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Let $\Omega$ be an open bounded domain with a boundary $\partial\Omega$. Consider the following Neumann eigenvalue problem for Laplacian: find $(\phi_n,\lambda_n)\in H^1(\Omega)\times \mathbb{R}$ \begin{align*} -\Delta \phi_n& = \lambda_n \phi_n\quad \mbox{in }\Omega,\\ \frac{\partial \phi_n}{\partial \nu} & = 0 \quad \mbox{on }\partial\Omega. \end{align*}

Now by spectral theory (see the notes at here https://faculty.math.illinois.edu/~laugesen/595Lectures.pdf), it is known that the sequence of eigenfunctions $\phi_n$ (ordered nondecreasingly by the eigenvalues, with multiplicity counted) can be taken to be a complete orthonormal basis in $L^2(\Omega)$, and also forms a complete orthogonal basis in $H^1(\Omega)$. Thus any function $u\in H^1(\Omega)$ can be expanded into \begin{equation*} u = \sum_{n=1}^\infty (u,\phi_n)_{L^2(\Omega)}\phi_n\quad \mbox{in } L^2(\Omega), \end{equation*} and this expansion holds also in $H^1(\Omega)$, since $u\in H^1(\Omega)$ by assumption.

I am puzzled over the fact that all the eigenfunctions $\phi_n$ have zero Neumann boundary condition, so any $n$-term truncation $u_n$ $$ u_n = \sum_{i=1}^n(u,\phi_n)_{L^2(\Omega)}\phi_n $$ has a zero Neumann boundary condition. However, a function $u$ in $H^1(\Omega)$ may not have a zero Neumann boundary condition. How shall one understand the convergence in $H^1(\Omega)$ ?

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  • $\begingroup$ Thanks for the comment. I would also expect so but it is a result proved on page 37 of the note (in the link), and I do not find any problem with the proof. $\endgroup$ – user118240 Jan 15 '18 at 13:11
  • $\begingroup$ you can also think of the simpler case with the Dirichelt boundary condition eigenfunctions; so they are zero on the boundary. Yet any $L^2(\Omega)$ function can be written as an infinite sum. $\endgroup$ – Math604 Jan 18 '18 at 15:04
  • $\begingroup$ Thank you for your comments. I completely agree with it: just like all the sines $\{\sin i\pi x\}_{i=1}^\infty$ are complete in $L^2(0,1)$. $\endgroup$ – user118240 Jan 19 '18 at 17:12
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The boundary has measure=0, and so there is no contradiction, because the convergence takes place in $L^2$ (or in $H^1$). The convergence is not pointwise. It is probably best to convince yourself of this for an interval $\Omega=(0,L)$ in $\mathbb R$.

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