Brezis' exercise 6.24.3: prove that $(0,1) \subset \rho(T)$ IFF $\langle Tu, u \rangle \leq |T u|^2$ for all $u\in H$

Question

Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. Let $\| \cdot\|$ be the operator norm on $\mathcal L(E)$. Let $I:E \to E$ be the identity map. For $T \in \mathcal L(E)$,

• we denote by $N(T)$ its kernel and by $R(T)$ its range.
• we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
• for $\lambda \in EV(T)$, the set $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.

I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Assume that $T \in \mathcal L(H)$ is self-adjoint.

1. Prove that the following properties are equivalent:

• (i) $T$ is positive, i.e., $\langle Tu, u \rangle \ge 0$ for all $u\in H$.
• (ii) $\sigma (T) \subset [0, \infty)$.

2. Prove that the following properties are equivalent:

• (iii) $\|T\| \leq 1$ and $(T u, u) \geq 0$ for all $u\in H$,
• (iv) $0 \leq \langle Tu, u \rangle \leq |u|^2$ for all $u\in H$,
• (v) $\sigma (T) \subset [0,1]$,
• (vi) $\langle Tu, u \rangle \geq |T u|^2$ for all $u\in H$.

3. Prove that the following properties are equivalent:

• (vii) $\langle Tu, u \rangle \leq |T u|^2$ for all $u\in H$,
• (viii) $(0,1) \subset \rho(T)$.

There are possibly subtle mistakes that I could not recognize in below attempt of (3). Could you please have a check on it?

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Let $Q(t) := t^2-t$ for $t \in \mathbb R$. Because $T$ is self-adjoint, we get $Q(T)$ is self-adjoint and $|T u|^2 = \langle T^2u, u \rangle$. Then (vii) is equivalent to $\langle Q(T)u, u \rangle$ for all $u \in H$, i.e., $Q(T)$ is positive. By (1), (vii) is equivalent to $\sigma (Q(T)) \subset [0, \infty)$. By exercise 6.22.5, $\sigma(Q(T)) =Q(\sigma(T))$. Then (vii) is equivalent to $Q ( \sigma (T) ) \subset [0, \infty)$.

Notice that $Q(t) \in [0, \infty) \iff t \in (-\infty, 0] \cup [1, +\infty)$. Then (vii) is equivalent to $\sigma (T) \subset (-\infty, 0] \cup [1, +\infty)$. By definition, $\rho(T) = \mathbb R \setminus \sigma(T)$. Then (vii) is equivalent to $(0, 1) \subset \rho (T)$. This completes the proof.