Why are Dirichlet eigenfunctions real-analytic?

Question

Consider the eigenvalue problem $\Delta u + \lambda u = 0$ on some bounded domain $\Omega \subset \Bbb R^d$ with smooth boundary, with Dirichlet data $u|_{\partial\Omega} =0$. It is known that solutions $u$ are real-analytic inside $\Omega$.

This is supposed to be a simple fact, but I couldn't find a reference/proof that shows this (that I could understand). I only found references for more advanced stuff, like quasilinear equations with non-analytic coefficients, etc... Anyway, most references I've read try to generalize, which makes it harder to read - I'm not interested in a more general case.

Why is $u$ real-analytic? Does it follow from the Cauchy-Kowalewsky theorem (and if so, how)? Is there a simple proof of this (only for the above PDE, not in general)?

Answer 1

The statement really is: if $f$ is a distribution on an open subset $\Omega$ of $\mathbb R^n$ and $\Delta f=\lambda f$ then $f$ is real analytic. To prove it:

1: find a distribution $G$ on $\mathbb R^n$ such that $(\Delta-\lambda)G=\delta$. This can be done explicitly via Fourier transform, and the result is a real-analytic function except at $0\in\mathbb R^n$.

2: for any $x\in\Omega$ take a compactly-supported smooth function $\psi$ on $\Omega$ which is equal to $1$ in a neighbourhood of $x$. Let $g=(\Delta-\lambda)(\psi f)$. We have $g=0$ on the open subset where $\psi$ is locally constant. As $g$ has compact support, $\psi f=G\ast g$.

3: as $G$ is real analytic except at $0$, or equivalently, $G$ is holomorphic in a small neighbourhood of $\mathbb R^n-\{0\}$ in $\mathbb C^n-\{0\}$, the convolution $G\ast g$ is real analytic in the complement of the support of $g$, as there $(G\ast g)(x)=\langle g(y),G(x-y)\rangle$, we can allow $x$ to be complex and differentiate w.r.t. $x$.

Answer 2

Here is a very simple proof assuming you know about the Sobolev embedding theorems.

Sobolev embedding for domains $$ \| u\|_{L^{\infty}(\Omega)} \leq C_{\Omega,d,s} \| u \|_{H^s_0(\Omega)} \tag{*}$$ if $s > s_0 = d/2$. The space $H^s_0$ is the $L^2$ Sobolev space with $s$ derivatives vanishing on the boundary. For $s$ a positive integer it is the closure of $C^\infty_0(\Omega)$ under the norm $$ \|u\|_{H^s(\Omega)}^2 = \|u\|_{L^2(\Omega)}^2 + \sum_{|\alpha| \leq s} \|\partial^\alpha u\|_{L^2(\Omega)}^2 $$ where $\alpha$ are multi-indices.

A priori estimate We integrate by parts $$ \int_{\Omega} u \triangle u = \int_{\Omega} - |\nabla u|^2 $$ assuming Dirichlet boundary conditions. So for Dirichlet eigenfunction we have $$ \|\nabla u \|_{L^2(\Omega)}^2 = \lambda \|u\|_{L^2(\Omega)}^2 $$ (this is basically the definition from the Rayleigh quotient of the eigenfunction anyway). This implies that $$ \| u\|_{H^s}^2 \leq \|u\|_{L^2} + \lambda \|u\|_{H^{s-1}}^2$$ which implies $$ \| u\|_{H^s}^2 \leq (1 + \lambda + \cdots + \lambda^s) \|u\|_{L^2}^2 $$ and that for multi-indices $\alpha$, we have $$ \sum_{|\alpha| = k} \| \partial^\alpha u\|_{H^s}^2 \leq \lambda^{k}(1 + \lambda + \cdots + \lambda^s) \|u\|_{L^2}^2 \tag{**}$$

Finishing the proof

The equations ($*$) and ($**$) combined with $s > s_0$ gives $$ \sum_{|\alpha| = k} \| \partial^\alpha u\|_{L^\infty(\Omega)}^2 \leq C_{d,s,\Omega, \lambda}' \lambda^{k} \|u\|_{L^2(\Omega)}^2 $$ where we absorbed the factor $(1 + \cdots + \lambda^s)$ into the constant factor; this shows

1. The function is smooth, since all higher derivatives are bounded and
2. The Taylor coefficients grow exponentially in $k$, and hence the Taylor series converge in an open neighbourhood of every point, and hence the function is real analytic.

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The same argument can be adapted to show the same result for many other linear elliptic operators.