Hi I am looking at part 3 of the proof in Evans Chapter 6. I have difficulty understanding "Furthermore from (6) and (7) we see that $(\lambda_k^{-1/2} w_k)$ is an orthonormal subset of $H_0^1(U)$. It is known, from the previous results of the chapter, $(w_k)_{k=1}^{\infty}$ is an orthonormal basis of $L^2(U)$, $w_k\in H_0^1(U)$ is an eigenfunction corresponding to $\lambda_k.$
The bilinear form is defined as $$ B[u,v]:=\int_{U} a^{ij} D^j u D^i v+b^i D^i u v+cuv .$$ Given $u$ is a weak solution, the following also holds for the eigenvalue problem $$ B[u,v]=(\lambda u,v)\,\,\,\forall\,\,\,v\in H_0^1(U)$$ I understand this is really a matter of checking definition. My confusions two-fold. I do not see, by diving $w_k$ by $\lambda_k^{1/2},$
1) Why we get an orthonormal subset? This means?? 2) Why this orthonormal subset is in $H_0^1(U)?$
Now the most important questions of mine.
A. It is true that $(w_k)_{k=1}^{\infty}$ forms an orthogonal basis of $H_0^1(U)$, in view of the Galerkin approximation of weak solutions in Chapt 7. How to verify this?
B. In view of the proof below, $(w_k)_{k=1}^{\infty}$ cannot be orthonormal basis of $H_0^1(U).$ (Though the scaled version does.) Can someone give a proof, perhaps a contradictory argument?
I am looking for a proof that make use of the definition of $H_0^1$ inner product and integration by parts, rather than inferring from (6) and (7).
The previous theorem (on page 355) shows that $w_k \in H^1_0(U)$. Identities $(6)$ and $(7)$ tell you that $\{ w_k \}_{k=1}^{\infty}$ is an orthogonal collection and the length of each $w_k$ with respect to $B$ is $||w_k||_{B} = \sqrt{B[w_k,w_k]} = \sqrt{\lambda_k}$. Thus, if we normalize $w_k$ we get an orthonormal collection of elements in $H^1_0(U)$ with respect to $B$.