Even trace moments transfer from the operator norm

Symmetry makes $A$ self-adjoint. By [F1], write $A=U\mathrm{diag}({\mu }_1,\dots ,{\mu }_r)U^{\ast }$ with real ${\mu }_j$ and unitary $U$. Matrix multiplication gives $\mathrm{Tr}(A^{2p})={\sum }_i{\sum }_j{\mu }_j^{2p}\mathrm{\mid }{U}_{ij}{\mathrm{\mid }}^2={\sum }_j{\mu }_j^{2p}$ because each column has squared norm one. Each summand is nonnegative and at most $\mathrm{\parallel }A{\mathrm{\parallel }}^{2p}$ by [F2]. Summation proves the deterministic bound without using Choice.

Let $f$ be a finite real-valued measurable function on a measure space $(\mathrm{\Omega },\mathcal{F},\mu )$. By [F4], $\mathrm{\mid }f\mathrm{\mid }$ is measurable. For $a<0$, $\{f^2>a\}=\mathrm{\Omega }$; for $a\ge 0$, $\{f^2>a\}=\{\mathrm{\mid }f\mathrm{\mid }>\sqrt{a}\}$. Thus every strict superlevel of $f^2$ is measurable, so [F4] makes $f^2$ measurable. If $g$ is another finite real-valued measurable function, [F4] and the identity $fg=((f+g{)}^2-(f-g{)}^2)\mathrm{/}4$ then make $fg$ measurable. Induction gives measurability of every finite product and finite sum of products of such functions.

Now assume Choice and use the Gaussian coordinates in the Statement. By [F6], each $Z_{ij}$ is a finite real-valued measurable random variable; every entry of $Y$ is one of these coordinates. Expanding matrix powers gives $\mathrm{Tr}(Y^{2p})={\sum }_{i_0,\dots ,i_{2p-1}=1}^r{\prod }_{k=0}^{2p-1}{Y}_{i_k{i}_{k+1}}$, where $i_{2p}=i_0$. This finite sum of finite products is measurable by step 1.2. The same step applied to the measurable norm from [F2] makes $\mathrm{\parallel }Y{\mathrm{\parallel }}^{2p}$ measurable.

Apply step 1.1 pointwise to $Y$. Steps 1.1 and 2.1 give nonnegative measurable functions with $0\le \mathrm{Tr}(Y^{2p})\le r\mathrm{\parallel }Y{\mathrm{\parallel }}^{2p}$. By [F2], $\mathbb{E}\mathrm{\parallel }Y{\mathrm{\parallel }}^{2p}<\mathrm{\infty }$, and [F3] gives $0\le \mathbb{E}\mathrm{Tr}(Y^{2p})\le r\mathbb{E}\mathrm{\parallel }Y{\mathrm{\parallel }}^{2p}<\mathrm{\infty }$. Taking the increasing positive $2p$th root gives exactly the stated expectation inequality.