Eigenbasis expansion in the form norm

Orthonormality in the form. For $j,k$ use symmetry of $a$ and the eigenrelation [F1] with $v=e_k$: $a(e_j,e_k)={\lambda }_j(e_j,e_k{)}_{L^2}={\lambda }_j{\delta }_{jk}$. Hence $a_{\mu }(e_j,e_k)=({\lambda }_j+\mu )(e_j,e_k{)}_{L^2}=({\lambda }_j+\mu ){\delta }_{jk},$ so the family $f_j:=({\lambda }_j+\mu {)}^{-1\mathrm{/}2}{e}_j$ (well defined by [F2]) is orthonormal in the inner product $a_{\mu }$ on $H_0^1(\mathrm{\Omega })$.

Parseval in $L^2$. Since $\{e_j\}$ is a Hilbert basis of $L^2(\mathrm{\Omega })$, [F3] gives $f={\sum }_j(f,e_j{)}_{L^2}{e}_j$ in $L^2(\mathrm{\Omega })$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2={\sum }_j\mathrm{\mid }(f,e_j{)}_{L^2}{\mathrm{\mid }}^2$ for every $f\in L^2(\mathrm{\Omega })$, which is claim 2.

Completeness in the form. Let $u\in H_0^1(\mathrm{\Omega })$ satisfy $a_{\mu }(u,f_j)=0$ for every $j$. Then $a_{\mu }(u,e_j)=({\lambda }_j+\mu )(u,e_j{)}_{L^2}=0$, so $(u,e_j{)}_{L^2}=0$ for every $j$; since $\{e_j\}$ is a Hilbert basis of $L^2(\mathrm{\Omega })$, $u=0$ as an $L^2$ class, hence $u=0$. Thus $(f_j)$ is a complete orthonormal family in the Hilbert space $(H_0^1(\mathrm{\Omega }),a_{\mu })$, and by [F3] for every $u\in H_0^1(\mathrm{\Omega })$ the net of finite partial sums of ${\sum }_j{a}_{\mu }(u,f_j)f_j$ converges to $u$ in the $a_{\mu }$ norm, with $a_{\mu }(u,u)={\sum }_j\mathrm{\mid }{a}_{\mu }(u,f_j){\mathrm{\mid }}^2\mathrm{.}$ Since $a_{\mu }(u,f_j)=\overline{a_{\mu }(f_j,u)}=\overline{({\lambda }_j+\mu {)}^{-1\mathrm{/}2}{a}_{\mu }(e_j,u)}=\overline{({\lambda }_j+\mu {)}^{1\mathrm{/}2}(e_j,u{)}_{L^2}}=({\lambda }_j+\mu {)}^{1\mathrm{/}2}(u,e_j{)}_{L^2}$, the partial sums are ${\sum }_j(u,e_j{)}_{L^2}{e}_j$ and claim 1 follows; the $H_0^1$ norm and the $a_{\mu }$ norm are equivalent by [F2].

Claim 3. Let $u\in H_0^1(\mathrm{\Omega })\subseteq L^2(\mathrm{\Omega })$. By claim 2 applied to $u$, $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2={\sum }_j\mathrm{\mid }(u,e_j{)}_{L^2}{\mathrm{\mid }}^2$, and by claim 1 $a_{\mu }(u,u)={\sum }_j({\lambda }_j+\mu )\mathrm{\mid }(u,e_j{)}_{L^2}{\mathrm{\mid }}^2$ with both sides finite. Subtracting $\mu$ times the first identity from the second gives $a(u,u)=a_{\mu }(u,u)-\mu \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2={\sum }_j{\lambda }_j\mathrm{\mid }(u,e_j{)}_{L^2}{\mathrm{\mid }}^2\mathrm{.}$ This series is absolutely convergent: since ${\lambda }_j\to +\mathrm{\infty }$, only finitely many ${\lambda }_j$ are negative, and for all remaining indices $0\le {\lambda }_j\mathrm{\mid }(u,e_j{)}_{L^2}{\mathrm{\mid }}^2\le ({\lambda }_j+\mu )\mathrm{\mid }(u,e_j{)}_{L^2}{\mathrm{\mid }}^2$, whose sum is finite by claim 1.