# Question about an exercise in Brezis' Functional analysis

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.

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. $\lambda \in \rho(U)$.
• The first inequality is just the reverse triangle inequality along with the fact that $\|Uv\|-\lambda \|v\|\geq 0$ for $\lambda\in (-1,1)$, right? Commented Dec 11, 2014 at 15:58
• Reverse triangle inequality, then (vii'), which says $\lVert Uv\rVert \geqslant \lVert v\rVert$. We need a $c > 0$, namely $c = 1-\lvert\lambda\rvert$, to apply theorem 2.20. Commented Dec 11, 2014 at 16:05