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.
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$\begingroup$ The inequality should be $\sum_j \lambda_j (u,e_j)^2 <\infty$. $\endgroup$ – Giorgio Metafune Oct 16 at 18:08

$\begingroup$ @ Giorgio Metafune. There was a mistake in the question. I have to prove that:$$ \sum_{j\in\mathbb{N}}(u,e_j)_{L^2(\Omega)}^2\lambda_j^s<+\infty.$$ $\endgroup$ – inoc Oct 16 at 18:23

$\begingroup$ Then this follows from the cases $s=0,1$. $\endgroup$ – Giorgio Metafune Oct 16 at 18:30

$\begingroup$ can you give me the details please ? $\endgroup$ – inoc Oct 16 at 18:31

2$\begingroup$ $s=0$ is just Bessel inequality for $u$. Now expand for $u$ with comapct support in $\Omega$ $$\Delta u=\sum_j (\Delta u, e_j)e_j=\sum_j (u,\Delta e_j)e_j=\sum_j \lambda_j (u,e_j)e_j$$ and use the identity $$\int_{\Omega} \nabla u ^2=\int_{\Omega} u\Delta u=\sum_j \lambda_j  (u,e_j)^2$$ by Parseval identity. By density this equality extends to $H^1_0$. $\endgroup$ – Giorgio Metafune Oct 16 at 18:38
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As Giorgio Metafune commented, the result follows by the endpoint cases $s=0,1$ and he proved these 2 cases.
 Case $s=0$. Here we only use that $\{e_j\}$ is an orthonormal sequence the Bessel inequality gives $\sum (u,e_j)^2_{L^2(\Omega)}\leq \u\_{L^2(\Omega)}^2$.
 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 $u\in H_0^1(\Omega)$.
 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.
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