boundedness of inverse (Evans PDE)

Question

First, I have difficulty understanding why $$\|u_k\|_{L^2(U)}>k\|f_k\|_{L^2(U)}$$ is being assumed in theorem 6 chapter 6.2 Evans.

Second, the last sentence of the proof says (30) implies $\|u\|_{L^2(U)}=1.$ Why?

Third, $f_k\to 0$ in $L^2(U).$ Why? Any clarification is appreciated.

Answer 1

You want to show:

There is $C>0$ so that $$\|u\|_{L^2(U)} \le C \| f\|_{L^2(U)}$$ for all $u \in H^1(U), f\in L^2(U)$ which satisfy the equation (weakly).

In the book, the author argue by contradiction, assuming that the statement is false. Thus:

For all $C>0$, there is $u \in H^1(U), f\in L^2(U)$ which satisfy the equation (weakly) and $$\|u\|_{L^2(U)} > C \| f\|_{L^2(U)}. $$

Of course $u,f$ might depends on $C$. Since it's true for all $C$, by choosing $C = k, k\in \mathbb N$, you come up with $u_k, f_k$ as stated in the book.