Proof of Theorem 2 from Theorem 1. Fix an indexed zero-sum family and choose an order attaining its minimum \(\beta\); this exists because there are only finitely many permutations. Relabel the vectors in that order. Put \[A_j=\sum_{i=1}^jv_i, \qquad B_j=\sum_{i=1}^j\varepsilon_i v_i, \qquad 0\le j\le N,\] where the signs come from Theorem 1. Hence \(\lVert A_j\rVert_2\le\beta\) and \(\lVert B_j\rVert_2\le C\sqrt d\). The positive-sign and negative-sign selections from an original prefix sum respectively to \[P_j=\frac{A_j+B_j}{2},\qquad M_j=\frac{A_j-B_j}{2}.\] Place all positive-sign indices in their original order, followed by all negative-sign indices in reverse original order. A partial sum in the first block is some \(P_j\). In the second block, the unused indices are the negative-sign indices of an original prefix. Since the total sum is zero, the current partial sum is therefore some \(-M_j\). This also covers either block being empty. Every new partial sum has norm at most \((\beta+C\sqrt d)/2\). By minimality, \[\beta\le\frac{\beta+C\sqrt d}{2}, \qquad\text{and hence}\qquad \beta\le C\sqrt d.\] This permutes indices and so preserves all multiplicities. ◻

Proof of Theorem 1. Write the prescribed sequence as \(v_1,\ldots,v_n\in\mathbb R^d\), with \(\lVert v_i\rVert_2\le1\). The hypotheses give \(d,n\ge1\).

Proof. The generator of \(W\) is \(\mathop{\mathrm{tr}}(Q\nabla^2)\). Let \(\tau_D=\inf\{s\ge0:W(s)\notin D\}\), and let \[P_s^D h(x)=\mathbb E_x[h(W(s))\mathbf 1_{\{s<\tau_D\}}]\] be its killed semigroup. The corresponding nonnegative self-adjoint operator is the Dirichlet realization of \(-\mathop{\mathrm{tr}}(Q\nabla^2)\), with form \(\int_D\nabla g^TQ\nabla g\) on \(H_0^1(D)\). These standard Dirichlet facts apply here because bounded convex domains are connected and Lipschitz; see, for example, (Davies and Simon 1984, sec. 2 and Appendix C), after the linear change of coordinates for \(Q\) and the time normalization above. The killed representation can also be obtained first on smooth interior domains and then by exhaustion: their compactly supported smooth functions exhaust the form domain, while any continuous path remaining in \(D\) on a compact time interval has compact image in \(D\).

Write \(\lambda=\lambda_Q(D)\). Compactness of \(H_0^1(D)\hookrightarrow L^2(D)\) and the spectral theorem give \(\lVert P_s^D\rVert_{2\to2}=e^{-s\lambda}\) and a first eigenfunction \(\varphi\) that can be chosen nonnegative by taking the absolute value of a Rayleigh minimizer. Interior elliptic regularity and the strong maximum principle make \(\varphi\) strictly positive at every point of \(D\). It is bounded: free heat kernel domination and \(\varphi=e^\lambda P_1^D\varphi\) give this directly by Cauchy–Schwarz. Consequently, \[e^{-T\lambda}\varphi(x) =P_T^D\varphi(x) \le \lVert\varphi\rVert_\infty\, \mathbb P_x\{\tau_D>T\}.\] For the reverse bound, let \(p_Q\) denote the free heat kernel. For \(T\ge1\), the semigroup property and free kernel domination yield \[\begin{align*} \mathbb P_x\{\tau_D>T\} &\le \lVert p_Q(1,x,\cdot)\rVert_2 \lVert P_{T-1}^D\mathbf 1_D\rVert_2\\ &\le \lVert p_Q(1,x,\cdot)\rVert_2\,|D|^{1/2} e^{-(T-1)\lambda}. \end{align*}\] Both prefactors are finite for the fixed \(D,Q,x\). Taking logarithms and dividing by \(T\) proves (16). Finally, a nonempty open convex set symmetric about \(0\) contains \(0\) as an interior point. ◻

Proof. Replace each interval by a finite grid. All the constraints are then closed centrally symmetric convex conditions on one finite-dimensional centered Gaussian vector. Represent that vector as a linear image of a standard Gaussian on its linear support; thus a possibly singular covariance creates no difficulty. Royen’s Gaussian correlation inequality (Royen 2014), iterated over the finitely many events, gives the desired product bound on the grids. Now refine nested grids whose unions are dense in the respective intervals. Closedness of each \(K_{ij}\) and path continuity identify the decreasing limit with the stated event. Continuity of probability from above proves the assertion. ◻

Proof of Lemma 5. The set \(K\) is again a bounded open centrally symmetric convex neighborhood of \(0\). Take Brownian motion \(W\) of covariance rate \(2Q\) started at \(0\). On \([0,T]\), impose the closed constraints \[W(s)\in\overline{D_*D},\qquad |m_iW(s)|\le\delta/2\quad(1\le i\le n).\] Their simultaneous occurrence implies survival in \(K\): \(\overline{D_*D}\subset 2D_*D\), since \(0\) is interior to the convex set \(D\), and the half-threshold slabs lie inside the open full-threshold slabs. By Lemma 7, and then by replacing each closed-event probability by its smaller open-event probability, \[\begin{align*} &\mathbb P_0\{W([0,T])\subset K\}\\ &\quad\ge \mathbb P_0\{W([0,T])\subset D_*D\} \prod_{i=1}^n \mathbb P_0\{|m_iW(s)|<\delta/2\text{ for all }0\le s\le T\}. \end{align*}\] For a nonzero row \(m_i\), the process \(m_iW\) is one-dimensional Brownian motion with covariance rate \(2c_i\), where \(c_i=m_iQm_i^T>0\). The first Dirichlet eigenfunction of \(-c_i\,d^2/dx^2\) on \((-\delta/2,\delta/2)\) is \(\cos(\pi x/\delta)\), with eigenvalue \(\pi^2c_i/\delta^2\). Lemma 6 gives the corresponding survival exponent. A zero row gives the identically zero process, so its probability is one and its cost is zero.

Take negative logarithms, divide by \(T\), and let \(T\to\infty\). Lemma 6 applies to \(K\) and \(D_*D\), and the change of variables in the Rayleigh quotient gives \(\lambda_Q(D_*D)=D_*^{-2}\lambda_Q(D)\). This proves (6). The argument used only inclusions between closed and open events; it requires no assertion about equality of their survival rates. ◻

Proof. Let \(\mathcal F\) be the nonnegative functions in \(H^1(\mathbb R^m)\) that vanish almost everywhere outside \(K\) and have \(L^2\) norm one, and put \(E_i(f)=\int|\partial_i f|^2\). Consider the set of simultaneous upper bounds \[\mathcal A=\{u\in\mathbb R^m:E_i(f)\le u_i\text{ for every }i \text{ for some }f\in\mathcal F\}.\] It is nonempty and is closed under addition of nonnegative vectors. It is also convex. For \(f,g\in\mathcal F\) and \(0\le t\le1\), the function \(h=(tf^2+(1-t)g^2)^{1/2}\) belongs to \(\mathcal F\) and satisfies \[|\partial_i h|^2\le t|\partial_i f|^2+(1-t)|\partial_i g|^2 \quad\text{a.e.}\] This follows from the Sobolev chain rule for the Euclidean norm and Cauchy–Schwarz.

To see that \(\mathcal A\) is closed, let \(u^{(j)}\in\mathcal A\) converge to \(u\), and choose corresponding \(f_j\in\mathcal F\). Their \(H^1\) norms are bounded. Since all supports lie in a fixed bounded set, weak \(H^1\) compactness and compactness in \(L^2\) give a subsequence converging weakly in \(H^1\) and strongly in \(L^2\) to a member \(f\) of \(\mathcal F\). Lower semicontinuity gives \(E_i(f)\le u_i\) for every \(i\).

If \(c=\kappa^2(1,\ldots,1)\) did not belong to \(\mathcal A\), strict separation would give a nonzero \(w\in\mathbb R^m\) with \[w\cdot c<\inf_{u\in\mathcal A}w\cdot u.\] The upper-set property forces every \(w_i\) to be nonnegative. For \(\epsilon>0\) let \(Q_\epsilon=\mathop{\mathrm{diag}}(w_i+\epsilon)\). By (17), taking absolute values and normalizing does not change the Rayleigh infimum, so \[\inf_{u\in\mathcal A}w\cdot u =\inf_{f\in\mathcal F}\sum_iw_iE_i(f) \le\lambda_{Q_\epsilon}(K) \le\kappa^2\sum_i(w_i+\epsilon).\] Letting \(\epsilon\downarrow0\) contradicts separation. This argument also covers zero components of \(w\). Thus \(c\in\mathcal A\), as required. ◻

Proof. The endpoints are immediate, so suppose \(0<\alpha<1\) and write \(D_\alpha=(1-\alpha)D_0+\alpha D_1\). Fix \(0<\eta<1\), set \(r=1-\eta\), and put \[A_i=r\overline{D_i},\qquad A_\alpha=(1-\alpha)A_0+\alpha A_1.\] These are compact convex sets, and \(A_\alpha\subset D_\alpha\): indeed, \(A_\alpha\subset r\overline{D_\alpha}\), which is compactly contained in \(D_\alpha\).

Let \(W\) start at zero with Brownian covariance rate \(2Q\). On a finite grid of positive times in \([0,T]\), its values form a Gaussian vector. Gaussian log-concavity (Prékopa 1973, Theorem 2) shows that \[\mathbb P\{W\text{ lies in }A_\alpha\text{ on the grid}\} \ge\prod_{i=0}^1 \mathbb P\{W\text{ lies in }A_i\text{ on the grid}\}^{\alpha_i}, \qquad (\alpha_0,\alpha_1)=(1-\alpha,\alpha).\] Here the Minkowski average of the two grid events is contained in the grid event for \(A_\alpha\). Increase the grids to a dense subset of \([0,T]\). Closedness of the \(A_i\) and continuity of Brownian paths allow passage to the decreasing limits of these events. Since \(rD_i\subset A_i\) and \(A_\alpha\subset D_\alpha\), it follows that \[\mathbb P_0\{W([0,T])\subset D_\alpha\} \ge\prod_{i=0}^1 \mathbb P_0\{W([0,T])\subset rD_i\}^{\alpha_i}.\] Lemma 6, followed by the scaling of the Dirichlet quotient, now gives \[\lambda_Q(D_\alpha) \le r^{-2}\bigl((1-\alpha)\lambda_Q(D_0)+\alpha\lambda_Q(D_1)\bigr).\] Finally let \(\eta\downarrow0\). ◻

Proof. Lift \(K\) to the bounded open convex set \[B=\{(y,s)\in\mathbb R^m\times\mathbb R:y+se\in K,\ |s|<\ell\}.\] Let \(P\) be its projection onto \(\mathbb R^m\). For \(y\in P\), its vertical fiber \(I_y=\{s:(y,s)\in B\}\) is a nonempty open interval. Write \(L(y)=|I_y|\). Convexity gives \[(1-t)I_y+tI_z\subset I_{(1-t)y+tz}, \qquad L((1-t)y+tz)\ge(1-t)L(y)+tL(z).\] Thus \(L\) is positive, concave and continuous on the open convex set \(P\). The vertical Steiner symmetrization of \(B\) is \[B^*=\{(y,s):y\in P,\ |s|<L(y)/2\}.\] This set is bounded, open and convex. Symmetry of \(K\) implies \(L(-y)=L(y)\), so \(B^*\) is symmetric separately in \(y\) and in \(s\). For \(s\ge0\) put \[D_s=\{y\in P:L(y)>2s\},\qquad b=\sup\{s\ge0:D_s\ne\varnothing\}.\] Then \(0<b\le\ell\). Each \(D_s\) with \(0\le s<b\) is a nonempty bounded open convex set symmetric about zero, and these domains decrease as \(s\) increases.

The section \(D_1\) is the target because each of its points permits the required shifted signed step. Indeed, for any \(y\in D_1\), write \(I_y=(a,b_0)\). Its length exceeds two. Choose \(c\in(a+1,b_0-1)\); then \[[c-1,c+1]\subset I_y,\qquad |c|<\ell-1=\frac32.\] For \(|q|\le\delta\), we have \(|c-q|<7/4<2\). If \(q\le c\), then \(q+1\) lies in \([c-1,c+1]\); if \(q\ge c\), then \(q-1\) lies in that segment. Thus one of \(q+1,q-1\) lies in the original open fiber, giving membership in \(K\) itself. It remains to prove that \(D_1\) is nonempty and has the same spectral bounds as \(K\).

Choose \(f\) from Lemma 8, and define \[\Psi(y,s)=(2\ell)^{-1/2}f(y+se)\mathbf1_{(-\ell,\ell)}(s).\] Translation invariance and Fubini give \[ \|\Psi\|_2=1,\qquad \int|\partial_{y_i}\Psi|^2=\int|\partial_i f|^2\le\kappa^2. \tag{18}\] Only derivatives in the horizontal variable \(y\) will be used.

Rearrange each nonnegative fiber \(\Psi(y,\cdot)\) into an even function nonincreasing in \(|s|\), denoted \(\Psi^*(y,\cdot)\). A jointly measurable version is obtained from the measurable superlevel lengths \(\mu(y,u)=|\{s:\Psi(y,s)>u\}|\) by replacing each superlevel set with its centered interval of length \(\mu(y,u)\). This preserves the \(L^2\) norm. Since \(\Psi\) vanishes almost everywhere outside \(B\), its rearrangement vanishes almost everywhere outside \(B^*\).

For nonnegative \(a,b\in L^2(\mathbb R)\), the layer-cake formula gives \[\int ab =\int_0^\infty\!\int_0^\infty |\{a>u\}\cap\{b>v\}|\,du\,dv \le\int a^*b^*.\] The intersection is bounded by the smaller of the two superlevel measures, and centered intervals attain that bound. Since rearrangement preserves the individual norms, \(\|a^*-b^*\|_2\le\|a-b\|_2\). Applying this to the fibers at \(y\) and \(y+he_i\), then integrating over \(y\), bounds the horizontal difference quotients of \(\Psi^*\) by those of \(\Psi\). Their weak limits yield \[ \int|\partial_{y_i}\Psi^*|^2\le\int|\partial_i f|^2\le\kappa^2. \tag{19}\] Sobolev slicing and Fubini therefore imply that, for almost every \(s\), \(\Psi^*(\cdot,s)\in H^1(\mathbb R^m)\) and vanishes almost everywhere outside \(D_{|s|}\). Each nonzero such slice belongs to \(H_0^1(D_{|s|})\) by (17).

We next show that the rearranged density puts enough mass at heights above one. Its mean absolute height is \[ \begin{split} M&=\iint |s|\Psi^*(y,s)^2\,dy\,ds\\ &=\frac1{8\ell}\int_{-\ell}^{\ell}\int_{-\ell}^{\ell} \int_{\mathbb R^m}\min\{f(y+re)^2,f(y+r'e)^2\}\,dy\,dr\,dr'. \end{split} \tag{20}\] Indeed, a centered interval of length \(u\) has integral of \(|s|\) equal to \(u^2/4\). Apply this to the superlevel intervals of \(\Psi^{*2}\) and use Tonelli; the factor \((2\ell)^{-1}\) in \(\Psi^2\) gives the displayed normalization. For nonnegative translates \(u,v\) of \(f\), both of norm one, \[\int\min(u^2,v^2) =1-\tfrac12\|u^2-v^2\|_1 \ge1-\|u-v\|_2.\] Moreover, \[\|f(\cdot+re)-f(\cdot+r'e)\|_2 \le |r-r'|\,\|\partial_e f\|_2\le2\ell\kappa.\] It follows from (20) that \[ M\ge\frac\ell2(1-2\ell\kappa)=\frac{19}{16}>1. \tag{21}\] Under the probability density \(\Psi^{*2}(y,s)\), we have \(|s|<b\) almost surely; hence \(M<b\). In particular, \(D_1\) is nonempty.

It remains to pass the derivative bounds to this particular section. Fix a positive definite diagonal \(Q\) and put \(h_Q(s)=\lambda_Q(D_s)\) for \(0\le s<b\). This function is finite and nondecreasing. Convexity of \(B^*\), domain monotonicity and Lemma 9 give \[\begin{split} h_Q((1-t)s+tr) &\le\lambda_Q((1-t)D_s+tD_r)\\ &\le(1-t)h_Q(s)+t h_Q(r), \end{split}\] so \(h_Q\) is convex. Give \(s\) the probability density \(\rho(s)=\int \Psi^*(y,s)^2\,dy\). Testing the Rayleigh quotient with each nonzero slice and then integrating, we obtain \[\int h_Q(|s|)\rho(s)\,ds \le\iint\langle Q\nabla_y\Psi^*,\nabla_y\Psi^*\rangle \le\kappa^2\mathop{\mathrm{tr}}Q.\] This also proves the integrability needed for Jensen’s inequality. Since \(1<M<b\), monotonicity and Jensen give \[\lambda_Q(D_1)=h_Q(1)\le h_Q(M) \le\int h_Q(|s|)\rho(s)\,ds\le\kappa^2\mathop{\mathrm{tr}}Q.\] The construction of \(D_1\) is independent of \(Q\), so every required spectral bound is preserved. Taking \(T_e(K)=D_1\) proves the lemma. ◻

Proof of Proposition 3. Write the prescribed coordinate directions as \(e_{j_1},\ldots,e_{j_s}\). Construct domains backwards by \[K_s=K,\qquad K_{i-1}=T_{e_{j_i}}(K_i) \quad(i=s,s-1,\ldots,1).\] Lemma 10 preserves all domain and spectral hypotheses. Each \(K_i\) therefore contains zero, and we may start with \(x^{(0)}=0\in K_0\). Suppose \(x^{(i-1)}\in K_{i-1}\), and set \[q=q_i(x^{(i-1)}_{j_1},\ldots,x^{(i-1)}_{j_{i-1}}).\] Since \(q\in[-\delta,\delta]\), the lemma supplies \(\varepsilon_i\in\{-1,1\}\) with \[x^{(i)}=x^{(i-1)}+(q+\varepsilon_i)e_{j_i}\in K_i.\] Induction gives \(x^{(s)}\in K_s=K\) and the required signs. Distinctness of the coordinate directions ensures that later steps do not alter the preceding coefficient coordinates on which a shift depends. The backwards domains may depend on the entire input sequence, as allowed in the proposition. ◻

Proof. The function \(u\mapsto\log w(e^u)=\min\{au,-u\}\) is \(a\)-Lipschitz. Write \(r=y/x\). For \(r<1\), the two endpoint lower bounds give \[w(u)\ge \max\{w(x)(u/x)^a,\,w(y)(y/u)^a\},\qquad y\le u\le x.\] The two expressions cross inside \([y,x]\), by the same Lipschitz bound. Integrating them against \(du/u\) yields \[p(x)-p(y)\ge \frac{w(x)+w(y)-2\sqrt{w(x)w(y)}\,r^{a/2}}a.\] For \(r>1\), integration of the minimum of the corresponding endpoint upper bounds on \([x,y]\) gives \[p(y)-p(x)\le \frac{2\sqrt{w(x)w(y)}\,r^{a/2}-w(x)-w(y)}a.\] Using \(2\sqrt{w(x)w(y)}\le w(x)+w(y)\) in either case, we obtain \[ \mathcal J(x,y)\ge\frac{w(x)+w(y)}2 B(r), \qquad B(r)=r-1-\frac{r^{a/2}-1}{a/2}. \tag{31}\] If \(b=a/2=3/4\), then \(B(1)=B'(1)=0\) and \(B''(r)=(1-b)r^{b-2}>0\). Moreover \(B(0+)=1/b-1>0\) and \(B(r)/r\longrightarrow1\) at infinity. Thus, with an absolute positive constant, \[B(r)\ge c \begin{cases} 1,&0<r\le\tfrac12,\\ (r-1)^2,&\tfrac12\le r\le2,\\ r,&r\ge2. \end{cases}\] The function \(rB(1/r)\) satisfies the same bounds, after changing \(c\). Applying (31) with the arguments reversed therefore gives these bounds for both left sides of (30).

If \(f_0(x)\ne f_0(y)\), at least one argument belongs to \([1/2,4]\). When \(1/2\le r\le2\), both arguments lie in \([1/4,8]\), where \(w\) is bounded below and \(1+y\) is bounded above. The Lipschitz bound \(|f_0(x)-f_0(y)|\le2x|r-1|\) applies. When \(r\le1/2\), one \(w\) value is bounded below and \(y\le4\). When \(r\ge2\), one \(w\) value is again bounded below, and either \(y\le4\) or \(y=rx\le4r\). The preceding three bounds, together with \(|f_0(x)-f_0(y)|\le1\), prove the result. If the cutoff values are equal, the right sides vanish and the result follows directly from (31). ◻

Proof. Let \(P_s\) be the coordinate projection onto the initial block and the first \(s\) new coordinates, and put \[Q_{\rm old}=P_s(Q/\theta)P_s, \qquad Q_{\rm new}=Q/\theta-Q_{\rm old}.\] Diagonality of \(Q\) makes these nonnegative matrices with disjoint coordinate supports. Since \(R_tP_s=\mathcal C R_s\), the support identities are \[ \begin{gathered} R_tR_s^\top=\mathcal C,\qquad R_tQ_{\rm old}R_s^\top=\mathcal C Y,\qquad R_tQ_{\rm old}R_t^\top=Z,\\ R_tQ_{\rm new}R_t^\top=V,\qquad F_sQ_{\rm new}=0. \end{gathered} \tag{33}\] We split the right weight into \(I_m+Q_{\rm old}\) and \(Q_{\rm new}\), and use the intermediate matrix \(\widetilde F=R_t^\top f_0(Z)R_t\) for the first part.

Proof. Follow the old and new parts separately through the interval \([s,t]\). Start with \((Z_*,V_*)=(Y,0)\). At each contraction, replace both matrices by their congruences under \(C_i\). At each gain, add the positive semidefinite gain to \(V_*\) only, interpolating linearly during this addition. At every stage \(D=Z_*+V_*\) is the matrix in the original contraction-and-gain path; at the endpoint \((Z_*,V_*)=(Z,V)\).

Define \[S=I_d+Z_*,\qquad A=S^{-1/2}V_*S^{-1/2},\qquad G=\mathop{\mathrm{tr}}A^a.\] The use of \(I_d+Z_*\) makes this quantity stable under contractions. We will show that its increases during gains are paid for by the \(p_1\) part of the energy.

Proof of Lemma 11. The total variation of \(\mathop{\mathrm{tr}}p\) along the separate contraction and gain steps from \(s\) to \(t\) is at most \(E\). Therefore \[\bigl|\mathop{\mathrm{tr}}p(Y)-\mathop{\mathrm{tr}}p(X)\bigr|\le E.\] Together with Lemma 14, this bounds the sum on the right of (32) by \[E+2\mathop{\mathrm{tr}}\bigl(p(X)-p(Z)\bigr) \le E+CE^{1/a}k_t.\] Since \(0\le E\le e_0\le1\), we have \(E\le E^{1/a}\). Lemma 13 now proves (29), including \(E=0\). ◻

Proof. The Gaussian density and its reciprocal are bounded on \(D\), so the Gaussian and Lebesgue \(H_0^1\) form domains agree. Integration by parts identifies the displayed form with the killed Ornstein–Uhlenbeck semigroup \(P_T^D\). Its \(L^2(\gamma_m)\) operator norm is \(e^{-T\lambda_Q^{\mathrm{OU}}(D)}\). The killed-semigroup representation used in Lemma 6 therefore gives \[\mathbb P\{U([0,T])\subset D\} =\langle\mathbf1_D,P_T^D\mathbf1_D\rangle_{L^2(\gamma_m)} \le\gamma_m(D)e^{-T\lambda_Q^{\mathrm{OU}}(D)}.\] This proves the first inequality. Equivalently, the representation follows from the Dirichlet Feynman–Kac formula after the conjugation below; its potential is bounded on \(D\) (Davies and Simon 1984).

Write \(\rho(x)=(2\pi)^{-m/2}e^{-|x|^2/2}\) and put \(g=h\sqrt\rho\). For \(h\in C_c^\infty(D)\), expansion and integration by parts give \[\int_D\langle Q\nabla h,\nabla h\rangle\,d\gamma_m =\int_D\left[ \langle Q\nabla g,\nabla g\rangle +\left(\tfrac14 x^TQx-\tfrac12\mathop{\mathrm{tr}}Q\right)g^2 \right]dx.\] The identity extends by form closure to \(H_0^1(D)\). Multiplication by \(\sqrt\rho\) maps this space bijectively to itself and preserves the corresponding \(L^2\) norms. Since the potential is at least \(-\mathop{\mathrm{tr}}Q/2\), taking Rayleigh infima proves the second inequality. ◻

Proof. We may omit indices with \(F_t=0\). By (26), every remaining index has \(k_t\ge1/4\). Divide these indices into the classes \[\mathcal I_j=\{t\in\mathcal I:K_j\le k_t<2K_j\}, \qquad K_j=2^j/4,\qquad j=0,1,\ldots.\] We first bound the expected supremum within one nonempty class, writing \(k=K_j\).

Stationarity, coisometry and (26) give \[ \mathbb E\lVert F_tU(\tau)\rVert^2 =\mathop{\mathrm{tr}}f_0(X_t)^2\le4k_t\le8k. \tag{45}\] For two indices in the class, Lemma 11, with the indices put in chronological order, implies \[ \mathbb E\lVert(F_t-F_s)U(\tau)\rVert^2 \le Ck|h_t-h_s|^{1/a}. \tag{46}\] Here we dropped the positive right weight in that lemma and used \(1+k_t\le6k\). For time increments at a fixed index, \[\operatorname{Cov}(U(\tau)-U(\sigma)) =2(I-e^{-Q|\tau-\sigma|})\preceq2|\tau-\sigma|Q,\] so that \[\begin{align*} \mathbb E\lVert F_t(U(\tau)-U(\sigma))\rVert^2 &\le2|\tau-\sigma|\mathop{\mathrm{tr}}(F_t^2Q)\\ &=2\theta|\tau-\sigma|\mathop{\mathrm{tr}}(X_tf_0(X_t)^2) \le Ck\theta|\tau-\sigma|. \tag{47}\end{align*}\] This trace identity uses \(R_tR_t^T=I\); it does not require \(Q\) and \(F_t\) to commute.

We give the elementary chaining argument for the two parameters. Choose \(H\) so that \(h_t\in[H,H+e_0]\) for every index under consideration, and put \(\alpha=(h_t-H)/e_0\) and \(\beta=\theta(\tau-u)\) in \([0,1]\). Split a mixed increment into an index increment and a time increment. Equations (46) and (47) bound its second moment by \[Ck\bigl(|\alpha-\alpha'|^{1/a}+|\beta-\beta'|\bigr),\] where \(\alpha\) is the normalized energy coordinate; the constant absorbs \(e_0\le1\). At depth \(l\), divide both parameter intervals into dyadic cells of width \(2^{-l}\) and choose a representative from each occupied product cell. There are at most \(C4^l\) representatives. The increment from each representative to that of its parent cell is a centered Gaussian vector with second moment at most \[v_l=Ck2^{-l/a}.\]

For use here, if \(G_1,\ldots,G_M\) are arbitrarily dependent centered Gaussian vectors and \(\mathbb E\lVert G_i\rVert^2\le v\), then \[ \mathbb E\max_{i\le M}\lVert G_i\rVert \le2\sqrt{v(\log M+1/2)}. \tag{48}\] Indeed, for \(v>0\), diagonalizing the covariance of \(G_i\) gives \(\mathbb E e^{\lVert G_i\rVert^2/(4v)}\le e^{1/2}\), because its eigenvalues are nonnegative and sum to at most \(v\). Bound the exponential of the maximum by the sum of the exponentials, then use Jensen’s inequality and Cauchy–Schwarz. The case \(v=0\) is immediate.

Apply (48) to the increments at each depth. Starting with (45) and summing the parent increments gives an absolute \(C_0\) such that \[ \mathbb E\sup_{\substack{t\in\mathcal I_j\\ u\le\tau\le u+\theta^{-1}}} \lVert F_tU(\tau)\rVert \le C\sqrt k\left(1+\sum_{l\ge1} 2^{-l/(2a)}(1+\sqrt l)\right) \le C_0\sqrt k. \tag{49}\] To justify convergence of the representatives, note that there are only finitely many distinct energy values. If \(h_t=h_s\), Lemma 11 gives \(F_t=F_s\) exactly. At sufficiently fine depth a target index therefore has the same filter as its representative, and the representative times converge to its target time. Continuity of \(U\) completes the passage to the supremum. Equivalently, one may first use finite time grids and then increase them to a dense set. Neither argument uses a lower bound on the positive gaps between energy values. In particular, the number of indices does not enter (49).

It remains to control all the \(k\) classes at once. On a finite time grid the supremum \(S_j\) in (49) is a maximum of Gaussian vector norms. Represent the joint Gaussian vectors as linear images of one standard Gaussian vector. Each corresponding linear map has norm at most one because its output covariance is \(F_t^2\preceq I\), by (26); consequently \(S_j\) is a \(1\)-Lipschitz function of that standard Gaussian vector. Equivalently, it is a supremum of scalar Gaussian functionals, each with variance at most one.

Markov’s inequality and (49) give \(\mathbb P\{S_j\le2C_0\sqrt{K_j}\}\ge1/2\). Borell’s Gaussian enlargement inequality (Borell 1975, Theorem 3.1), applied to this sublevel set, gives \[\mathbb P\{S_j>2C_0\sqrt{K_j}+r\} \le1-\Phi(r)\le e^{-r^2/2},\qquad r\ge0,\] where \(\Phi\) is the standard normal distribution function. The same bound holds for the full time supremum by increasing dense grids and path continuity. Choose \(c_1>2C_0\). The probability that any constraint in class \(j\) exceeds \(c_1\sqrt{k_t}\) is at most \[\exp\bigl(-(c_1-2C_0)^2K_j/2\bigr).\] With \(A=(c_1-2C_0)^2/8\), the sum of these bounds is at most \[\sum_{j\ge0}e^{-A2^j} \le\frac{e^{-A}}{1-e^{-A}}\le\tfrac12\] once \(c_1\) is an absolute constant large enough that \(A\ge\log3\). This proves (44). Restricting the time interval can only increase its probability. ◻

Proof of Proposition 4. Fix \(d,n,Q\) and the entire recursion. Let \(\Theta\) be the finite set of dyadic scales for which some \(F_t^\theta\) is nonzero. If \(q_{\min}\) and \(q_{\max}\) are the smallest and largest diagonal entries of \(Q\), then coisometry gives \(q_{\min}I\preceq L_t\preceq q_{\max}I\), and the support of \(f_0\) gives \[ \Theta\subset[q_{\min}/4,2q_{\max}], \qquad \sum_{\theta\in\Theta}\theta\le4q_{\max}\le4\mathop{\mathrm{tr}}Q. \tag{50}\] At each scale, partition the sequence indices by the half-open intervals \[[r e_0,(r+1)e_0),\qquad r=0,1,\ldots,\] containing \(E_\theta(0,t)\). These are the energy intervals to which Lemma 16 applies. There are at most \(C(1+E_\theta(0,n))\) nonempty intervals, including any repetitions of an energy value in the same interval.

For each fixed \(t\), the scale constraints control the original state norm. To see this, diagonalize \(L_t\) and write \(y=R_tU(\tau)\) in this eigenbasis, with eigenvalues \(\lambda_i>0\). Each \(\lambda_i\) admits a dyadic scale with \(\lambda_i/\theta\in[1,2]\), where \(f_0=1\); that scale belongs to \(\Theta\). Since \(R_t^T\) is an isometry, \[\begin{align*} \sum_{\theta\in\Theta}\lVert F_t^\theta U(\tau)\rVert^2 &=\sum_i |y_i|^2\sum_{\theta\in\Theta} f_0(\lambda_i/\theta)^2 \ge\lVert R_tU(\tau)\rVert^2. \tag{51}\end{align*}\] Thus all the scale constraints at a given time imply, by (25), \[ \lVert R_tU(\tau)\rVert^2 \le c_1^2\sum_{\theta\in\Theta}k_t(\theta) \le c_1^2 C_\phi d \qquad(0\le t\le n), \tag{52}\] with an absolute \(C_\phi\).

We also impose coordinate bounds. A stationary scalar Ornstein–Uhlenbeck process of rate \(q>0\), restricted to an interval of length at most \(1/q\), stays in \([-2,2]\) with probability at least an absolute \(p_0>0\). By stationarity and time rescaling, it suffices to consider rate one on \([0,1]\). In distribution that process is \[e^{-s}\bigl(Z+B(e^{2s}-1)\bigr),\qquad 0\le s\le1,\] where \(Z\) is standard normal and \(B\) is an independent standard Brownian motion. The event \(|Z|\le1\) and \(\sup_{r\le e^2-1}|B(r)|\le1\) has a fixed positive probability and implies the asserted bound.

Choose absolute constants \[A_0>c_1\sqrt{C_\phi},\qquad B_0>2,\] and let \(\mathcal D\) be the domain in Proposition 4 with these constants. It is bounded, open, centrally symmetric and convex, and contains a neighborhood of zero. The closed constraints (52) and \(|U_j(\tau)|\le2\) lie strictly inside \(\mathcal D\).

For a fixed \(T>0\), cover \([0,T]\) by at most \(1+T\theta\) consecutive closed intervals of length at most \(1/\theta\) at scale \(\theta\). For every such interval and every nonempty energy interval impose the event in Lemma 16. Also cover \([0,T]\) by at most \(1+TQ_{jj}\) intervals of length at most \(1/Q_{jj}\) for coordinate \(j\) and impose its box event. All these events are closed symmetric convex constraints on linear evaluations of the centered Gaussian path. There are finitely many events for this fixed input and \(T\). Lemma 7 therefore multiplies their probability lower bounds. In particular, its finite-grid argument and the continuity of the paths apply even when some constraints are duplicates. The strict slack just chosen ensures that their intersection is an event of survival in the open domain \(\mathcal D\).

It follows that \[\begin{align*} -\log\mathbb P\{U([0,T])\subset\mathcal D\} &\le C\sum_{\theta\in\Theta} (1+T\theta)(1+E_\theta(0,n)) +C\sum_{j=1}^m(1+TQ_{jj})\\ &\le A(Q,R,d,n)+C'T\mathop{\mathrm{tr}}Q. \tag{53}\end{align*}\] Here (28) and (50) bound the coefficient of \(T\) by an absolute constant times \(\mathop{\mathrm{tr}}Q\), and \[A(Q,R,d,n) =C\left[\sum_{\theta\in\Theta}(1+E_\theta(0,n))+m\right]<\infty.\] This additive constant may depend on the input and on the smallest entry of \(Q\); no uniform bound on it is required.

By Lemma 15, \[\lambda_Q^{\mathrm{OU}}(\mathcal D) \le C'\mathop{\mathrm{tr}}Q+\frac{A(Q,R,d,n)}{T}.\] Let \(T\) tend to infinity while keeping the input fixed. This removes the additive constant, including the number of spectral scales. The second inequality in (43) now gives \[\lambda_Q(\mathcal D)\le(C'+1/2)\mathop{\mathrm{tr}}Q.\] Thus \(H_0=C'+1/2\) is absolute. The constants \(A_0,B_0,H_0\) are independent of \(d,n,Q\) and of the recursion, as required. ◻

Proof. For \(1\le p\le2\), the norm comparisons \[\lVert x\rVert_2\le\lVert x\rVert_p \le d^{1/p-1/2}\lVert x\rVert_2\] allow direct application of Theorems 1 and 2, followed by the bound \(C d^{1/p}\) on all prefixes. For \(2\le p\le\infty\), use \[\lVert x\rVert_p\le\lVert x\rVert_2 \le d^{1/2-1/p}\lVert x\rVert_p.\] Divide the input vectors by \(d^{1/2-1/p}\), apply the corresponding Euclidean theorem, and scale back to obtain \(C d^{1-1/p}\). Scaling preserves the zero sum, and in each case the one signing or permutation controls all prefixes. At \(p=\infty\) this gives \(Cd\).

For the lower bounds, let \(1\le p\le2\). The ordered standard basis \(e_1,\ldots,e_d\) has final signed sum of norm \(d^{1/p}\) for every signing. For the ordering problem, take \(e_1,\ldots,e_d\) and \(d\) copies of \(w=-\tfrac1d(1,\ldots,1)\). This indexed family sums to zero and lies in the unit ball, since \(\lVert w\rVert_p=d^{1/p-1}\le1\). If \(d\ge2\), consider any permutation and its prefix just after the \(k=\lfloor d/2\rfloor\)-th basis vector. If \(t\) copies of \(w\) have appeared, put \(a=t/d\in[0,1]\). The prefix sum \(s\) has \(k\) coordinates equal to \(1-a\) and \(d-k\) equal to \(-a\). Since both \(k\) and \(d-k\) are at least \(d/3\), convexity gives \[\lVert s\rVert_p^p =k(1-a)^p+(d-k)a^p \ge\frac d3\bigl((1-a)^p+a^p\bigr) \ge\frac d3\,2^{1-p}.\] Thus \(\lVert s\rVert_p\ge\tfrac12(2d/3)^{1/p} \ge\tfrac13d^{1/p}\), for every permutation. When \(d=1\), the family is \((1,-1)\) and either permutation has a prefix of norm one. ◻

Proof. For an arbitrary \(k\times n\) array \(B=(b_i^j)\) in the Euclidean unit ball \(B_2^d\), with no zero-sum condition, list its entries column by column: \[v_{(i-1)k+j}=b_i^j \qquad(1\le i\le n,\ 1\le j\le k).\] Apply Theorem 1 to this sequence of length \(kn\). With \(\varepsilon_i^j=\varepsilon_{(i-1)k+j}\), every complete-column prefix satisfies \[\left\lVert\sum_{j=1}^k\sum_{i=1}^m\varepsilon_i^j b_i^j\right\rVert_2 = \left\lVert\sum_{\ell=1}^{km}\varepsilon_\ell v_\ell\right\rVert_2 \le C\sqrt d.\] This is the column-major reduction in the proof of (Bárány 2024, Lemma 2.1), with the present signed-prefix bound in place of its dimension-linear bound. Thus Bárány’s matrix signed constant satisfies \(V(B_2^d)\le C\sqrt d\), uniformly in \(k,n\). Here \(V\) allows an entrywise signing of an arbitrary array, whereas the row-permutation constant \(U\) ranges over total-zero arrays and controls the same aggregate prefixes. His transference theorem (Bárány 2024, Theorem 2.2) gives \(U(B_2^d)\le2V(B_2^d)\) for every number of rows. Shorter auxiliary arrays may be padded with trailing zero columns when using the bound uniformly in the number of columns. Since \(B_2^d\) is a centrally symmetric convex body, this gives the asserted bound. ◻