Why are Dirichlet eigenfunctions real-analytic? 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)?
 A: 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 


*

*The function is smooth, since all higher derivatives are bounded and

*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. 



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