# $0 ≤ A ≤ I.$ $\iff$ ⟨ψ|A|ψ⟩ ∈ [0, 1] for every unit vector $|ψ⟩ ∈ H$

Given two operators $$A$$and$$B$$, where $$A ≤ B$$ means the operator $$B − A$$ is positive semidefnite.

(i) $$0 ≤ A ≤ I.$$

(ii) ⟨ψ|A|ψ⟩ ∈ [0, 1] for every unit vector $$|ψ⟩ ∈ H$$ are equivalent

I am having trouble proving $$(ii) \implies (i)$$

If I take any non zero vector $$|\phi⟩$$, then pluging in $$|\phi⟩|/\|\phi|\|$$is aunit vector which implies $$0\le ⟨\phi|A|\phi⟩\le \|\phi|\| ^2$$

Then I wanted to use $$(d)\iff (a)$$ in the proposition below to conclude that A is positive semidefnite, but for that first I need to prove that A is hermitian and I can't figure out how to do that. Any idea?

Proposition For a Hermitian operator A ∈ L(H), the following fve conditions are equivalent:

(a) A is positive semidefnite.

(b) A = $$B^†B$$ for an operator B ∈ L(H).

(c) A = $$B^†B$$ for an operator B ∈ L(H, K) and some Hilbert space K.

(d) ⟨ψ|A|ψ⟩ ≥ 0 for every |ψ⟩ ∈ H.

(e) Tr[AC] ≥ 0 for every C ∈ PSD(H)

Observe that \begin{align*} \langle \psi | A |\psi\rangle&\leq 1\\ &= \langle \psi | \psi \rangle \end{align*} therefore $$\langle \psi | A-I | \psi \rangle \leq 0$$. You also have $$0\leq \langle \psi|A|\psi\rangle$$ and both for all $$\psi$$ unitary which you can now scale. Therefore $$0\preccurlyeq A\preccurlyeq I$$
• I did manage to arrive to that expression as well, but I don't know how you can conclude that the matrix is hermitian from that. You need A and I-A to be hermitian before applying the proposition $(d)\implies (a)$For that it sufficies to see that A is hermitian but I am stuck there Commented Feb 23 at 7:02
• @darkside The proof that $A$ is Hermitian is well-known. See math.stackexchange.com/a/1587131 for instance. Commented Feb 23 at 7:22