongly Elliptic Systems and Boundary Integral Equations:

If $\Omega$ is a Lipschitz domain, there is a bounded trace operator $\gamma\colon H^s(\Omega) \to H^{s-1/2}(\partial\Omega)$ for any $1/2 < s < 3/2$. Moreover, if $1/2 < s \leq 1$, one has $$H^s_0(\Omega) = \{u \in H^s(\Omega) : \gamma u = 0\}.$$

In particular, this implies that $H^1_0(\Omega)$ is precisely the set of $H^1(\Omega)$ functions that vanish on the boundary.

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    $\begingroup$ True and upvoted, but in fact there is a large gap between the standard example mentioned by Michael Renardy (punctured ball, the issue being that the isolated point has capacity 0) and domains with Lipschitz boundary. There is a lot of research about rough domains and for example a recent result by Arendt and ter Elst states that a $H^1(\Omega)$-function has a well-defined and unique trace $u_{|\partial \Omega}\in L^2(\partial \Omega)$ iff for every Borel subset $B$ of $\partial \Omega$ relative capacity =0 implies Hausdorff measure =0. $\endgroup$
    – Delio Mugnolo
    Commented Jun 9, 2016 at 6:53

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