> [!definition] > > Let $V$ be an [[Inner Product|inner product space]] and $A$ be a [[Self-Adjoint Operator|Hermitian operator]]. Define $A \ge 0$ and say that $A$ is **semipositive** if $\angles{Ax, x} \ge 0$ for all $x \in V$, and define $A \ge B$ if $A - B \ge 0$. > [!theorem] > > Let $A$ be a semipositive operator, then $A$ is self-adjoint. > > *Proof*. Firstly, since $\angles{Ax, x}$ is always real, > $ > \angles{Ax, x} = \angles{x, Ax} = \ol{\angles{A^*x, x}} = \angles{A^*x, x} > $ > where $\angles{(A - A^*)x, x} = 0$ for all $x \in V$. Thus $A - A^* =0$. > [!theorem] > > Let $A$ and $B$ be self-adjoint operators. Suppose that $A$ commutes with $B$ and $A$ is [[Semipositive Operator|semipositive]], then $AB^2$ is semipositive. > > *Proof*. > $ > \angles{AB^2x, x} = \angles{ABx, Bx} \ge 0 > $ > [!theorem] > > Let $p \in \real[x]$ be a polynomial with $p(t) \ge 0$ for all $t \in [\alpha, \beta]$, and $\alpha \le A \le \beta$ be a Hermitian operator, then $p(A) \ge 0$ for all $t \in [\alpha, \beta]$. > > *Proof*. Since $p$ is a non-negative sum of squares of polynomials and $(A - \alpha)$ and $(B - \beta)$ are both semipositive, $p(A)$ is semipositive. > [!theorem] > > Let $p \in \real{[x]}$ and $[\alpha, \beta]$ be an interval, define > $ > \norm{p} = \sup_{t = [\alpha, \beta]}\abs{p(t)} > $ > then for any Hermitian operator $\alpha \le A \le \beta$, $\norm{p(A)} \le \norm{p}$. > > *Proof*. Let $q = \norm{p} \pm p$, then $q \ge 0$ on $[\alpha, \beta]$, and $q(A) \ge 0$. So $\norm{p(A)} \le \norm{p}$.