Proof, using trigonometric moment sequences, [Theorem 2.6, Po02]. Assume without loss of generality that $\norm{T}_{B(H)}< 1$. Using Proposition 41.7.2, it is sufficient to show that the Toeplitz matrix

is positive for all $n \in \natp$. To this end, let

then for each $k \in \natz$, $R^{k} = \sum_{j = 1}^{n-k}E_{(j+k)j}\otimes T^{k}$. In particular, $R^{n+1}= 0$ and

As $\norm{R}_{M_n(B(H))}= \norm{T}_{B(H)}< 1$, by recognising the Neumann series and using the resolvent identity,

$\square$

Proof, using von Neumann’s inequality. Let $p = \sum_{j = 0}^{n} \lambda_{j} z^{j}, q = \sum_{j = 0}^{n} \mu_{j} z^{j} \in \complex[z]$ such that $p + \ol q \ge 0$, then

so $\lambda_{j} = (\lambda_{j} + \mu_{j})/2 = \mu_{j}$ for all $1 \le j \le n$. After rebalancing $\lambda_{0}$ and $\mu_{0}$, assume without loss of generality that $q = p$.

Let $c = \normn{p + \ol p}_{C(\mathbb{T}; \complex)}/2$, then $\normn{p + \ol p - c}_{C(\mathbb{T}; \complex)}= \norm{p + \ol p}_{C(\mathbb{T}; \complex)}/2$. By von Neumann’s inequality,

As $p(T) + p(T)^{*} - cI$ is self-adjoint,

$\square$