In Brezis' Functional analysis, Sobolev spaces and partial differential eqautions, exercise 6.24 (3) asks to prove that for a self-adjoint operator $T\in \mathcal{L}(H)$, $H$ a Hilbert space, the following properties are equivalent
(vii) $(Tu, u)\leq |Tu|^2, u\in H$
(viii) $(0,1)\subset \rho(T)$
where $\rho(T)$ denotes the resolvent set of $T$. The book gives the following solution
Set $U=2T-I$. Clearly (vii) is equivalent to
$$\text{(vii')} \;\;\;\;\;\;\;\;\;|u|\leq |Uu| \; \;\;\;\;\;\forall u\in H$$
Applying Theorem 2.20 we see that (vii)$\Rightarrow (-1,+1)\subset \rho(U)=2\rho(T)-1$. Thus (vii)$\Rightarrow$ (viii).
and then it proceeds to prove the converse, which I can understand perfectly. The part I don't get is how applying Theorem 2.20 gives $(-1,+1)\subset \rho (U)$ (everything else is clear). Theorem 2.20 states
Theorem 2.20: Let $A:D(A)\subset E\rightarrow F$ be an unbounded linear operator that is densely defined and closed. The following properties are equivalent:
(a) $A$ is surjective, i.e. $R(A)=F$,
(b) there is a constant $C$ such that
$$\|v\|\leq C\|A^{\ast}v\| \;\; \forall v\in D(A^{\ast}),$$
(c) $N(A^{\ast})=\{0\}$ and $R(A^{\ast})$ is closed.
If someone could explain to me how Theorem 2.20 is applied or even give another solution for the (vii)$\Rightarrow$ (viii) part, I would be grateful.
Best Answer
Remember that $T$, and hence $U = 2T-I$, is self-adjoint by assumption.
Now, for $\lambda \in (-1,1)$, by using (vii'), we have
$$\lVert (\lambda I - U)v\rVert \geqslant \lVert Uv\rVert - \lambda \lVert v\rVert \geqslant (1-\lvert\lambda\rvert)\cdot \lVert v\rVert \tag{$\ast$}$$
for all $v\in H = D(U)$. But $(\ast)$ is just condition b) of theorem 2.20, and the equivalence of that with conditions a) and c) shows that $\lambda I - U$ is invertible, i.e. $\lambda \in \rho(U)$.