Evans says this can be estimate by energy estimates,but this is true if α＜β,what about α≥β,could the estimate be valid again (screenshot transcribed below)
THEOREM 6 (Boundedness of the inverse). If $\lambda \notin \Sigma$, there exists a constant $C$ such that
$$
\|u\|_{L^{2}(U)} \leq C\|f\|_{L^{2}(U)} \tag{29}
$$
whenever $f \in L^{2}(U)$ and $u \in H_{0}^{1}(U)$ is the unique weak solution of
$$
\left\{\begin{array}{cl}
L u=\lambda u+f & \text { in } U \\
u=0 & \text { on } \partial U
\end{array}\right.
$$
The constant $C$ depends only on $\lambda, U$ and the coefficients of $L .$
This constant will blow up if $\lambda$ approaches an eigenvalue.Proof. If not, there would exist sequences $\left\{f_{k}\right\}_{k=1}^{\infty} \subset L^{2}(U)$ and $\left\{u_{k}\right\}_{k=1}^{\infty} \subset$ $H_{0}^{1}(U)$ such that
$$
\left\{\begin{aligned}
L u_{k} &=\lambda u_{k}+f_{k} & & \text { in } U \\
u_{k} &=0 & & \text { on } \partial U
\end{aligned}\right.
$$
in the weak sense, but
$$
\left\|u_{k}\right\|_{L^{2}(U)}>k\left\|f_{k}\right\|_{L^{2}(U)} \quad(k=1, \ldots)
$$
As we may with no loss suppose $\left\|u_{k}\right\|_{L^{2}(U)}=1,$ we see $f_{k} \rightarrow 0$ in $L^{2}(U)$. According to the usual energy estimates the sequence $\left\{u_{k}\right\}_{k=1}^{\infty}$ is bounded in $H_{0}^{1}(U) .$ Thus…
The energy estimate is below (screenshot):
THEOREM 2 (Energy estimates). There exist constants $\alpha, \beta>0$ and $\gamma \geq 0$ such that
$$
|B[u, v]| \leq \alpha\|u\|_{H_{0}^{1}(U)}\|v\|_{H_{0}^{1}(U)}\tag{i}
$$
and
$$
\beta\|u\|_{H_{0}^{1}(U)}^{2} \leq B[u, u]+\gamma\|u\|_{L^{2}(U)}^{2}\tag{ii}
$$
Could someone help me, thanks a lot.
Best Answer
You don't need to use (i), (ii) is enough and gives \begin{align} \|u\|_{H^1_0(U)}^2 &\le \frac1\beta B[u,u] + \frac\gamma\beta \|u\|_{L^2(U)}^2 \\&= \frac1\beta (u,f)_{L^2(U)} + \frac\gamma\beta\|u\|^2_{L^2(U)} \\&\le \frac1\beta\|u\|_{L^2(U)}\|f\|_{L^2(U)}+\frac\gamma\beta\|u\|_{L^2(U)}^2 \\&=\frac1\beta\|f\|_{L^2(U)}+\frac\gamma\beta \end{align} where we used the definition of a weak solution $B[u,v]=(f,v)_{L^2(U)}$ valid for all $v\in H^1_0(U)$ (in particular valid for $v=u$).