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

Question

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?

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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)

Answer 1

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 $0\preccurlyeq A\preccurlyeq I$