# A question about series involving a Sobolev functions

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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• The inequality should be $\sum_j \lambda_j |(u,e_j)|^2 <\infty$. – Giorgio Metafune Oct 16 at 18:08
• @ Giorgio Metafune. There was a mistake in the question. I have to prove that:$$\sum_{j\in\mathbb{N}}(u,e_j)_{L^2(\Omega)}^2\lambda_j^s<+\infty.$$ – inoc Oct 16 at 18:23
• Then this follows from the cases $s=0,1$. – Giorgio Metafune Oct 16 at 18:30
• can you give me the details please ? – inoc Oct 16 at 18:31
• $s=0$ 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 $H^1_0$. – Giorgio Metafune Oct 16 at 18:38

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.