Proof. Work on the support and set \(B=\sqrt P\). The derivative of the squaring map at \(B\) is \[\mathcal L_B(X)=BX+XB.\] This is the square-root instance of the classical matrix derivative calculus (Daletskii 1957, sec. 1, Equation (1)). In an eigenbasis of \(P\), this map multiplies entry \((i,j)\) by \(\sqrt{p_i}+\sqrt{p_j}>0\). The inverse function theorem therefore gives a smooth square root whose derivative \(X\) in direction \(K\) satisfies \(BX+XB=K\). Differentiating \(\mathop{\mathrm{tr}}(B^3)\) and using cyclicity, \[d\Phi_P(K)=3\mathop{\mathrm{tr}}(B^2X)=\frac32\mathop{\mathrm{tr}}(BK).\] A second differentiation gives \[d^2\Phi_P(K,K) =\frac32\langle K,\mathcal L_B^{-1}K\rangle_{\mathrm{HS}} =b_P(K),\] where \(\langle X,Y\rangle_{\mathrm{HS}}=\mathop{\mathrm{tr}}(X^*Y)\). The Hessian is nonnegative. This proves convexity on positive definite matrices; replacing each matrix by itself plus \(\eta\mathrm{Id}\) and letting \(\eta\downarrow0\) proves convexity on the closed cone. ◻

Proof. For \(x\in\ker Q\), the inequalities \(0\le\langle Px,x\rangle\le a\langle Qx,x\rangle=0\) imply \(Px=0\). Thus \(\ker Q\subseteq\ker P\), which proves the support assertion.

First suppose \(P,Q\) are positive definite. Inversion reverses positive definite order: conjugating \(0<X\le Y\) by \(X^{-1/2}\), inverting the resulting matrix at least \(\mathrm{Id}\), and conjugating back gives \(Y^{-1}\le X^{-1}\). The spectral integral \[\sqrt X=\frac1\pi\int_0^\infty X(X+t\mathrm{Id})^{-1}\,\frac{\mathrm dt}{\sqrt t}\] then proves monotonicity of the positive square root (Pedersen 1972, Theorem, p. 309), since \(X(X+t\mathrm{Id})^{-1}=\mathrm{Id}-t(X+t\mathrm{Id})^{-1}\). Consequently \(\sqrt P\le\sqrt a\,\sqrt Q\).

For a positive semidefinite matrix \(A\), left and right multiplication by \(A\) are positive semidefinite self-adjoint operators on the full Hilbert–Schmidt matrix space: their quadratic forms are \(\mathop{\mathrm{tr}}(X^*AX)\) and \(\mathop{\mathrm{tr}}(X^*XA)\). Hence \[\mathcal L_{\sqrt P}\le\sqrt a\,\mathcal L_{\sqrt Q}, \qquad \mathcal L_{\sqrt P}^{-1} \ge a^{-1/2}\mathcal L_{\sqrt Q}^{-1}.\] Taking the quadratic form at \(K\) and using Lemma 2 proves (6).

In the semidefinite case, \(a\ge1\) gives \(P+\eta\mathrm{Id}\le a(Q+\eta\mathrm{Id})\). Apply the positive definite result and let \(\eta\downarrow0\). In a full eigenbasis of either matrix, every entry of \(K\) meeting its kernel is zero. Formula (5) therefore gives convergence to the two supported quadratic forms in (6). ◻

Proof. For \(x\in\ker T\), positivity of \(T+U\) and \(T-U\), whose quadratic forms both vanish at \(x\), implies \(Ux=0\). For \(\lvert t\rvert<1\), \[(1-\lvert t\rvert)T\le T+tU\le(1+\lvert t\rvert)T\le2T.\] Thus \(T+tU\) has the same support as \(T\). The function \(f(t)=\Phi(T+tU)\) satisfies, by Lemmas 2 and 3, \[f''(t)=b_{T+tU}(U)\ge 2^{-1/2}b_T(U).\] Subtracting \(b_T(U)t^2/(2\sqrt2)\) leaves a convex function on \((-1,1)\), continuous on \([-1,1]\). Its midpoint inequality is (7). If \(T=0\), all terms are zero. ◻

Proof. The identity \(\ker T_v=\ker R_v\) proves the support assertion. We first place the three quadratic forms on one space, then bound a cubic trace in an eigenbasis.

Proof. If \(a,b\) are the Hilbert–Schmidt norms of the two edges incident to \(v\), then \[e_v(\mathbf I) =\mathop{\mathrm{tr}}(T_v^{3/2}) \le(\mathop{\mathrm{tr}}T_v)^{3/2} =(a^2+b^2)^{3/2} \le\sqrt2(a^3+b^3).\] For an edge \(W_v\), its probability normalization gives \[ \lVert W_v\rVert_{\mathrm{HS}}^3 =\left(\frac1{\lvert I_v\rvert\lvert I_{v+1}\rvert} \int_{I_v\times I_{v+1}}\lvert F_v\rvert^2\right)^{3/2} \le \frac1{\lvert I_v\rvert\lvert I_{v+1}\rvert} \int_{I_v\times I_{v+1}}\lvert F_v\rvert^3. \tag{22}\] Every edge occurs in two row stars. Also, every ordered pair of intervals in \(\mathcal D_k\) determines exactly one admissible third interval, by XOR. Multiplication by \(2^{2k}=\lvert I_v\rvert\lvert I_{v+1}\rvert\) and summation of (22) therefore prove (20). These sums are finite: compact support leaves finitely many nonzero edge matrices, each belonging to one admissible triple.

Deleting the last bit of each child index shows that every admissible child has a unique admissible parent. As \(2^{2k}/4=2^{2(k-1)}\), the child part in the definition of \(\delta\) sums to \(J_{k-1}\), proving (21). ◻

Proof of Theorem 1 for step functions. The case \(S=\varnothing\) is immediate. Otherwise set \(k_-=\min S\), \(k_+=\max S\), and refine the common atom level, if necessary, so that \(m\le k_--1\). By (16), Proposition 6, and Lemma 7, \[\begin{align*} \sum_{k\in S}\sum_{\mathbf I\in\mathcal A_k}\lvert L_{\mathbf I}(F_0,F_1,F_2)\rvert &\le\frac{10\sqrt2}{3} \sum_{k=k_-}^{k_+}(J_{k-1}-J_k)\\ &\le\frac{10\sqrt2}{3}J_{k_--1} \le\frac{40}{3}\sum_{v=0}^2\lVert F_v\rVert_3^3. \end{align*}\] The insertion of missing scales is valid because each energy difference is nonnegative. If every input has nonzero norm, apply this estimate to \(F_v/\lVert F_v\rVert_3\). Trilinearity of each local integral gives (3), with constant \(40\). If any norm is zero, the local integrals vanish. ◻

Proof. Define the nonnegative edge averages \[a_v(\mathbf I)= \frac1{\lvert I_v\rvert\lvert I_{v+1}\rvert} \int_{I_v\times I_{v+1}}\lvert F_v\rvert^3.\] Hölder’s inequality on the product probability space of the three intervals, with the Haar signs bounded by one, gives \[\lvert L_{\mathbf I}(F_0,F_1,F_2)\rvert \le2^{2k}\prod_{v=0}^2 a_v(\mathbf I)^{1/3}.\] Apply Hölder again, now to the sum over \(\mathcal A_k\) with weight \(2^{2k}\). The uniqueness of the third interval yields \[\sum_{\mathbf I\in\mathcal A_k}2^{2k}a_v(\mathbf I)=\lVert F_v\rVert_3^3,\] proving (23). These bounds also justify all integrals and absolute sums for \(L^3\) inputs.

For two input triples \(F,G\), expand the difference of local integrals into the three terms with inputs \[(F_0-G_0,F_1,F_2),\quad (G_0,F_1-G_1,F_2),\quad (G_0,G_1,F_2-G_2).\] Estimate each term by (23). If \(G_v\to F_v\) in \(L^3\), the sum of these three product bounds tends to zero. This is the asserted \(\ell^1\) continuity. ◻

Completion of Theorem 1. Bounded compactly supported dyadic step functions are dense in \(L^3(\mathbb R_+^2)\). To see this, first approximate an \(L^3\) function by a continuous compactly supported function, then take its averages on sufficiently small dyadic squares. Uniform continuity and a common bounded support give convergence in \(L^3\).

For arbitrary real \(F_v\in L^3\), choose such step functions \(F_v^{(n)}\to F_v\) in \(L^3\). Assuming \(S\ne\varnothing\), refine their common atom levels, without changing the functions, to satisfy \(m_n\le\min S-1\). Apply the step-function estimate to these inputs. Lemma 8, summed over the fixed finite set \(S\), gives convergence of their local integrals in \(\ell^1\). The inequality \(\lvert \lvert a\rvert-\lvert b\rvert\rvert\le\lvert a-b\rvert\) permits passage to the absolute sum in (3), and norm convergence handles its right-hand side. This proves the theorem and its stated extension to real \(L^3\) inputs. The case \(S=\varnothing\) remains immediate.

For any coefficients \(\lvert \epsilon_{\mathbf I}\rvert\le1\), the triangle inequality gives the corresponding bound for \[\sum_{k\in S}\sum_{\mathbf I\in\mathcal A_k} \epsilon_{\mathbf I}L_{\mathbf I}(F_0,F_1,F_2). \qedhere\] ◻