PDE Evans, 6.5.1, Theorem 2

Question

You can find the proof here:Theorem $2$ (Variational principle for the principal eigenvalue)

1. In the screenshot step 3. is missing where he shows that (8) converges also in the $_0^1()$ norm. Why do we need this convergence? Is it that we can pull out the limes in [,] in step 4. where we use that [,] is continuous as an inner product only on $^{1}_{0}()$?
2. In the theorem it says that $u$ is positive but he shows in step 6. that it is positive or negative. What am I missing here?
3. So $u=u^{+}-u^{-}$ and with strong maximum principal we get $u^{\pm}>0$ or $u^{\pm}=0$ in $U$. What does this mean for $u^{+}-u^{-}$ in the case where both are positive $u^{\pm}>0$? Does it mean that we have $u^{+} >u^{-}$ or $u^{+} < u^{-}$? How do we follow from this that $u$ either positive or negative?

Answer 1

1. Yes. The convergence of the series in $H^1_0(U)$ is the reason why pulling out limits is allowed when it comes to compute $B[u,u]$.
2. Thesis (ii) claims the existence of a positive function satisfying several properties, one of which is solving (10). In order to prove (ii), it is first shown that each nontrivial solution $u$ of (10) satisfies $u>0$ or $u<0$. Then, notice that the eigenfunction $w_1$ defined at the beginning of the theorem proof solves (10) and therefore after step 7 is proved we know $w_1>0$ or $w_1<0$. Two are the possibilities: $w_1$ satisfies (ii) or $-w_1$ satisfies (ii). In any case, (ii) is proved.
3. $u^+ := \max\{u,0\}$ while $u^-:=\max\{-u,0\}$. So, if you fix a point $x$ in $U$, at most one of $u^+(x), u^-(x)$ is positive.