Finite-fourth-moment covariance spectral confinement

Orient the upper-left block or its transpose with $m_h$ rows and $N_h$ columns, and use F1 to center and truncate it. In the rest of the proof write $n=N_h$, $p=m_h$, $y_h=p\mathrm{/}n\to y\in (0,1]$, and $Y$ for this centered normalized matrix. F2 gives every fixed-length bound for $T_j$, in particular $\mathrm{\parallel }Y\mathrm{\parallel }=O(1)$. The diagonal laws give $D=\mathrm{diag}(Y{Y}^{\mathsf{T}})\to I$ and $E=\mathrm{diag}(Y^{\mathsf{T}}Y)\to yI$ in operator norm. Set $T=T_1$ and $T_0=I$. All limits below are along the original sequence index $h$.

For compatible alternating rectangular matrices, write $A_1\odot \cdots \odot A_s$ for the path sum imposing $i_r\mathrm{\ne }{i}_{r+2}$ at every available position (including endpoints). Its norm is at most $2^{s-1}\prod \mathrm{\parallel }{A}_i\mathrm{\parallel }$. Here is a local proof rather than an appeal to Appendix A. Expand at the first forbidden reversal: $A_1\odot \cdots \odot A_s=A_1(A_2\odot \cdots \odot A_s)-\mathrm{diag}(A_1{A}_2)(A_3\odot \cdots \odot A_s)+(A_1\circ A_2^{\mathsf{T}}\circ A_3)\odot A_4\odot \cdots \odot A_s\mathrm{.}$

The correction restores the additional exclusion lost when the first reversal is contracted. This identity follows by fixing the indices $i_1=i_3$, then excluding $i_2=i_4$; for $s=2$ it is simply deletion of the diagonal. Rectangular diagonal matrices have the compatible rectangular identity embedding where necessary. The diagonal term has norm at most $\mathrm{\parallel }{A}_1\mathrm{\parallel }\mathrm{\parallel }{A}_2\mathrm{\parallel }$. The fused Schur product has norm at most $\mathrm{\parallel }{A}_1\mathrm{\parallel }\mathrm{\parallel }{A}_2\mathrm{\parallel }\mathrm{\parallel }{A}_3\mathrm{\parallel }$: it is a compression of $A_1\otimes A_2^{\mathsf{T}}\otimes A_3$ by the coordinate isometries $e_a\mapsto e_a\otimes e_a\otimes e_a$ and the analogous column isometry. The tensor-product norm bound follows by factoring a tensor operator into maps acting on one coordinate at a time; each such map is an orthogonal direct sum of copies of that factor and therefore has norm at most its norm, by summing the squared block norms. With $c_1=1$, $c_2=2$, induction gives $c_s\le c_{s-1}+2c_{s-2}\le 2^{s-1}$.

Define $Y^{[q]}$ to have entries $n^{-q\mathrm{/}2}\mathrm{\mid }x{\mathrm{\mid }}^{q-1}x$ for odd $q$ and $n^{-q\mathrm{/}2}\mathrm{\mid }x{\mathrm{\mid }}^q$ for even $q$. The strong-law inputs give $\mathrm{\parallel }{Y}^{[2]}{\mathrm{\parallel }}_F^2=O(1)$ and $\mathrm{\parallel }{Y}^{[q]}{\mathrm{\parallel }}_F^2\le (\max \mathrm{\mid }x\mathrm{\mid }\mathrm{/}\sqrt{n}{)}^{2q-4}{n}^{-2}\sum \mathrm{\mid }x{\mathrm{\mid }}^4\to 0$ for $q>2$. By F2, $\mathrm{\parallel }Y{\mathrm{\parallel }}^2\le \mathrm{\parallel }T\mathrm{\parallel }+\mathrm{\parallel }\mathrm{diag}(Y{Y}^{\ast })\mathrm{\parallel }=O(1)$. Thus every fixed-length restricted product with at least one factor $Y^{[q]}$, $q>2$, is $o(1)$ in norm.

Centering comparison is now legitimate without circularity: F2 proves the graph estimate for centered truncations and bounds their $\mathrm{\parallel }Y\mathrm{\parallel }=O(1)$ using their row law; the mean matrix has norm $\sqrt{p}\mathrm{\mid }E{x}_{11}^{\mathrm{t}\mathrm{r}}\mathrm{\mid }=o(1)$ by F1. Expand the difference of fixed-length restricted products multilinearly and apply the norm inequality. There are $2^{2j}-1$ terms, not the book's printed $2j-1$.

Let $D=\mathrm{diag}(Y{Y}^{\ast })\to I$, $E=\mathrm{diag}(Y^{\ast }Y)\to yI$ in norm. Let $B_j$ be the restricted product of length $2j-1$ starting with $Y^{\ast }$ and alternating thereafter. For $j\ge 2$, the deterministic restricted-product identity above gives $T_j=Y{B}_j-D{T}_{j-1}+Y^{[3]}\odot Y^{\ast }\odot \cdots \odot Y^{\ast }=Y{B}_j-T_{j-1}+o(1)\mathrm{.}$

For $j=1$, $Y{B}_1=T_1+D=T_1+T_0+o(1)$, so the unified relation is $Y{B}_j=T_j+T_{j-1}+o(1)$ for all $j\ge 1$. Similarly $B_{j+1}=Y^{\ast }{T}_j-E{B}_j+(Y^{[3]}{)}^{\ast }\odot Y\odot \cdots \odot Y^{\ast }\mathrm{.}$

For $j=1$ the final correction is simply $(Y^{[3]}{)}^{\ast }$, which is $o(1)$; this case must not be silently omitted. All restricted products have fixed length and bounded norm; the fused corrections vanish by the estimates above. Therefore $T_{j+1}=Y{B}_{j+1}-D{T}_j+o(1)=(Y{Y}^{\ast }-D)T_j-yY{B}_j+o(1)=T{T}_j-y{T}_j-y{T}_{j-1}+o(1)\mathrm{.}$

Errors are in operator norm, for each fixed $j$. The recurrence holds with $T_0=I$ and includes $j=1$.

Put $H=T-yI$. Define $P_0(z)=1$, $P_1(z)=z+y$, and $P_{j+1}(z)=z{P}_j(z)-y{P}_{j-1}(z)$. Since $\mathrm{\parallel }H\mathrm{\parallel }=O(1)$, induction in the recurrence gives $T_j=P_j(H)+o(1)$ for each fixed $j$. Explicitly $P_j(z)=y^{j\mathrm{/}2}[U_j\left(\frac{z}{2\sqrt{y}}\right)+\sqrt{y}{U}_{j-1}\left(\frac{z}{2\sqrt{y}}\right)],$ where $U_{-1}=0$, $U_0=1$, $U_{j+1}(t)=2t{U}_j(t)-U_{j-1}(t)$. This explicit formula follows immediately by checking the two initial values and recurrence.

For real $\mathrm{\mid }t\mathrm{\mid }\ge 1+ϵ$, write $\mathrm{\mid }t\mathrm{\mid }=\cosh a$ and $\rho =e^a>1$. The recurrence directly gives $U_j(\cosh a)=\sinh ((j+1)a)\mathrm{/}\sinh a$ and $U_j(-t)=(-1{)}^j{U}_j(t)$. Hence $\mathrm{\mid }{U}_j(t)\mathrm{\mid }\ge {\rho }^j$ and $\mathrm{\mid }{U}_{j-1}(t)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{U}_j(t)\mathrm{\mid }\le {\rho }^{-1}$, the last inequality following from $\sinh (ja)\mathrm{/}\sinh ((j+1)a)<e^{-a}$. Therefore $\mathrm{\mid }{U}_j(t)+\sqrt{y}{U}_{j-1}(t)\mathrm{\mid }\ge (1-\sqrt{y}\mathrm{/}\rho ){\rho }^j\ge (1-{\rho }^{-1}){\rho }^j$. This is strictly positive uniformly for $\mathrm{\mid }t\mathrm{\mid }\ge 1+ϵ$, including $y=1$. If $\mathrm{\parallel }H\mathrm{\parallel }>2\sqrt{y}+ϵ$ infinitely often, the finite-dimensional spectral theorem and spectral mapping for a polynomial give, for every fixed $j$, $y^{j\mathrm{/}2}(1-{\rho }_{ϵ}^{-1}){\rho }_{ϵ}^j\le \mathrm{lim\hspace{0.17em}sup}\mathrm{\parallel }{P}_j(H)\mathrm{\parallel }\le (2j+1)(j+1)y^{(j-1)\mathrm{/}2}$ with a fixed ${\rho }_{ϵ}>1$.

The inequality in step 11.1 is impossible for all positive integers $j$: after dividing by $y^{j\mathrm{/}2}$, its left side grows exponentially in $j$ while the right side is the polynomial $(2j+1)(j+1)y^{-1\mathrm{/}2}$. Thus almost surely ${\mathrm{lim\hspace{0.17em}sup}}_h\mathrm{\parallel }H\mathrm{\parallel }\le 2\sqrt{y}$. Since $Y{Y}^{\mathsf{T}}-(1+y)I-H=D-I\to 0$, the centered matrices satisfy the stated confinement. All fixed-$j$ estimates hold simultaneously by intersecting countably many probability-one events.

Let $A_h$ denote the original oriented upper-left block divided by $\sqrt{N_h}$. F1 gives $\mathrm{\parallel }{A}_h-Y_h\mathrm{\parallel }\to 0$ through eventual truncation equality and the explicit constant-mean norm, while F2 gives $\mathrm{\parallel }{Y}_h\mathrm{\parallel }=O(1)$ and hence $\mathrm{\parallel }{A}_h\mathrm{\parallel }=O(1)$. Thus $\mathrm{\parallel }{A}_h{A}_h^{\mathsf{T}}-Y_h{Y}_h^{\mathsf{T}}\mathrm{\parallel }\le (\mathrm{\parallel }{A}_h\mathrm{\parallel }+\mathrm{\parallel }{Y}_h\mathrm{\parallel })\mathrm{\parallel }{A}_h-Y_h\mathrm{\parallel }\to 0$. The proved centered confinement therefore passes to the original matrices. Since the argument allowed $y_h=1$ and orientation changes, it includes the square-aspect boundary.