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.
- 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) $\sigma (T) \subset [0, \infty)$.
- 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 $u\in H$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
We need the following result (in the same book), i.e.,
Proposition 6.9. Let $T \in \mathcal{L}(H)$ be a self-adjoint operator. Set $$ m=\inf _{\substack{u \in H \\|u|=1}} \langle T u, u \rangle \quad \text { and } \quad M=\sup _{\substack{u \in H \\|u|=1}} \langle T u, u \rangle . $$ Then $\sigma(T) \subset [m, M]$ and $\{m, M\} \subset \sigma(T)$. Moreover, $\|T\|=\max \{|m|,|M|\}$.
(i) $\implies$ (ii): By Proposition 6.9, $\sigma(T) \subset [m, \infty)$. Because $T$ is positive, $m \ge 0$. The claim then follows.
(ii) $\implies$ (i): By Proposition 6.9, $m \in \sigma(T)$. Because $\sigma (T) \subset [0, \infty)$, we get $m \ge 0$. The claim then follows.
2.
(iii) $\implies$ (iv): We have $\langle Tu, u \rangle \le M |u|^2$. By Proposition 6.9, $|M| \le \|T\| \le 1$. The claim then follows.
(iv) $\implies$ (v): We have $0 \le m \le M \le 1$. By Proposition 6.9, $\sigma(T) \subset [m, M]$. The claim then follows.
(v) $\implies$ (vi): Because $T$ is self-adjoint, $|Tu|^2 = \langle T^2u, u \rangle$. Let $Q (t) := t-t^2$ for $t \in \mathbb R$. We need to prove $\langle Q(T)u, u \rangle \ge 0$ for all $u\in H$, i.e., $Q(T)$ is positive. Clearly, $Q(T)$ is self-adjoint. By (1), it suffices to prove $\sigma(Q(T)) \subset [0, \infty)$. By exercise 6.22.5, $\sigma(Q(T)) =Q(\sigma(T))$. The claim then follows.
(vi) $\implies$ (iii): By definition of operator norm, $\|T\| = \sup_{\substack{u \in H \\|u|=1}} |Tu|$ and thus $\|T\|^2 = \sup_{\substack{u \in H \\|u|=1}} |Tu|^2$. Then $\|T\|^2 \le \sup_{\substack{u \in H \\|u|=1}} \langle T u, u \rangle = M$. By Proposition 6.9, $|M| \le \|T\|$. Then $\|T\|^2 \le \|T\|$ and thus $\|T\| \le 1$.
