Evans tell us there exists an orthonormal basis $\{w_k\}_{k=1}^\infty$ of $L^2(U)$ in the following theorem 1. Whether this is basis a maximal orthonormal set in $L^2(U)$ (and why?).
In the following theorem 2 , the author tells us $\left \{\frac{\omega _k}{ \lambda _k^{\frac{1}{2}}}\right\}_{k=1}^{\infty}$ is in fact an orthonormal basis of $H_0^1(U)$. Then he shows $$B[\omega _k,u]=\lambda _k(\omega_k,u)=0, (k=1,…)$$ force $u≡0$, as $\{\omega _k\}_{k=1}^{\infty}$ is basis of $L^2(U)$ .(Why this forces $u$≡0?).
Could someone give me some details, thank you !
Theorem is typed as follow.
THEOREM 1 (Eignevalues of symmetric elliptic operators).
$\quad$(i) Each eigenvalue of $L$ is real.
$\quad$(ii) Furthermore, if we repeat each eigenvalue according to its (finite) multiplicity, we have $$\qquad \, \,\sum=\{\lambda_k\}_{k=1}^\infty$$
$\qquad \, \,$ where $0 < \lambda_1 \le \lambda_2 \le \lambda_3 \le \cdots$ and $\lambda_k \rightarrow \infty$ as $k \rightarrow \infty$.
$\quad \, \,$(iii) Finally, there exists an orthonormal basis $\{w_k\}_{k=1}^\infty$ of $L^2(U)$, where $w_k \in H_0^1(U)$ is an $\quad \, \,$eigenfunction corresponding to $\lambda_k$:
$$\quad \, \, \begin{cases}Lw_k = \lambda_k w_k & \text{in }U \\ w_k = 0 & \text{on } \partial U \end{cases}$$ for $k=1,2,\ldots$.
Best Answer
The existence of an orthonormal eigenbasis follows from the spectral theorem for compact self-adjoint operators applied to the inverse $L^{-1}$ of the elliptic operator.
The $\lambda_k$ are all positive and therefore $\lambda_k (w_k, u) = 0$ implies that the inner product $(w_k, u)$ is zero for all $k$. Since $\{w_k\}$ are an orthonormal basis, $\{w_k\}$ is complete, which means exactly that if $(w_k, u) = 0$ for all $k$, then $u$ must be the zero element of the vector space $L^2(U)$. Thus $u = 0$ in $L^2(U)$; in particular $u = 0$ a.e. and thus $u = 0$ in $H^1_0(U)$.