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we consider the following eigenvalue problem for the Laplacian $$ - \Delta w\left( x \right) = \lambda w\left( x \right),\,x \in \left( {0,1} \right),\,w\left( 0 \right) = w\left( 1 \right) = 0.$$ By direct calculation, we can find $${w_j}\left( x \right) = \sqrt 2 \sin \left( {j\pi x} \right),\,{\lambda _j} = {j^2}{\pi ^2},\,j \in \mathbb{N}.$$ Morever, we have $\left\{ {{w_j}} \right\}_{j = 1}^\infty $ is Hilbert orthonormal base of ${L^2}\left( {0,1} \right)$ and $\left\{ {\frac{{{w_j}}}{{\sqrt {{\lambda _j}} }}} \right\}_{j = 1}^\infty $ is Hilbert orthonormal base of $H_0^1\left( {0,1} \right)$ with inner product $${\left\langle {u,v} \right\rangle _{H_0^1}} = \left\langle {{u_x},{v_x}} \right\rangle = \int_0^1 {{u_x}\left( x \right){v_x}\left( x \right)dx} .$$

Now, i try to prove that for every ${u_0} \in {H^2}\left( {0,1} \right) \cap H_0^1\left( {0,1} \right)$ there exist sequence $\left\{ {{u_{0m}}} \right\}_{k = 1}^\infty $ have form ${u_{0m}} = \sum\limits_{j = 1}^m {\alpha _j^m{w_j}} $ such that $$\mathop {\lim }\limits_{m \to \infty } \left\| {{u_{0m}} - {u_0}} \right\|_{{H^2}\left( {0,1} \right) \cap H_0^1\left( {0,1} \right)}^2 = \mathop {\lim }\limits_{m \to \infty } \left( {\left\| {\Delta \left( {{u_{0m}} - {u_0}} \right)} \right\|_{{L^2}\left( {0,1} \right)}^2 + \left\| {\nabla \left( {{u_{0m}} - {u_0}} \right)} \right\|_{{L^2}}^2} \right) = 0.$$ For this purpose, i contruct sequence $\left\{ {{{\tilde w}_j}} \right\}_{j = 1}^\infty $ define by ${{\tilde w}_j} = \frac{{{w_j}}}{{\sqrt {\lambda _j^2 + {\lambda _j}} }}$. From direct calculation, we have $${\left\langle {{{\tilde w}_i},{{\tilde w}_j}} \right\rangle _{{H^2}\left( {0,1} \right) \cap H_0^1\left( {0,1} \right)}} = {\delta _{ij}},\,\forall i,\,j \in \mathbb{N}.$$ But i cant prove that $\left\{ {{{\tilde w}_j}} \right\}_{j = 1}^\infty $ is Hilbert orthonormal base of ${H^2}\left( {0,1} \right) \cap H_0^1\left( {0,1} \right)$.

This problem has occurred when i proved the existence of weak solution of nonliner wave equation by galerkin's method.

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    $\begingroup$ This question would have been more suitable for MSE. As for the actual question, you can obtain the desired statement by noting that (for example) $C_0^{\infty}(0,1)$ is contained in the closed linear span of the $\widetilde{w}_j$ because the Fourier coefficients of such a function decay rapidly. $\endgroup$ – Christian Remling Sep 11 at 20:29

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