Let $\Omega\subset\mathbb{R}^n$ open, bounded and smooth. Let $\lambda\_j$ and $e\_j$, $j\in\mathbb{N}$, be the eigenvalue and the corresponding eigenfunctions of the Laplacian operator $-\Delta$ in $\Omega$ with zero Dirichlet boundary data on $\partial\Omega$. We suppose that: $|| e\_j ||\_{L^2(\Omega)}=1$. Let $s\in(0,1)$. Let $u\in H\_0^1(\Omega)$, i want to prove that:
$$ \sum\_{j\in\mathbb{N}}(u,e\_j)\_{L^2(\Omega)}^2\lambda\_j^s<+\infty, $$
where:
$$(u,e\_j)\_{L^2(\Omega)}=\int\_\Omega e\_ju\,dx. $$
I have no idea to go on, any help would be appreciated.



A question about series involving a Sobolev functions
As Giorgio Metafune commented, the result follows by the endpoint cases $s=0,1$ and he proved these 2 cases.


1. Case $s=0$. Here we only use that $\{e\_j\}$ is an orthonormal sequence the Bessel inequality gives $\sum (u,e\_j)^2\_{L^2(\Omega)}\leq \|u\|\_{L^2(\Omega)}^2$.
2. Case $s=1$. We combine the density of $C\_c^\infty(\Omega)\subset H\_0^1(\Omega)$ and the fact that $\{e\_j\}\_{j\in \mathbb{N}}$ forms an $L^2$ basis. If $u\in C\_c^\infty(\Omega)$ we can (take the classical second derivative) and using 2 integration by parts we have
$$-\Delta u=\sum\_j (-\Delta u, e\_j)e\_j=\sum\_j (u,-\Delta e\_j)e\_j=\sum\_j \lambda\_j (u,e\_j)e\_j.$$
With another integration by part and the Paserval identity we obtain
$$
\|\nabla u\|\_{L^2(\Omega)}^2=\int\_{\Omega} u(-\Delta u) \,dx= \int\_\Omega \sum\_{j\in \mathbb{N}} \lambda\_j (u,e\_j) u e\_j \,dx=\sum\_{j\in \mathbb{N}} \lambda\_j (u,e\_j)^2\_{L^2(\Omega)}
$$
By density we conclude that this is true for all $u\in H\_0^1(\Omega)$.
3. Case $s\in (0,1)$. We can bound the sum as
$$\sum\_{j=1}^\infty (u,e\_j)\_{L^2(\Omega)}^2 \lambda\_j^s \leq \sum\_{j=N+1}^\infty (u,e\_j)\_{L^2(\Omega)}^2 \lambda\_j + \sum\_{j=1}^N (u,e\_j)\_{L^2(\Omega)}^2$$
where we take $N$ such that $\lambda\_j\leq 1$ when $j\leq N$. For simplicity I used the fact that $\lambda\_j \to +\infty$, but we really do not need this fact and we can split in 2 series with $j$ running in the sets $A=\{j: \lambda\_j<1\}$ and its complement.


