Expected extreme eigenvalues from finite Gaussian comparison

For a real symmetric matrix $A$, [F3] applied with $b=\mathrm{\parallel }A\mathrm{\parallel }$ gives $\mathrm{\mid }{v}^{T}Av\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }$ for every real unit $v$. Thus the two signed suprema are finite, since $S$ is nonempty. Put $f_{\pm }(z)={\sup }_{v\in S}(\pm v^{T}T(z)v)$. For each $v\in S$, [F3] and [F4] give $\mathrm{\mid }{v}^T(T(z)-T(w))v\mathrm{\mid }\le \sqrt{2}\mathrm{\parallel }z-w{\mathrm{\parallel }}_2$. Bounding each signed quadratic value at $z$ by its value at $w$ plus this constant, and taking suprema, gives $f_{\pm }(z)\le f_{\pm }(w)+\sqrt{2}\mathrm{\parallel }z-w{\mathrm{\parallel }}_2$. Interchanging $z,w$ proves $\mathrm{\mid }{f}_{\pm }(z)-f_{\pm }(w)\mathrm{\mid }\le \sqrt{2}\mathrm{\parallel }z-w{\mathrm{\parallel }}_2$. In particular both functions are continuous.

For the comparison vector $g$ in [F1], Cauchy–Schwarz gives ${\sup }_{v\in S}2v^{T}g=2\mathrm{\parallel }g{\mathrm{\parallel }}_2$: if $g\mathrm{\ne }0$ take $v=g\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_2$, and if $g=0$ every value is zero. The triangle inequality in [F3] gives $\mathrm{\mid }2\mathrm{\parallel }a{\mathrm{\parallel }}_2-2\mathrm{\parallel }b{\mathrm{\parallel }}_2\mathrm{\mid }\le 2\mathrm{\parallel }a-b{\mathrm{\parallel }}_2$, so [F6] makes this supremum and its square measurable as functions of the Borel vector $g$. By [F7], each $g_i^2$ is integrable with expectation one. Applying [F8] to the finite sum gives $\mathbb{E}(2\mathrm{\parallel }g{\mathrm{\parallel }}_2{)}^2=4{\sum }_{i=1}^r\mathbb{E}{g}_i^2=4r$. Cauchy–Schwarz in [F4], with the constant one, now gives $\mathbb{E}(2\mathrm{\parallel }g{\mathrm{\parallel }}_2)\le 2\sqrt{r}$, establishing integrability of the comparison supremum.

Choice implies Countable Choice by restricting to countable families. The measurable Gaussian coordinates and [F6] therefore make $Z$ a Borel random vector, $f_{\pm }$ Borel functions, and ${\lambda }_{\pm }(Y)=f_{\pm }(Z)$ measurable real random variables. The absolute-value and squaring functions on $\mathbb{R}$ are continuous (use $\mathrm{\mid }\mathrm{\mid }s\mathrm{\mid }-\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }s-t\mathrm{\mid }$ and $\mathrm{\mid }{s}^2-t^2\mathrm{\mid }=\mathrm{\mid }s-t\mathrm{\mid }\mathrm{\mid }s+t\mathrm{\mid }$), so [F6] also makes their compositions measurable. By [F4], Cauchy–Schwarz applied to $\mathrm{\parallel }Y\mathrm{\parallel }$ and the constant one gives $\mathbb{E}\mathrm{\parallel }Y\mathrm{\parallel }\le (\mathbb{E}\mathrm{\parallel }Y{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Since $\mathrm{\mid }{\lambda }_{\pm }(Y)\mathrm{\mid }\le \mathrm{\parallel }Y\mathrm{\parallel }$ by step 1.1, monotonicity in [F8] proves integrability of both extreme values.

Fix a realization of $Y$ and put $M=\max ({\lambda }_{+}(Y),{\lambda }_{-}(Y))$. For any unit $v$, $M\ge \max (v^{T}Yv,-v^{T}Yv)=\mathrm{\mid }{v}^{T}Yv\mathrm{\mid }\ge 0$. Scaling real vectors gives $\mathrm{\mid }{x}^{T}Yx\mathrm{\mid }\le M\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, also when $x=0$. For $u=x+iy$, real symmetry cancels the imaginary cross terms, so $u^{\ast }Yu=x^{T}Yx+y^{T}Yy$ and $\mathrm{\mid }{u}^{\ast }Yu\mathrm{\mid }\le M(\mathrm{\parallel }x{\mathrm{\parallel }}_2^2+\mathrm{\parallel }y{\mathrm{\parallel }}_2^2)=M\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. Thus $-MI⪯Y⪯MI$ in [F3], which gives $\mathrm{\parallel }Y\mathrm{\parallel }\le M$. Step 1.1 gives $M\le \mathrm{\parallel }Y\mathrm{\parallel }$, proving equality.

Apply [F2] with its larger-increment process equal to $H_v=2v^{T}g$ and its smaller-increment process equal to $Q_v=v^{T}Yv$. The common index sphere has dimension $r\ge 1$, centering, continuity and increment ordering hold by [F1], and measurable integrable suprema were verified in steps 2.1 and 1.2. Hence $\mathbb{E}{\lambda }_{+}(Y)\le \mathbb{E}(2\mathrm{\parallel }g{\mathrm{\parallel }}_2)\le 2\sqrt{r}$. Also ${\lambda }_{+}(Y)\ge e_1^{T}Y{e}_1=Y_{11}$, whose mean is zero by [F7]. Their difference is an integrable nonnegative random variable by step 2.1 and [F8], so [F8] gives $\mathbb{E}{\lambda }_{+}(Y)\ge 0$.

By [F5], $Z\sim N_N(0,I)$ and for every $u\in {\mathbb{R}}^N$ the scalar $u^{T}Z$ has law $N(0,\mathrm{\parallel }u{\mathrm{\parallel }}_2^2)$. The vector $-Z$ is Borel measurable by [F6], since negation is continuous. Its projection $u^T(-Z)=(-u{)}^{T}Z$ has that same scalar law, including $u=0$. Since $N=r(r+1)\mathrm{/}2\ge 1$ and Choice is assumed, Cramér–Wold in [F5] identifies the Borel laws of $Z$ and $-Z$. For every Borel $B\subseteq \mathbb{R}$, the set $f_{+}^{-1}(B)$ is Borel by step 2.1; [F9] therefore gives $\mathbb{P}(f_{+}(Z)\in B)=\mathbb{P}(f_{+}(-Z)\in B)$. Linearity of $T$ yields $f_{+}(-z)=f_{-}(z)$ pointwise. Consequently ${\lambda }_{+}(Y)$ and ${\lambda }_{-}(Y)$ have the same law. Their integrability from step 2.1 and the expectation formula in [F9], applied to the identity function on $\mathbb{R}$, give equality of their expectations. Together with steps 2.2 and 3.1 this proves every assertion.