Proof. For each \(k=0,\dots,j-1\), a direct computation on the parent interval \(I_{n,k}\) gives \[\begin{align} &f(t_{n+1,2k})\big(g(t_{n+1,2k+1})-g(t_{n+1,2k})\big)+f(t_{n+1,2k+1})\big(g(t_{n+1,2k+2})-g(t_{n+1,2k+1})\big)\\ &-f(t_{n,k})\big(g(t_{n,k+1})-g(t_{n,k})\big)\\ = \, &\big(f(t_{n+1,2k+1})-f(t_{n+1,2k})\big)\big(g(t_{n+1,2k+2})-g(t_{n+1,2k+1})\big). \end{align}\] Summing over \(k=0,\dots,j-1\) proves the result. ◻

Proof. Consider a left child increment across \(I_{n,k}^L=I_{n+1,2k}\). Since the endpoints \(t_{n+1,2k}\) and \(t_{n+1,2k+1}\) are level-\((n+1)\) dyadic points, all Faber–Schauder levels \(m\ge n+1\) vanish at both endpoints. Hence, only levels \(m\le n\) contribute; for \(0 \le m \le n\), by the child-wise sign above, \[e_{m,\kappa(m;n,k)}(t_{n+1,2k+1}) - e_{m,\kappa(m;n,k)}(t_{n+1,2k}) = 2^{-(n+1)}2^{\frac{m}{2}}\varepsilon_m^L(n,k).\] Therefore, with the affine level \(m=-1\), we obtain \[f_{n+1,L}(k)=2^{-(n+1)}\theta_{-1,0}^f+\sum_{m=0}^{n}2^{-(n+1)}2^{\frac{m}{2}}\varepsilon_m^L(n,k)\theta_{m,\kappa(m;n,k)}^f.\] By Minkowski’s inequality, \[\bigg(\sum_{k\in I_n}\big|f_{n+1,L}(k)\big|^r\bigg)^{\frac{1}{r}}\le \bigg(\sum_{k\in I_n}\big|2^{-(n+1)}\theta_{-1,0}^f\big|^r\bigg)^{\frac{1}{r}}+\sum_{m=0}^{n}\bigg(\sum_{k\in I_n}\Big|2^{-(n+1)}2^{\frac{m}{2}}\theta_{m,\kappa(m;n,k)}^f\Big|^r\bigg)^{\frac{1}{r}}.\] From Definition 2, the affine-level contribution is \[\bigg(\sum_{k\in I_n}\big|2^{-(n+1)}\theta_{-1,0}^f\big|^r\bigg)^{\frac{1}{r}}=2^{-(n+1)}2^{\frac{n}{r}}\big|\theta_{-1,0}^f\big|=\frac{1}{2}K_r(n+1)\zeta_{f}^{(r)}(-1).\] This is precisely the reason for the normalization factor \(2^{1-\frac{1}{r}}\) in the affine level: with this choice, the affine contribution appears as the \(m=-1\) term in the same convolutional estimate as the levels \(m\ge0\).

For fixed \(0\le m\le n\), every coefficient \(\theta_{m,\ell}^f\) appears exactly \(2^{n-m}\) times as \(k\) ranges over \(I_n\). Thus \[\begin{align} \bigg(\sum_{k\in I_n}\Big|2^{-(n+1)}2^{\frac{m}{2}}\theta_{m,\kappa(m;n,k)}^f\Big|^r\bigg)^{\frac{1}{r}} &=\bigg(2^{n-m}2^{-r(n+1)}2^{\frac{mr}{2}}\sum_{\ell\in I_m}\big|\theta_{m,\ell}^f\big|^r\bigg)^{\frac{1}{r}}\\ &=\frac{1}{2}K_r(n-m)\zeta_{f}^{(r)}(m). \end{align}\] Summing over \(m = -1, 0, \dots, n\) gives \[\bigg(\sum_{k\in I_n}\big|f_{n+1,L}(k)\big|^r\bigg)^{\frac{1}{r}}\le \frac{1}{2}\sum_{m=-1}^{n}K_r(n-m)\zeta_{f}^{(r)}(m)=\frac{1}{2}\,\mathcal{A}^{(r)}_f(n).\] The proof for right-child increments is identical, using the signs \(\varepsilon_m^R(n,k)\). ◻

Proof. By Hölder’s inequality and Lemma 2, \[\big|S_{n+1}^L(t_{n,j};f,g)-S_n^L(t_{n,j};f,g)\big|\le \bigg(\sum_{k\in I_n}\big|f_{n+1,L}(k)\big|^p\bigg)^{\frac{1}{p}}\bigg(\sum_{k\in I_n}\big|g_{n+1,R}(k)\big|^q\bigg)^{\frac{1}{q}}.\] Applying Lemma 3 to \(f\) with exponent \(p\) and to \(g\) with exponent \(q\) proves the estimate. ◻

Proof. Fix \(t\in\mathbb{T}\). For \(M>N\ge\ell(t)\), \[\big|S_M^L(t;f,g)-S_N^L(t;f,g)\big|\le \sum_{n=N}^{M-1}\big|S_{n+1}^L(t;f,g)-S_n^L(t;f,g)\big|.\] Since \(t\in\mathbb{T}_n\) for every \(n\ge\ell(t)\), Proposition 1 gives \[\big|S_M^L(t;f,g)-S_N^L(t;f,g)\big|\le \frac{1}{4}\sum_{n=N}^{M-1}\mathcal{A}^{(p)}_f(n) \, \mathcal{A}^{(q)}_g(n).\] By 17 , the right-hand side tends to \(0\) as \(M, N \to \infty\). Hence, \(\big(S_n^L(t;f,g)\big)_{n\ge\ell(t)}\) is Cauchy and therefore converges. ◻

Proof. Define the shifted sequence \(\widetilde{\zeta}_f^{p}=(\widetilde{\zeta}_f^{p}(j))_{j\ge0}\) by \[\widetilde{\zeta}_f^{p}(j):=\zeta_{f}^{(p)}(j-1), \qquad j\ge0.\] Then \(\widetilde{\zeta}_f^{p}\in\ell^p(\mathbb{N}_0)\) and, for every \(n\ge0\), \[\mathcal{A}^{(p)}_f(n)=(K_p*\widetilde{\zeta}_f^{p})_{n+1}.\] Hence, by Young’s convolution inequality and 12 , \[\label{ineq:Ap-bound} \Vert\mathcal{A}^{(p)}_f\Vert_{\ell^p(\mathbb{N}_0)} \le \Vert K_p*\widetilde{\zeta}_f^{p} \Vert_{\ell^p(\mathbb{N}_0)} \le \Vert K_p\Vert_{\ell^1(\mathbb{N}_0)} \Vert\widetilde{\zeta}_f^{p}\Vert_{\ell^p(\mathbb{N}_0)} = \Vert K_p\Vert_{\ell^1(\mathbb{N}_0)} \Vert\zeta_{f}^{(p)}\Vert_{\ell^p(\mathbb{N}_{-1})}.\tag{18}\] Thus \(\mathcal{A}^{(p)}_f\in\ell^p(\mathbb{N}_0)\). Similarly, \(\mathcal{A}^{(q)}_g\in\ell^q(\mathbb{N}_0)\). Therefore, by Hölder’s inequality, \[\label{ineq:A95pA95q-l1} \sum_{n=0}^{\infty} \mathcal{A}^{(p)}_f(n) \, \mathcal{A}^{(q)}_g(n) \le \Vert\mathcal{A}^{(p)}_f\Vert_{\ell^p(\mathbb{N}_0)} \, \Vert\mathcal{A}^{(q)}_g\Vert_{\ell^q(\mathbb{N}_0)} < \infty.\tag{19}\] The conclusion follows from Theorem 3. ◻

Proof. We first prove 21 . Fix \(N\ge0\) and \(j=0,\dots,2^N-1\). Since \(t_{N,j},t_{N,j+1}\in\mathbb{T}_N\), the level-\(N\) left sum over \([t_{N,j},t_{N,j+1}]\) is exactly \(f(t_{N,j})\big(g(t_{N,j+1})-g(t_{N,j})\big)\). By telescoping the refinements from level \(N\) to infinity, \[I_{f,g}^{\mathbb{T}}(t_{N,j+1})-I_{f,g}^{\mathbb{T}}(t_{N,j}) = f(t_{N,j})\big(g(t_{N,j+1})-g(t_{N,j})\big)+\sum_{m=N}^{\infty}\Delta_m(t_{N,j},t_{N,j+1}),\] where \[\Delta_m(a,b):=\big(S_{m+1}^L(b;f,g)-S_m^L(b;f,g)\big)-\big(S_{m+1}^L(a;f,g)-S_m^L(a;f,g)\big).\] By Lemma 2, \[\Delta_m(t_{N,j},t_{N,j+1}) = \sum_{k:\,I_{m,k}\subset[t_{N,j},t_{N,j+1}]} f_{m+1,L}(k)g_{m+1,R}(k).\] Hence, by Hölder’s inequality and Lemma 3, \[\big|\Delta_m(t_{N,j},t_{N,j+1})\big| \le \sum_{k\in I_m}\big|f_{m+1,L}(k)\big|\big|g_{m+1,R}(k)\big| \le \frac{1}{4} \mathcal{A}^{(p)}_f(m) \, \mathcal{A}^{(q)}_g(m).\] As in the proof of Corollary 1, the assumptions \(\zeta_{f}^{(p)}\in\ell^p(\mathbb{N}_{-1})\) and \(\zeta_{g}^{(q)}\in\ell^q(\mathbb{N}_{-1})\) imply \(\mathcal{A}^{(p)}_f \mathcal{A}^{(q)}_g \in\ell^1(\mathbb{N}_0)\), therefore the series \[R_{N, j} := \sum_{m=N}^{\infty}\Delta_m(t_{N,j},t_{N,j+1})\] is absolutely convergent and 21 follows.

We next prove 22 . Fix \(t\in\mathbb{T}\) and define \(t_n=2^{-n}\lfloor2^nt\rfloor\). If \(t_{n+1}=t_n\), then the corresponding increment is zero, and we set \(R_n(t)=0\). If \(t_{n+1}>t_n\), then \([t_n,t_{n+1}]\) is one level-\((n+1)\) dyadic interval. Telescoping from level \(n+1\) to infinity gives \[I_{f,g}^{\mathbb{T}}(t_{n+1})-I_{f,g}^{\mathbb{T}}(t_n) = f(t_n)\big(g(t_{n+1})-g(t_n)\big)+\sum_{m=n+1}^{\infty}\Delta_{m,n}(t),\] where \[\Delta_{m,n}(t):=\big(S_{m+1}^L(t_{n+1};f,g)-S_m^L(t_{n+1};f,g)\big)-\big(S_{m+1}^L(t_n;f,g)-S_m^L(t_n;f,g)\big).\] We define \[R_n(t):=\sum_{m=n+1}^{\infty}\Delta_{m,n}(t).\] For \(m\ge n+1\), Lemma 2 gives \[\Delta_{m,n}(t)=\sum_{k:\,I_{m,k}\subset[t_n,t_{n+1}]}f_{m+1,L}(k)g_{m+1,R}(k).\] For fixed \(m\), the intervals \([t_n,t_{n+1}]\), \(n=N,\dots,m-1\), are pairwise disjoint up to endpoints. Therefore the corresponding index sets \(\{k\in I_m:I_{m,k}\subset[t_n,t_{n+1}]\}\) are disjoint. Hence, by Hölder’s inequality and Lemma 3, \[\sum_{n=N}^{m-1}\big|\Delta_{m,n}(t)\big| \le \sum_{k\in I_m}\big|f_{m+1,L}(k)\big|\big|g_{m+1,R}(k)\big| \le \frac{1}{4} \mathcal{A}^{(p)}_f(m) \, \mathcal{A}^{(q)}_g(m).\] Tonelli’s theorem yields \[\sum_{n=N}^{\infty}\big|R_n(t)\big| \le \sum_{m=N}^{\infty}\sum_{n=N}^{m-1}\big|\Delta_{m,n}(t)\big| \le \frac{1}{4}\sum_{m=N}^{\infty} \mathcal{A}^{(p)}_f(m) \, \mathcal{A}^{(q)}_g(m) = \frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] ◻

Proof. Fix \(N\ge0\) and \(u\in[t_{N,k},t_{N,k+1}]\). Consider the level-\(N\) dyadic interpolation \[\Pi_Nh(v):=h(0)+\sum_{m=-1}^{N-1}\sum_{\ell\in I_m}\theta_{m,\ell}^h e_{m,\ell}(v), \qquad v\in[0,1].\] Since the Schauder tail vanishes at level-\(N\) dyadic points, \[\Pi_Nh(t_{N,k})=h(t_{N,k}), \qquad \Pi_Nh(t_{N,k+1})=h(t_{N,k+1}).\] Moreover, \(\Pi_Nh\) is affine on \([t_{N,k},t_{N,k+1}]\). Hence \[\big|\Pi_Nh(u)-\Pi_Nh(t_{N,k})\big| \le \big|\Pi_Nh(t_{N, k+1})-\Pi_Nh(t_{N,k})\big| = \big|h(t_{N,k+1})-h(t_{N,k})\big|,\] and the same estimate holds with \(t_{N,k}\) replaced by \(t_{N,k+1}\). The increment \(h(t_{N,k+1})-h(t_{N,k})\) is the sum of the two child increments over the level-\(N\) interval, so Lemma 3 gives \[\big|h(t_{N,k+1})-h(t_{N,k})\big| \le \mathcal{A}^{(r)}_h(N).\] For the tail term, at each level \(m\ge N\), at most one Faber–Schauder tent is nonzero at \(u\), and its height is at most \(2^{-\frac{m}{2}-1}\). Therefore \[\label{ineq:N-dyadic-interpol} \big|h(u)-\Pi_Nh(u)\big| = \bigg|\sum_{m=N}^{\infty}\sum_{\ell \in I_m} \theta_{m,\ell}^h \, e_{m,\ell}(u)\bigg| \le \frac{1}{2}\sum_{m=N}^{\infty}2^{-\frac{m}{2}}\sup_{\ell\in I_m}\big|\theta_{m,\ell}^h\big| = \frac{1}{2}H_h(N).\tag{26}\] Combining these estimates proves the two estimates 23 .

If \(|u-v|\le2^{-N}\), then \(u\) and \(v\) lie either in the same level-\(N\) dyadic interval or in two adjacent level-\(N\) dyadic intervals. In both cases, there is a level-\(N\) dyadic point \(z\) such that \(u\) and \(v\) each lie in a level-\(N\) interval having \(z\) as one endpoint. Hence \[\big|h(u)-h(v)\big| \le 2\mathcal{A}^{(r)}_h(N)+H_h(N).\] Taking the supremum over all such \(u,v\) proves the oscillation estimate 24 .

Now fix \(t\in\mathbb{T}\cap[0,1)\) and \(n\ge0\). Applying the first endpoint estimate at level \(n\) to the interval containing \(t\) gives \[\big|h(t)-h(t_n)\big| \le \mathcal{A}^{(r)}_h(n)+\frac{1}{2}H_h(n).\] Taking the \(\ell^r\) norm over \(n\ge N\) and then the supremum over \(t\in\mathbb{T}\cap[0,1)\) proves 25 . ◻

Proof. The claim is trivial at \(t=1\), since \(t_N=1\) for all \(N\). Fix \(t\in\mathbb{T}\cap[0,1)\). Since \(t_n=2^{-n}\lfloor2^nt\rfloor\) satisfies \(t_n\uparrow t\) and \(t_n=t\) for all sufficiently large \(n\), we have \[F(t)-F_N(t)=\sum_{n=N}^{\infty}\big(F(t_{n+1})-F(t_n)\big).\] By Lemma 4 (ii), \[F(t_{n+1})-F(t_n)=f(t_n)\big(g(t_{n+1})-g(t_n)\big)+R_n(t),\] with \[\sum_{n=N}^{\infty} \big|R_n(t)\big| \le \frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] Therefore \[F(t)-F(t_N)-f(t)\big(g(t)-g(t_N)\big) = \sum_{n=N}^{\infty}\big(f(t_n)-f(t)\big)\big(g(t_{n+1})-g(t_n)\big)+\sum_{n=N}^{\infty}R_n(t).\] Each nonzero increment \(g(t_{n+1})-g(t_n)\) is a left-child increment of \(g\) at level \(n+1\). Hence Lemma 3 gives \[\big|g(t_{n+1})-g(t_n)\big|\le \frac{1}{2}\mathcal{A}^{(q)}_g(n).\] By Hölder’s inequality and Lemma 5, \[\sum_{n=N}^{\infty}\big|f(t_n)-f(t)\big|\big|g(t_{n+1})-g(t_n)\big|\le \frac{1}{2}\Vert\mathcal{A}^{(q)}_g\Vert_{\ell^q_{\ge N}}\bigg(\Vert\mathcal{A}^{(p)}_f\Vert_{\ell^p_{\ge N}}+\frac{1}{2}\Vert H_f\Vert_{\ell^p_{\ge N}}\bigg).\] Combining the last two estimates proves for every \(N\ge0\) and every \(t\in\mathbb{T}\), \[\begin{align} \Big|F(t)-F(t_N)&-f(t)\big(g(t)-g(t_N)\big)\Big| \label{eq:projection-bound-on-T} \\ &\le \frac{1}{2}\Vert\mathcal{A}^{(q)}_g\Vert_{\ell^q_{\ge N}}\bigg(\Vert\mathcal{A}^{(p)}_f\Vert_{\ell^p_{\ge N}}+\frac{1}{2}\Vert H_f\Vert_{\ell^p_{\ge N}}\bigg)+\frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}. \nonumber \end{align}\tag{27}\] Finally, \[|F(t)-F_N(t)|\le \big|F(t)-F(t_N)-f(t)\big(g(t)-g(t_N)\big)\big|+\|f\|_{\infty}\omega_g(2^{-N}).\] Taking the supremum over \(t\in\mathbb{T}\) gives \[\begin{align} \sup_{t\in\mathbb{T}}\big|F(t)&-F_N(t)\big| \\ &\le \frac{1}{2}\Vert \mathcal{A}^{(q)}_g \Vert_{\ell^q_{\ge N}}\bigg(\Vert \mathcal{A}^{(p)}_f \Vert_{\ell^p_{\ge N}}+\frac{1}{2}\Vert H_f\Vert_{\ell^p_{\ge N}}\bigg) + \frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}} + \|f\|_{\infty}\omega_g(2^{-N}). \end{align}\] The right-hand side converges to \(0\) as \(N\to\infty\), since \(\mathcal{A}^{(p)}_f\in\ell^p\), \(\mathcal{A}^{(q)}_g \in\ell^q\), \(H_f\in\ell^p\), \(\mathcal{A}^{(p)}_f \mathcal{A}^{(q)}_g\in\ell^1\), and \(g\) is uniformly continuous. ◻

Proof. By Lemma 4 (i), \[F(t_{N,j+1})-F(t_{N,j})=f(t_{N,j})\big(g(t_{N,j+1})-g(t_{N,j})\big)+R_{N,j},\] where \[|R_{N,j}|\le\frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] The first term is bounded by \(\|f\|_{\infty}\omega_g(2^{-N})\). Taking the maximum over \(j\) gives \[\max_{0\le j<2^N}\big|F(t_{N,j+1})-F(t_{N,j})\big|\le\|f\|_{\infty}\omega_g(2^{-N})+\frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] The right-hand side tends to zero because \(g\) is uniformly continuous and \(\mathcal{A}^{(p)}_f \mathcal{A}^{(q)}_g\in\ell^1\). ◻

Proof of Theorem 4. For \(t\in\mathbb{T}\), set \[F(t):=I_{f,g}^{\mathbb{T}}(t), \qquad F_N(t):=F(2^{-N}\lfloor2^Nt\rfloor).\] By Lemma 6, \[\sup_{u\in\mathbb{T}}\big|F(u)-F_N(u)\big|\xrightarrow{N\to\infty}0.\] By Lemma 7, \[\max_{0\le j<2^N}\big|F(t_{N,j+1})-F(t_{N,j})\big|\xrightarrow{N\to\infty}0.\] Let \(\varepsilon>0\). Choose \(N\) large enough so that \[2\sup_{u\in\mathbb{T}}\big|F(u)-F_N(u)\big|+\max_{0\le j<2^N}\big|F(t_{N,j+1})-F(t_{N,j})\big|<\varepsilon.\] If \(s,t\in\mathbb{T}\) and \(|s-t|<2^{-N}\), then \[s_N:=2^{-N}\lfloor2^Ns\rfloor, \qquad t_N:=2^{-N}\lfloor2^Nt\rfloor\] are either equal or adjacent level-\(N\) dyadic points. Hence \[\big|F_N(s)-F_N(t)\big|=\big|F(s_N)-F(t_N)\big|\le\max_{0\le j<2^N}\big|F(t_{N,j+1})-F(t_{N,j})\big|.\] Therefore \[\big|F(s)-F(t)\big|\le\big|F(s)-F_N(s)\big|+\big|F_N(s)-F_N(t)\big|+\big|F_N(t)-F(t)\big|<\varepsilon.\] This proves uniform continuity of \(F=I_{f,g}^{\mathbb{T}}\) on \(\mathbb{T}\). Since \(\mathbb{T}\) is dense in \([0,1]\) and \(\mathbb{R}\) is complete, the standard extension theorem for uniformly continuous maps on dense subsets gives a unique continuous extension of \(F\) to \([0,1]\). ◻

Proof. By Minkowski’s inequality, \[\begin{align} \Vert H_f\Vert_{\ell^p} &= \bigg\Vert\sum_{m=0}^{\infty} 2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta_{m,k}^f\big| \mathbf{1}_{\{0,\dots,m\}}\bigg\Vert_{\ell^p} \\ &\le \sum_{m=0}^{\infty} \Big( 2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta_{m,k}^f\big| \Big) \big\Vert\mathbf{1}_{\{0,\dots,m\}}\big\Vert_{\ell^p} = \sum_{m=0}^{\infty}(m+1)^{\frac{1}{p}} 2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta_{m,k}^f\big|. \end{align}\] The \(m=0\) term is finite, and \((m+1)^{\frac{1}{p}}\le2^{\frac{1}{p}}m^{\frac{1}{p}}\) holds for \(m\ge1\). Hence 28 implies \(\Vert H_f\Vert_{\ell^p}<\infty\). The conclusion follows from Theorem 4. ◻

Proof. For \(t\in\mathbb{T}\), bilinearity of the finite dyadic left sums gives \[S_n^L(t;\alpha f_1+\beta f_2,g) = \alpha S_n^L(t;f_1,g)+\beta S_n^L(t;f_2,g).\] Letting \(n\to\infty\) shows that \(I_{\alpha f_1+\beta f_2,g}^{\mathbb{T}}(t)\) exists and satisfies \[I_{\alpha f_1+\beta f_2,g}^{\mathbb{T}}(t) = \alpha I_{f_1,g}^{\mathbb{T}}(t)+\beta I_{f_2,g}^{\mathbb{T}}(t), \qquad t\in\mathbb{T}.\] The right-hand side has a continuous extension to \([0,1]\), namely \(\alpha I_{f_1,g}+\beta I_{f_2,g}\). Hence, \(I_{\alpha f_1+\beta f_2,g}\) exists on \([0,1]\) and the identity holds everywhere on \([0, 1]\) by density of \(\mathbb{T}\).

The proof of linearity in the second variable is identical, using the linearity of the finite dyadic left sums in the second variable.

The bilinearity identities for interval integrals follow by subtracting the corresponding identities at \(s\) and at \(t\). Additivity follows directly from the definition of the interval integral: \[\int_r^s f\,dg+\int_s^t f\,dg = \big(I_{f,g}(s)-I_{f,g}(r)\big)+\big(I_{f,g}(t)-I_{f,g}(s)\big) = I_{f,g}(t)-I_{f,g}(r)= \int_r^t f\,dg.\] The identities \(\int_s^s f\,dg=0\) and \(\int_s^t f\,dg=-\int_t^s f\,dg\) are immediate. ◻

Proof. By Corollary 1, both integral maps \(I_{f,g}^{\mathbb{T}}\) and \(I_{g,f}^{\mathbb{T}}\) are well-defined on \(\mathbb{T}\).

We first prove the integration-by-parts identity on \(\mathbb{T}\). Fix \(t\in\mathbb{T}\) and let \(n\ge\ell(t)\). At level \(n\), the finite dyadic sums satisfy \[S_n^L(t;f,g)+S_n^L(t;g,f) = f(t)g(t)-f(0)g(0) - \sum_{k=0}^{2^nt-1}\big(f(t_{n,k+1})-f(t_{n,k})\big) \big(g(t_{n,k+1})-g(t_{n,k})\big).\] It remains to show that the last term converges to zero. By Hölder’s inequality, \[\begin{align} &\bigg|\sum_{k=0}^{2^nt-1}\big(f(t_{n,k+1})-f(t_{n,k})\big)\big(g(t_{n,k+1})-g(t_{n,k})\big)\bigg| \\ & \qquad \qquad \qquad \le \bigg(\sum_{k\in I_n}\big|f(t_{n,k+1})-f(t_{n,k})\big|^p\bigg)^{\frac{1}{p}} \bigg(\sum_{k\in I_n}\big|g(t_{n,k+1})-g(t_{n,k})\big|^q\bigg)^{\frac{1}{q}}. \end{align}\] Since \(f(t_{n,k+1})-f(t_{n,k})=f_{n+1,L}(k)+f_{n+1,R}(k)\), Lemma 3 gives \[\bigg(\sum_{k\in I_n}\big|f(t_{n,k+1})-f(t_{n,k})\big|^p\bigg)^{\frac{1}{p}} \le \mathcal{A}^{(p)}_f(n),\] and a similar bound for \(g\). Hence, \[\bigg|\sum_{k=0}^{2^nt-1}\big(f(t_{n,k+1})-f(t_{n,k})\big)\big(g(t_{n,k+1})-g(t_{n,k})\big)\bigg| \le \mathcal{A}^{(p)}_f(n) \, \mathcal{A}^{(q)}_g(n).\] As in the proof of Corollary 1, the conditions \(\zeta_{f}^{(p)}\in\ell^p(\mathbb{N}_{-1})\) and \(\zeta_{g}^{(q)}\in\ell^q(\mathbb{N}_{-1})\) imply \[\mathcal{A}^{(p)}_f\,\mathcal{A}^{(q)}_g\in\ell^1(\mathbb{N}_0), \qquad \text{and in particular} \qquad \mathcal{A}^{(p)}_f(n)\mathcal{A}^{(q)}_g(n)\xrightarrow{n\to\infty}0.\] Letting \(n\to\infty\) in the finite dyadic sums identity gives \[I_{f,g}^{\mathbb{T}}(t)+I_{g,f}^{\mathbb{T}}(t)=f(t)g(t)-f(0)g(0), \qquad t\in\mathbb{T}.\]

Now define \(I_{g,f}\) by 29 . This function is continuous on \([0,1]\) and agrees with \(I_{g,f}^{\mathbb{T}}\) on the dense set \(\mathbb{T}\). Hence, it is the unique continuous extension of \(I_{g,f}^{\mathbb{T}}\). The displayed integration-by-parts identities follow immediately. ◻

Proof. If \(u\in\mathbb{T}\), the estimate is exactly 27 applied to the pair \((h,g)\).

For a general \(u\in[0,1)\), consider \(u_m=2^{-m}\lfloor2^m u\rfloor\) for \(m\ge N\). Then \(u_m\in\mathbb{T}\), \(u_m\to u\), and \[2^{-N}\lfloor2^Nu_m\rfloor=u_N.\] Applying the estimate already proved for dyadic points to \(u_m\) gives \[\begin{align} \Big|I_{h,g}(u_m)-I_{h,g}(u_N)-h(u_m)\big(g(u_m)-g(u_N)\big)\Big| \le \frac{1}{2}\Vert\mathcal{A}^{(q)}_g \Vert_{\ell^q_{\ge N}} L^{(p)}_h(N) + \frac{1}{4} \big\Vert \mathcal{A}^{(p)}_h \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}. \end{align}\] Letting \(m\to\infty\) and using the continuity of \(I_{h,g}\), \(h\), and \(g\) gives the estimate at \(u\). The case \(u=1\) is trivial because \(u_N=1\) and the left-hand side is zero. ◻

Proof. Set \(h:=f-f(\xi)\). Then \[\zeta_{h}^{(p)}=\zeta_{f}^{(p)}, \qquad \mathcal{A}^{(p)}_h=\mathcal{A}^{(p)}_f, \qquad H_h=H_f.\] By the bilinearity of Faber–Schauder integrals, \[\int_s^t f\,dg-f(\xi)\big(g(t)-g(s)\big) = \int_s^t h\,dg.\] Consider \(s_N:=2^{-N}\lfloor2^Ns\rfloor\) and \(t_N:=2^{-N}\lfloor2^Nt\rfloor\). Since \(s\le t\) and \(t-s\le2^{-N}\), the dyadic points \(s_N\) and \(t_N\) are either equal or adjacent level-\(N\) dyadic points. We decompose \[\int_s^t h\,dg = \big(I_{h,g}(t)-I_{h,g}(t_N)\big) + \big(I_{h,g}(t_N)-I_{h,g}(s_N)\big) + \big(I_{h,g}(s_N)-I_{h,g}(s)\big).\]

We first estimate the two boundary terms. By Lemma 8, applied at \(t\) and at \(s\), \[\Big|I_{h,g}(t)-I_{h,g}(t_N)-h(t)\big(g(t)-g(t_N)\big)\Big| + \Big|I_{h,g}(s)-I_{h,g}(s_N)-h(s)\big(g(s)-g(s_N)\big)\Big|\] is bounded by \[\Vert\mathcal{A}^{(q)}_g\Vert_{\ell^q_{\ge N}}L^{(p)}_f(N) + \frac{1}{2} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] Since \(s\le\xi\le t\) and \(t-s\le2^{-N}\), the oscillation estimate 24 in Lemma 5, applied with \(r=p\) to \(h\), gives \[|h(s)| = |f(s) - f(\xi)| \le2D^{(p)}_f(N), \qquad |h(t)| = |f(t) - f(\xi)| \le2D^{(p)}_f(N).\] The endpoint estimate 23 in Lemma 5, applied with \(r=q\) to \(g\), gives \[\big|g(s)-g(s_N)\big|\le D^{(q)}_g(N), \qquad \big|g(t)-g(t_N)\big|\le D^{(q)}_g(N).\] Hence, the product part of the two boundary terms is bounded by \[|h(t)|\big|g(t)-g(t_N)\big| + |h(s)|\big|g(s)-g(s_N)\big| \le 4D^{(p)}_f(N)D^{(q)}_g(N).\]

It remains to estimate the middle term \(I_{h,g}(t_N)-I_{h,g}(s_N)\). If \(s_N=t_N\), then it is zero. Otherwise we can assume that \(s_N\) and \(t_N\) are adjacent level-\(N\) dyadic points. Lemma 4 (i) gives \[I_{h,g}^{\mathbb{T}}(t_N)-I_{h,g}^{\mathbb{T}}(s_N) = h(s_N)\big(g(t_N)-g(s_N)\big)+R_N,\] where \[|R_N| \le \frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] We claim that \[|h(s_N)|=\big|f(s_N)-f(\xi)\big|\le2D^{(p)}_f(N).\] Indeed, since \(s_N\) and \(t_N\) are adjacent, if \(\xi\le t_N\), then \(\xi\) lies in the level-\(N\) interval with left endpoint \(s_N\), so the endpoint estimate 23 in Lemma 5 gives \[\big|f(s_N)-f(\xi)\big|\le D^{(p)}_f(N).\] If \(\xi\ge t_N\), then \[\big|f(s_N)-f(\xi)\big| \le \big|f(s_N)-f(t_N)\big|+\big|f(t_N)-f(\xi)\big| \le 2D^{(p)}_f(N).\] Thus, the claim holds. Also, \(g(t_N)-g(s_N)\) is the increment of \(g\) over one level-\(N\) dyadic interval, so Lemma 3 gives \[\big|g(t_N)-g(s_N)\big|\le \mathcal{A}^{(q)}_g(N).\] Therefore, the middle term is bounded by \[2D^{(p)}_f(N)\mathcal{A}^{(q)}_g(N) + \frac{1}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\]

Combining the boundary and middle estimates yields \[\bigg|\int_s^t h\,dg\bigg| \le 4D^{(p)}_f(N)D^{(q)}_g(N) + 2D^{(p)}_f(N)\mathcal{A}^{(q)}_g(N) + \Vert\mathcal{A}^{(q)}_g\Vert_{\ell^q_{\ge N}}L^{(p)}_f(N) + \frac{3}{4} \big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}}.\] Note that \[D^{(p)}_f(N) \le L^{(p)}_f(N), \qquad \mathcal{A}^{(p)}_f(N) \le \Vert\mathcal{A}^{(p)}_f\Vert_{\ell^p_{\ge N}} \le L^{(p)}_f(N),\] with similar inequalities for \(f, p\) replaced by \(g, q\). By Hölder’s inequality, \[\big\Vert \mathcal{A}^{(p)}_f \, \mathcal{A}^{(q)}_g \big\Vert_{\ell^1_{\ge N}} \le \Vert\mathcal{A}^{(p)}_f\Vert_{\ell^p_{\ge N}} \Vert\mathcal{A}^{(q)}_g\Vert_{\ell^q_{\ge N}} \le L^{(p)}_f(N)L^{(q)}_g(N),\] thus, we obtain \[\bigg|\int_s^t h\,dg\bigg| \le \Big(4+2+1+\frac{3}{4}\Big)L^{(p)}_f(N)L^{(q)}_g(N) \le 8L^{(p)}_f(N)L^{(q)}_g(N).\] ◻

Proof. It is straightforward to check, using the linearity of the Faber–Schauder coefficients and the subadditivity of the level energies and Schauder tails, that \(\mathcal{E}^p\) is a vector space and that \(\|\cdot\|_{\mathcal{E}^p}\) is a norm.

Let \((f_m)_{m\ge1}\) be a Cauchy sequence in \(\mathcal{E}^p\). Then \((f_m)\) is Cauchy in the uniform norm, and hence converges uniformly to some \(f\in C([0,1])\). For each fixed coefficient index \((n,k)\), the coefficient formulae 9 , 10 show that the map \(h\mapsto\theta_{n,k}^h\) is continuous with respect to the uniform norm. Hence \[\theta_{n,k}^{f_m-f} \longrightarrow 0 \qquad \text{as } m\to\infty\] for every fixed \((n,k)\).

We first prove the convergence of the level energy part. Fix \(m\) and \(N\ge0\). Since \(f_\ell\to f\) uniformly, \[\sum_{n=-1}^{N}\big(\zeta_{f_m-f}^{(p)}(n)\big)^p = \lim_{\ell\to\infty} \sum_{n=-1}^{N} \big(\zeta_{f_m-f_\ell}^{(p)}(n)\big)^p.\] Therefore \[\sum_{n=-1}^{N}\big(\zeta_{f_m-f}^{(p)}(n)\big)^p \le \liminf_{\ell\to\infty}\Vert\zeta_{f_m-f_\ell}^{(p)}\Vert_{\ell^p(\mathbb{N}_{-1})}^p.\] Since \((f_m)\) is Cauchy in \(\mathcal{E}^p\), the right-hand side tends to \(0\) as \(m\to\infty\), uniformly in \(N\). Letting \(N\to\infty\) gives \[\Vert\zeta_{f_m-f}^{(p)}\Vert_{\ell^p(\mathbb{N}_{-1})}\longrightarrow0.\]

We next prove the convergence of the Schauder tail part. For fixed \(m\), \(n\ge0\), and \(M\ge n\), \[\sum_{r=n}^{M}2^{-\frac{r}{2}}\sup_{k\in I_r}\big|\theta_{r,k}^{f_m-f}\big| = \lim_{\ell\to\infty} \sum_{r=n}^{M}2^{-\frac{r}{2}}\sup_{k\in I_r}\big|\theta_{r,k}^{f_m-f_\ell}\big|,\] hence \[\sum_{r=n}^{M}2^{-\frac{r}{2}}\sup_{k\in I_r}\big|\theta_{r,k}^{f_m-f}\big| \le \liminf_{\ell\to\infty}H_{f_m-f_\ell}(n).\] Letting \(M\to\infty\) yields \[H_{f_m-f}(n)\le\liminf_{\ell\to\infty}H_{f_m-f_\ell}(n), \qquad n\ge0.\] Therefore, for every \(N\ge0\), \[\sum_{n=0}^{N} \big(H_{f_m-f}(n)\big)^p \le \liminf_{\ell\to\infty}\sum_{n=0}^{N} \big(H_{f_m-f_\ell}(n)\big)^p \le \liminf_{\ell\to\infty}\Vert H_{f_m-f_\ell}\Vert_{\ell^p(\mathbb{N}_0)}^p.\] Since \((f_m)\) is Cauchy in \(\mathcal{E}^p\), the right-hand side tends to \(0\) as \(m\to\infty\), uniformly in \(N\). Letting \(N\to\infty\) gives \[\Vert H_{f_m-f}\Vert_{\ell^p(\mathbb{N}_0)}\longrightarrow0.\]

Combining uniform convergence, level energy convergence, and Schauder tail convergence, we obtain \[\|f_m-f\|_{\mathcal{E}^p}\longrightarrow0.\] In particular, \(f\in\mathcal{E}^p\), and \(\mathcal{E}^p\) is complete. ◻

Proof. For \(n\ge0\), the monotonicity of \(\ell^r\) norms gives \[\zeta_{f}^{(q)}(n) = 2^{-\frac{n}{2}}\bigg(\sum_{k\in I_n}\big|\theta^f_{n,k}\big|^q\bigg)^{\frac{1}{q}} \le 2^{-\frac{n}{2}}\bigg(\sum_{k\in I_n}\big|\theta^f_{n,k}\big|^p\bigg)^{\frac{1}{p}} = \zeta_{f}^{(p)}(n).\] The affine level is controlled by a constant depending only on \(p\) and \(q\). Hence \[\|\zeta_{f}^{(q)}\|_{\ell^q(\mathbb{N}_{-1})} \le C_{p,q}\|\zeta_{f}^{(p)}\|_{\ell^p(\mathbb{N}_{-1})}\] holds for some constant \(C_{p,q} > 0\) depending only on \(p, q\). Also, since \(p<q\), \[\|H_f\|_{\ell^q(\mathbb{N}_0)} \le \|H_f\|_{\ell^p(\mathbb{N}_0)}.\] Together with the uniform norm term, this proves the continuous embedding.

To see that the inclusion is strict, set \[a_n:=2^{-\frac{n}{q}}(n+1)^{-\frac{2}{q}}, \qquad n\ge0,\] and define \[f(t):=\sum_{n=0}^{\infty}\sum_{k\in I_n}2^{\frac{n}{2}}a_ne_{n,k}(t), \qquad t\in[0,1].\] The series converges uniformly. Indeed, since \[\Vert e_{n,k}\Vert_{\infty}\le \frac{1}{2} 2^{-\frac{n}{2}},\] and since at each point \(t\in[0,1]\) at most one Faber–Schauder function at level \(n\) is nonzero, the level-\(n\) contribution is bounded in the uniform norm by \(\frac{1}{2} a_n\). Since \(\sum_{n=0}^{\infty}a_n<\infty\), the series converges uniformly. Moreover, \[\big(\zeta_{f}^{(q)}(n)\big)^q = 2^{-\frac{nq}{2}}\sum_{k\in I_n}\big|2^{\frac{n}{2}}a_n\big|^q = 2^na_n^q = (n+1)^{-2},\] so \(\zeta_{f}^{(q)}\in\ell^q(\mathbb{N}_0)\). Also, \[H_f(n)=\sum_{m=n}^{\infty}a_m \le C_q2^{-\frac{n}{q}}(n+1)^{-\frac{2}{q}},\] and hence \(H_f\in\ell^q(\mathbb{N}_0)\). Thus \(f\in\mathcal{E}^q\).

On the other hand, \[\big(\zeta_{f}^{(p)}(n)\big)^p = 2^{-\frac{np}{2}}\sum_{k\in I_n}\big|2^{\frac{n}{2}}a_n\big|^p = 2^na_n^p = 2^{n(1-\frac{p}{q})}(n+1)^{-\frac{2p}{q}},\] which tends to infinity as \(n\to\infty\). Therefore \(\zeta_{f}^{(p)}\notin\ell^p(\mathbb{N}_0)\), and hence \(f\notin\mathcal{E}^p\). This proves that the inclusion is strict. ◻

Proof. By Proposition 10, we can choose conjugate exponents \(\bar p,\bar q>1\) with \(\bar p \ge p\), \(\bar q \ge q\) such that \(f\in\mathcal{E}^p\subset\mathcal{E}^{\bar p}\) and \(g\in\mathcal{E}^q\subset\mathcal{E}^{\bar q}\). Therefore, Theorem 4 gives the existence of \(I_{f,g}\) on \([0,1]\). Applying the same theorem to the pair \((g,f)\) with the conjugate exponents \((\bar q,\bar p)\) gives the existence of \(I_{g,f}\) on \([0,1]\). The remaining assertions follow directly from Proposition 5, Theorem 6, and Theorem 8, applied with the conjugate pair \((\bar p,\bar q)\). ◻

Proof. As in Corollary 3, choose conjugates \(\bar p,\bar q>1\) such that \(\bar p\ge p\), \(\bar q\ge q\), \(f\in\mathcal{E}^p\subset\mathcal{E}^{\bar p}\), and \(g\in\mathcal{E}^q\subset\mathcal{E}^{\bar q}\). Thus, the map is well-defined, and its bilinearity follows from Proposition 5. It remains to prove boundedness. By Theorem 8 with \(s=\xi=0\), we have for every \(t\in[0,1]\), \[\big|I_{f,g}(t)-f(0)\big(g(t)-g(0)\big)\big|\le 8L_f^{(p)}(0)L_g^{(q)}(0).\] Hence \[|I_{f,g}(t)|\le 2\|f\|_{\infty}\|g\|_{\infty}+8L_f^{(p)}(0)L_g^{(q)}(0).\] From the \(\ell^p\)-bound of \(\mathcal{A}^{(p)}_f\) in 18 , we have \[L_f^{(p)}(0)=\|\mathcal{A}^{(p)}_f\|_{\ell^p(\mathbb{N}_0)}+\frac{1}{2}\|H_f\|_{\ell^p(\mathbb{N}_0)}\le C_p\|f\|_{\mathcal{E}^p}, \qquad \text{with} \qquad C_p:=\|K_p\|_{\ell^1(\mathbb{N}_0)}+\frac{1}{2}.\] Similarly, \[L_g^{(q)}(0)\le C_q\|g\|_{\mathcal{E}^q}, \qquad C_q:=\|K_q\|_{\ell^1(\mathbb{N}_0)}+\frac{1}{2}.\] Taking the supremum over \(t\in[0,1]\) gives \[\|I_{f,g}\|_{\infty}\le \big(2+8C_pC_q\big)\|f\|_{\mathcal{E}^p}\|g\|_{\mathcal{E}^q}.\] ◻

Proof. Set \(h_N:=f-\Pi_N f\). Then the Faber–Schauder coefficients of \(h_N\) vanish at levels \(-1,0,\dots,N-1\), and agree with those of \(f\) at levels \(n\ge N\).

First, by the tail estimate 26 , for every \(t\in[0,1]\), \[|h_N(t)|\le \frac{1}{2}\sum_{m=N}^{\infty}2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta_{m,k}^f\big|=\frac{1}{2}H_f(N).\] Since \(H_f\in\ell^p(\mathbb{N}_0)\), we have \(H_f(N)\to0\), and hence \[\|h_N\|_{\infty}\xrightarrow{N\to\infty}0.\]

The level energy part satisfies \[\|\zeta_{h_N}^{(p)}\|_{\ell^p(\mathbb{N}_{-1})}^p=\sum_{n=N}^{\infty}\big(\zeta_{f}^{(p)}(n)\big)^p\xrightarrow{N\to\infty}0.\]

It remains to control the Schauder tail part. Since \(h_N\) has no coefficients below level \(N\), \[H_{h_N}(n)=H_f(\max\{n,N\}), \qquad n\ge0.\] Therefore \[\|H_{h_N}\|_{\ell^p(\mathbb{N}_0)}^p = \sum_{n=0}^{N-1} \big(H_f(N)\big)^p + \sum_{n=N}^{\infty} \big(H_f(n)\big)^p = N \big(H_f(N)\big)^p + \sum_{n=N}^{\infty} \big(H_f(n)\big)^p.\] The second term tends to zero because \(H_f\in\ell^p(\mathbb{N}_0)\). For the first term, note that \(H_f\) is nonincreasing. Hence \[\frac{N}{2} \big(H_f(N)\big)^p\le \sum_{n=\lfloor N/2\rfloor}^{N-1} \big(H_f(n)\big)^p \xrightarrow{N\to\infty} 0.\] Thus \(\|H_{h_N}\|_{\ell^p(\mathbb{N}_0)}\xrightarrow{N\to\infty}0\).

Combining the three estimates gives \(\|f - \Pi_N f\|_{\mathcal{E}^p}\to0\) as \(N \to \infty\). ◻

Proof. For \(f\in\mathcal{E}^p\), we have \[\|\mathscr C_p f\|_{\mathfrak e^p}=|f(0)|+\|\zeta_{f}^{(p)}\|_{\ell^p(\mathbb{N}_{-1})}+\|H_f\|_{\ell^p(\mathbb{N}_0)}\le \|f\|_{\mathcal{E}^p}.\] Hence \(\mathscr C_p\) is bounded.

Conversely, let \((c,a)\in\mathfrak e^p\). Since \(H_a(0)<\infty\), the series \[\sum_{n=0}^{\infty}\sum_{k\in I_n}a_{n,k}e_{n,k}(t), \qquad t \in [0, 1]\] converges uniformly on \([0,1]\), because at each level \(n\ge0\) at most one Faber–Schauder function \(e_{n,k}\) is nonzero at a given point and its height is at most \(2^{-\frac{n}{2}-1}\). Therefore \(\mathscr R_p(c,a)\) is a well-defined continuous function.

Moreover, \[\|\mathscr R_p(c,a)\|_{\infty}\le |c|+|a_{-1,0}|+\frac{1}{2}H_a(0)\le |c|+\|\zeta_a^{(p)}\|_{\ell^p(\mathbb{N}_{-1})}+\frac{1}{2}\|H_a\|_{\ell^p(\mathbb{N}_0)}.\] The Faber–Schauder coefficients of \(\mathscr R_p(c,a)\) are precisely \(a_{n,k}\), and its initial value is \(c\). Hence \[\|\mathscr R_p(c,a)\|_{\mathcal{E}^p}\le |c|+2\|\zeta_a^{(p)}\|_{\ell^p(\mathbb{N}_{-1})}+\frac{3}{2}\|H_a\|_{\ell^p(\mathbb{N}_0)}\le 2\|(c,a)\|_{\mathfrak e^p}.\] Thus \(\mathscr R_p\) maps \(\mathfrak e^p\) boundedly into \(\mathcal{E}^p\).

Finally, the identities \[\mathscr R_p\mathscr C_p f=f \quad \text{for} \quad f\in\mathcal{E}^p, \qquad \text{and} \qquad \mathscr C_p\mathscr R_p(c,a)=(c,a) \quad \text{for} \quad (c,a)\in\mathfrak e^p,\] follow from the uniqueness of the Faber–Schauder expansion. Therefore, \(\mathscr C_p\) is a bounded linear bijection with bounded inverse \(\mathscr R_p\). Since \(\mathcal{E}^p\) is Banach by Proposition 9, \(\mathfrak e^p\) is Banach as well, and \(\mathscr C_p\) is a Banach space isomorphism. ◻

Proof. Since \(\Pi_N f\) and \(\Pi_M g\) are continuous piecewise linear functions on \([0,1]\), the function \(\Pi_N f\) is continuous and \(\Pi_M g\) has bounded variation. Hence, the classical Riemann–Stieltjes integral \[\int_0^t \Pi_N f\,d(\Pi_M g)\] exists for every \(t\in[0,1]\).

We first show that this classical integral agrees with the Faber–Schauder integral of the pair \((\Pi_N f,\Pi_M g)\). Fix \(t\in\mathbb{T}\). For every \(m\ge\ell(t)\), the sum \(S_m^L(t;\Pi_N f,\Pi_M g)\) is a Riemann–Stieltjes sum over the dyadic partition of \([0,t]\), with mesh size \(2^{-m}\). Since \(\Pi_N f\) is continuous and \(\Pi_M g\) has bounded variation, these sums converge to the classical Riemann–Stieltjes integral. Therefore \[I_{\Pi_N f,\Pi_M g}^{\mathbb{T}}(t)=\int_0^t \Pi_N f\,d(\Pi_M g), \qquad t\in\mathbb{T}.\] Both sides are continuous functions of \(t\), so the identity extends to all \(t\in[0,1]\): \[I_{\Pi_N f,\Pi_M g}(t)=\int_0^t \Pi_N f\,d(\Pi_M g), \qquad t\in[0,1].\]

By Proposition 12, \[\Pi_N f \xrightarrow{N\to\infty} f \quad\text{in }\mathcal{E}^p, \qquad \Pi_M g \xrightarrow{M\to\infty} g \quad\text{in }\mathcal{E}^q.\] Using bilinearity and Proposition 11, we obtain \[\|I_{\Pi_N f,\Pi_M g}-I_{f,g}\|_{\infty} \le C_{p,q}\|\Pi_N f-f\|_{\mathcal{E}^p}\|\Pi_M g\|_{\mathcal{E}^q} + C_{p,q}\|f\|_{\mathcal{E}^p}\|\Pi_M g-g\|_{\mathcal{E}^q}.\] Since \(\Pi_M g\to g\) in \(\mathcal{E}^q\), the sequence \(\|\Pi_M g\|_{\mathcal{E}^q}\) is bounded. Hence, the right-hand side converges to \(0\) as \(N,M\to\infty\). This proves the result. ◻

Proof. Set \(h(t):=I_{f,g}(t)\) for \(t\in[0,1]\), and write \[P_{f,g}(N) := \sum_{n=N}^{\infty} \mathcal{A}_f^{(p)}(n) \, \mathcal{A}_g^{(q)}(n), \qquad N \ge 0.\] Since \(\mathcal{A}_f^{(p)}(n)\le L_f^{(p)}(n)\le C2^{-\rho n}\) and \(\mathcal{A}_g^{(q)}\in\ell^q(\mathbb{N}_0)\), Hölder’s inequality gives for some \(C_{p,q,\rho}>0\) \[P_{f,g}(N)\le C\sum_{n=N}^{\infty}2^{-\rho n}\mathcal{A}_g^{(q)}(n)\le C_{p,q,\rho}\,2^{-\rho N}\Vert\mathcal{A}_g^{(q)}\Vert_{\ell^q_{\ge N}}.\] Since \(N\mapsto\Vert\mathcal{A}_g^{(q)}\Vert_{\ell^q_{\ge N}}\) is nonincreasing, the same bound also gives \[\sum_{m=N}^{\infty}P_{f,g}(m) \le C_{p,q,\rho}2^{-\rho N}\Vert\mathcal{A}_g^{(q)}\Vert_{\ell^q_{\ge N}},\] possibly with a different constant \(C_{p,q,\rho}\). In particular, \[P_{f,g}\in\ell^q(\mathbb{N}_0), \qquad \sum_{N=0}^{\infty}\bigg(\sum_{m=N}^{\infty}P_{f,g}(m)\bigg)^q<\infty.\]

We first estimate the level energies of \(h\). For a level-\(n\) dyadic interval, consider three dyadic points \(t_{n,k}\), \(t_{n+1,2k+1}\), and \(t_{n,k+1}\). By Lemma 4, applied to the two level-\((n+1)\) intervals \([t_{n,k},t_{n+1,2k+1}]\) and \([t_{n+1,2k+1},t_{n,k+1}]\), there are remainders \(R_{n+1,2k}\) and \(R_{n+1,2k+1}\) such that \[h(t_{n+1,2k+1})-h(t_{n,k}) = f(t_{n,k})\big(g(t_{n+1,2k+1})-g(t_{n,k})\big)+R_{n+1,2k},\] and \[h(t_{n,k+1})-h(t_{n+1,2k+1}) = f(t_{n+1,2k+1})\big(g(t_{n,k+1})-g(t_{n+1,2k+1})\big)+R_{n+1,2k+1}.\] Moreover, from the proof of Lemma 4 we claim the level-wise estimate \[\label{level-wise-estimate} \bigg(\sum_{j\in I_N}\big|R_{N,j}\big|^q\bigg)^{\frac{1}{q}}\le \frac{1}{4} P_{f,g}(N), \qquad N\ge0.\tag{32}\] Indeed, for fixed \(m\ge N\), the level-\(m\) refinement contributions inside the level-\(N\) intervals form disjoint blocks. If \(\Delta_{m,j}\) denotes the contribution of level \(m\) inside \([t_{N,j},t_{N,j+1}]\), then \[R_{N,j}=\sum_{m=N}^{\infty}\Delta_{m,j}, \qquad \Delta_{m,j} := \sum_{k:\,I_{m,k}\subset[t_{N,j},t_{N,j+1}]} f_{m+1,L}(k)g_{m+1,R}(k).\] By Minkowski’s inequality, \[\bigg(\sum_{j\in I_N}|R_{N,j}|^q\bigg)^{\frac{1}{q}} \le \sum_{m=N}^{\infty}\bigg(\sum_{j\in I_N}|\Delta_{m,j}|^q\bigg)^{\frac{1}{q}}.\] For each fixed \(m\ge N\), set \(K_{m,j}:=\{k\in I_m:I_{m,k}\subset[t_{N,j},t_{N,j+1}]\}\) for \(j\in I_N\). The sets \(K_{m,j}\) are disjoint and their union is \(I_m\). For \(j\in I_N\), Hölder’s inequality on the block \(K_{m,j}\) gives \[|\Delta_{m,j}| \le \bigg(\sum_{k\in K_{m,j}}\big|f_{m+1,L}(k)\big|^p\bigg)^{\frac{1}{p}} \bigg(\sum_{k\in K_{m,j}}\big|g_{m+1,R}(k)\big|^q\bigg)^{\frac{1}{q}}.\] Write \[F_{m,j}:=\bigg(\sum_{k\in K_{m,j}}\big|f_{m+1,L}(k)\big|^p\bigg)^{\frac{1}{p}}, \qquad G_{m,j}:=\bigg(\sum_{k\in K_{m,j}}\big|g_{m+1,R}(k)\big|^q\bigg)^{\frac{1}{q}}.\] Then \[\bigg(\sum_{j\in I_N}|\Delta_{m,j}|^q\bigg)^{\frac{1}{q}} \le \bigg(\sum_{j\in I_N}F_{m,j}^qG_{m,j}^q\bigg)^{\frac{1}{q}} \le \Big( \sup_{j\in I_N}F_{m,j} \Big) \bigg(\sum_{j\in I_N}G_{m,j}^q\bigg)^{\frac{1}{q}}.\] Since \[\sup_{j\in I_N}F_{m,j}\le\bigg(\sum_{j\in I_N}F_{m,j}^p\bigg)^{\frac{1}{p}} = \bigg(\sum_{k\in I_m}\big|f_{m+1,L}(k)\big|^p\bigg)^{\frac{1}{p}},\] and \[\bigg(\sum_{j\in I_N}G_{m,j}^q\bigg)^{\frac{1}{q}} = \bigg(\sum_{k\in I_m}\big|g_{m+1,R}(k)\big|^q\bigg)^{\frac{1}{q}},\] Lemma 3 yields \[\bigg(\sum_{j\in I_N}|\Delta_{m,j}|^q\bigg)^{\frac{1}{q}} \le \frac{1}{4}\mathcal{A}_f^{(p)}(m) \, \mathcal{A}_g^{(q)}(m).\] Summing over \(m\ge N\) gives the claimed estimate 32 .

Since \[2^{-\frac{n}{2}} \theta_{n,k}^h = \big(h(t_{n+1,2k+1})-h(t_{n,k})\big)-\big(h(t_{n,k+1})-h(t_{n+1,2k+1})\big),\] we obtain \[\begin{align} 2^{-\frac{n}{2}}\theta_{n,k}^h = f(t_{n,k})\,2^{-\frac{n}{2}}\theta_{n,k}^g &+ \big(f(t_{n,k})-f(t_{n+1,2k+1})\big) \big(g(t_{n,k+1})-g(t_{n+1,2k+1})\big) \\ & + R_{n+1,2k}-R_{n+1,2k+1}. \end{align}\] Taking the \(\ell^q\) norm over \(k\in I_n\) and using Lemma 3 with the estimate 32 , we get \[\zeta_{h}^{(q)}(n)\le \Vert f\Vert_\infty\zeta_{g}^{(q)}(n)+\frac{1}{4} \mathcal{A}_f^{(p)}(n) \, \mathcal{A}_g^{(q)}(n) +\frac{1}{2} P_{f,g}(n+1).\] Since \(\zeta_{g}^{(q)}\in\ell^q(\mathbb{N}_{-1})\), \(\mathcal{A}_f^{(p)} \mathcal{A}_g^{(q)}\in\ell^q(\mathbb{N}_0)\), and \(P_{f,g}\in\ell^q(\mathbb{N}_0)\), it follows that \(\zeta_{h}^{(q)}\in\ell^q(\mathbb{N}_{-1})\).

It remains to estimate the Schauder tail of \(h\). From the same coefficient identity and the pointwise remainder estimate in Lemma 4, we have \[2^{-\frac{n}{2}} \sup_{k\in I_n} \big|\theta_{n,k}^h\big| \le \Vert f\Vert_\infty\,2^{-\frac{n}{2}}\sup_{k\in I_n}\big|\theta_{n,k}^g\big| + \frac{1}{4} \mathcal{A}_f^{(p)}(n) \, \mathcal{A}_g^{(q)}(n) + \frac{1}{2} P_{f,g}(n+1).\] Therefore, for every \(N\ge0\), \[H_h(N)\le \Vert f\Vert_\infty H_g(N)+\frac{1}{4} P_{f,g}(N)+\frac{1}{2}\sum_{n=N}^{\infty}P_{f,g}(n+1).\] The first term belongs to \(\ell^q(\mathbb{N}_0)\) because \(g\in\mathcal{E}^q\), and the last two terms belong to \(\ell^q(\mathbb{N}_0)\) by the estimate on \(P_{f,g}\) above. Hence, \(H_h\in\ell^q(\mathbb{N}_0)\). Since \(h\) is continuous and \(h(1)-h(0)\) is finite, the affine level is finite. Therefore \(h=I_{f,g}\in\mathcal{E}^q\). ◻

Proof. Let \(f\in C^\alpha([0,1])\). For \(n\ge0\) and \(k\in I_n\), the coefficient formula 10 gives \[\begin{align} \big|\theta_{n,k}^f\big| &= 2^{\frac{n}{2}}\big|2f(t_{n+1,2k+1})-f(t_{n,k})-f(t_{n,k+1})\big| \label{ineq:FS-coefficient-bound-Holder} \\ &\le 2^{\frac{n}{2}} \Big( \big|f(t_{n+1,2k+1})-f(t_{n,k})\big| + \big|f(t_{n+1,2k+1})-f(t_{n,k+1})\big| \Big) \le 2^{1-\alpha}[f]_\alpha 2^{n(\frac{1}{2}-\alpha)}. \nonumber \end{align}\tag{34}\] Therefore, for \(n\ge0\), \[\zeta_{f}^{(p)}(n) = \bigg(2^{-\frac{np}{2}}\sum_{k\in I_n}\big|\theta_{n,k}^f\big|^p\bigg)^{\frac{1}{p}} \le 2^{1-\alpha}[f]_\alpha 2^{-n(\alpha-\frac{1}{p})}.\] Since \(\alpha>1/p\), this gives \(\|\zeta_{f}^{(p)}\|_{\ell^p(\mathbb{N}_0)}<\infty\). The affine level is also finite, since \[\zeta_{f}^{(p)}(-1)=2^{1-\frac{1}{p}}\big|f(1)-f(0)\big|\le 2^{1-\frac{1}{p}}[f]_\alpha.\] Hence, \(\zeta_{f}^{(p)}\in\ell^p(\mathbb{N}_{-1})\).

It remains to estimate the Schauder tail. From the same coefficient bound, \[2^{-\frac{n}{2}}\sup_{k\in I_n}\big|\theta_{n,k}^f\big| \le 2^{1-\alpha}[f]_\alpha 2^{-\alpha n}.\] Thus \[H_f(n) = \sum_{m=n}^{\infty}2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta_{m,k}^f\big| \le 2^{1-\alpha}[f]_\alpha\sum_{m=n}^{\infty}2^{-\alpha m} = \frac{2^{1-\alpha}}{1-2^{-\alpha}}[f]_\alpha 2^{-\alpha n}.\] Hence \(H_f\in\ell^p(\mathbb{N}_0)\). Combining these estimates with Definition 5 proves the claimed bound. ◻

Proof. Since \(\alpha+\beta>1\), we may choose \(p>1\) such that \[1-\beta<\frac{1}{p}<\alpha.\] Let \(q\) be the conjugate exponent of \(p\). Then \[\frac{1}{q}=1-\frac{1}{p}<\beta.\] Hence \(\alpha p > 1\) and \(\beta q > 1\). Proposition 17 gives \(f\in\mathcal{E}^p\), \(g\in\mathcal{E}^q\), and Corollary 3 proves the existence of Faber–Schauder integrals \(I_{f,g}\) and \(I_{g, f}\) on \([0,1]\), along with the integration-by-parts identity and the dyadic Young–Loève estimate.

To show that the Faber–Schauder integrals agree with the Young integrals, let \(t\in\mathbb{T}\). For every \(n\ge\ell(t)\), the sum \(S_n^L(t;f,g)\) is a left Riemann sum over the dyadic partition of \([0,t]\), with mesh size \(2^{-n}\). Since \(\alpha+\beta>1\), the classical Young integral exists and these Riemann sums converge to it as \(n\to\infty\). Therefore \[I_{f,g}^{\mathbb{T}}(t)=\int_0^t f\,d^Yg, \qquad t\in\mathbb{T}.\] Both sides are continuous functions of \(t\) on \([0,1]\): the left-hand side by construction of \(I_{f,g}\), and the right-hand side by the classical Young theory. Since \(\mathbb{T}\) is dense in \([0,1]\), the identity extends to all \(t\in[0,1]\). The identity for \(I_{g,f}\) follows by the same argument, with the roles of \(f\) and \(g\) interchanged. ◻

Proof. The existence and continuity of the Faber–Schauder integrals follow from Corollary 3, while the existence and continuity of the Young integrals follow from the classical finite-variation Young theorem. For each \(t\in\mathbb{T}\), the dyadic left sums \(S_n^L(t;f,g)\) are Riemann–Stieltjes sums with mesh tending to zero; hence they converge both to \(I_{f,g}^{\mathbb{T}}(t)\) by definition and to \(\int_0^t f\,d^Yg\) by Young’s theorem. Therefore the two integrals agree on \(\mathbb{T}\), and by continuity on all of \([0,1]\). The argument for \(I_{g,f}\) is the same. ◻

Proof. We first check uniform convergence. At each point \(t\in[0,1]\), at most one Faber–Schauder function at level \(n\) is nonzero, and its height is at most \(2^{-\frac{n}{2}-1}\). Hence \[\bigg\|\sum_{k\in\Lambda_n}a_{n,k}e_{n,k}\bigg\|_{\infty} \le \frac{1}{2}\,2^{-\frac{n}{2}}A_n \le \frac{C_A}{2}2^{-\alpha n}.\] Since \(\alpha>0\), the series converges uniformly on \([0,1]\), and therefore \(f\in C([0,1])\).

By construction, the affine coefficient is \(\theta_{-1,0}^f=b\), and for \(n\ge0\) \[\theta_{n,k}^f= \begin{cases} a_{n,k}, & k\in\Lambda_n,\\ 0, & k\notin\Lambda_n. \end{cases}\] The affine level contributes only a finite quantity \[\zeta_{f}^{(p)}(-1)=2^{1-\frac{1}{p}}|b|.\] For \(n\ge0\), we have \[\big(\zeta_{f}^{(p)}(n)\big)^p = 2^{-\frac{np}{2}}\sum_{k\in I_n}\big|\theta_{n,k}^f\big|^p = 2^{-\frac{np}{2}}\sum_{k\in\Lambda_n}|a_{n,k}|^p \le 2^{-\frac{np}{2}}M_nA_n^p.\] Using the assumptions on \(A_n\) and \(M_n\), we obtain \[\big(\zeta_{f}^{(p)}(n)\big)^p \le C_MC_A^p\,2^{(\eta-p\alpha)n}.\] Since \(p\alpha>\eta\), this implies \(\zeta_{f}^{(p)} \in \ell^p(\mathbb{N}_{-1})\).

It remains to check the Schauder tail condition. For \(n\ge0\), \[H_f(n) = \sum_{m=n}^{\infty}2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta_{m,k}^f\big| \le \sum_{m=n}^{\infty}2^{-\frac{m}{2}}A_m \le C_A\sum_{m=n}^{\infty}2^{-\alpha m} = \frac{C_A}{1-2^{-\alpha}}2^{-\alpha n}.\] Hence \(H_f\in\ell^p(\mathbb{N}_0)\). Therefore \(f\in\mathcal{E}^p\). ◻

Proof. We first prove that \(f\in\mathcal{E}^p\). For each \(n\ge0\), the active coefficients of \(f\) satisfy \[\sup_{k\in\Lambda_n^f}\big|\sigma^f_{n,k}2^{n(\frac{1}{2}-\alpha)}\big|=2^{n(\frac{1}{2}-\alpha)}, \qquad \#\Lambda_n^f\le C_f2^{\eta_f n}.\] Proposition 19, applied with \(A_n=2^{n(\frac{1}{2}-\alpha)}\) and \(M_n=\#\Lambda_n^f\), gives \(f\in\mathcal{E}^p\), since \(p\alpha>\eta_f\). The same argument, with \(\beta\), \(\eta_g\), \(q\), and \(\Lambda_n^g\) in place of \(\alpha\), \(\eta_f\), \(p\), and \(\Lambda_n^f\), gives \(g\in\mathcal{E}^q\). Therefore, the conclusions about the Faber–Schauder integrals follow from Corollary 3.

We next show that \(f\) has no Hölder regularity exponent strictly larger than \(\alpha\). Suppose, for contradiction, that \(f\in C^\gamma([0,1])\) for some \(\gamma>\alpha\). Then, the coefficient estimate 34 in the proof of Proposition 17 gives \[\big|\theta^f_{n,k}\big|\le C_\gamma2^{n(\frac{1}{2}-\gamma)}, \qquad n\ge0, \quad k\in I_n,\] for some constant \(C_\gamma>0\). Since \(\Lambda_n^f\) is nonempty for every \(n\), choose \(k_n\in\Lambda_n^f\). Then \[2^{n(\frac{1}{2}-\alpha)}=\big|\theta^f_{n,k_n}\big|\le C_\gamma2^{n(\frac{1}{2}-\gamma)}.\] Equivalently, \(2^{n(\gamma-\alpha)}\le C_\gamma\) for all \(n\ge0\), which is impossible because \(\gamma>\alpha\). Hence, \(f\) is not \(\gamma\)-Hölder continuous for any \(\gamma>\alpha\). The proof for \(g\) is identical.

If the classical Hölder–Young condition were verifiable for this pair \((f, g)\), then there would exist exponents \(\gamma,\delta>0\) such that \[f\in C^\gamma([0,1]), \qquad g\in C^\delta([0,1]), \qquad \gamma+\delta>1.\] The preceding paragraph forces \(\gamma\le\alpha\) and \(\delta\le\beta\), hence \(\gamma+\delta\le\alpha+\beta\le1\), a contradiction. ◻

Proof. The uniform convergence of the two series and the energy-space membership \(f\in\mathcal{E}^p\), \(g\in\mathcal{E}^q\) follow from the same computation as in the proof of Proposition 10. The conclusions about the Faber–Schauder integrals follow from Corollary 3.

We now prove that the finite-variation Young condition cannot be satisfied by this pair. Let \(1<r<p\). Then \[\big(\zeta_{f}^{(r)}(n)\big)^r = 2^{-\frac{nr}{2}}\sum_{k\in I_n}\big|\theta^f_{n,k}\big|^r = 2^n a_n^r = 2^{n(1-\frac{r}{p})}(n+1)^{-\frac{2r}{p}} \xrightarrow{n\to\infty}\infty.\] Applying the coefficient characterization 16 with exponent \(r\), we get \(\|f\|_{r\text{-var}}=\infty\). If \(f\) had finite \(1\)-variation, then, since \(f\) is bounded, it would have finite \(r\)-variation for every \(r>1\), contradicting the preceding conclusion for any \(1<r<p\). Thus, \(f\) has infinite \(r\)-variation for every \(1\le r<p\). Similarly, \(g\) has infinite \(s\)-variation for every \(1\le s<q\).

Suppose now that the finite-variation Young’s theorem were applicable to this pair. Then there would exist \(r,s\ge1\) such that \(f\) has finite \(r\)-variation, \(g\) has finite \(s\)-variation, and \[\frac{1}{r}+\frac{1}{s}>1.\] The variation divergence proved above forces \(r\ge p\) and \(s\ge q\). Therefore \[\frac{1}{r}+\frac{1}{s}\le \frac{1}{p}+\frac{1}{q}=1,\] a contradiction, and no such variation exponents exist. ◻

Proof. For \(F\), we have \[\theta^F_{n,k}=2^{\frac{n}{2}}a_n\xi_{n,k}, \qquad n\ge0, \quad k\in I_n,\] and hence \[\mathbb{E}\Big[\big(\zeta_{F}^{(p)}(n)\big)^p\Big] = a_n^p\sum_{k\in I_n}\mathbb{E}|\xi_{n,k}|^p = \mathbb{E}|\xi_{0,0}|^p\,2^na_n^p = \mathbb{E}|\xi_{0,0}|^p\,(n+1)^{-2}.\] The affine coefficient vanishes, since all \(e_{n,k}\) with \(n\ge0\) vanish at \(0\) and \(1\). Therefore, by Fubini’s theorem, \(\zeta_{F}^{(p)}\in\ell^p(\mathbb{N}_{-1})\) almost surely. For the Schauder tail, the Gaussian tail bound and the union bound give \[\mathbb{P}\bigg(\max_{k\in I_n}|\xi_{n,k}|>C\sqrt n\bigg) \le 2^{n+1}e^{-\frac{C^2}{2}n}.\] Choosing \(C>0\) large enough that \(C^2>2\log2\), we obtain \[\sum_{n=1}^{\infty}\mathbb{P}\bigg(\max_{k\in I_n}|\xi_{n,k}|>C\sqrt n\bigg)<\infty.\] By the Borel–Cantelli lemma, almost surely, for all sufficiently large \(n\), \[\max_{k\in I_n}|\xi_{n,k}|\le C\sqrt n.\] Thus, almost surely, for all sufficiently large \(n\), \[\label{bound-stochastic-full} 2^{-\frac{n}{2}}\sup_{k\in I_n}\big|\theta^F_{n,k}\big| = a_n\max_{k\in I_n}|\xi_{n,k}| \le C\sqrt n\,2^{-\frac{n}{p}}(n+1)^{-\frac{2}{p}}.\tag{35}\] This bound implies the uniform convergence of the defining series, since at each point at most one Faber–Schauder function at level \(n\) is nonzero and has height at most \(2^{-n/2-1}\). Hence, \(F\) has continuous sample paths. Moreover, if \(F_N\) denotes the level-\(N\) partial sum, then \[\mathbb{E}\big|F(t)-F_N(t)\big|^2 \le \frac{1}{4}\sum_{n=N+1}^{\infty}a_n^2 \xrightarrow{N\to\infty}0 \qquad \text{for each fixed } t \in [0, 1].\] Thus, for every finite collection of times, the corresponding random vectors of \(F_N\) converge in \(L^2\) to those of \(F\). Since each \(F_N\) is a centered Gaussian process, the limit \(F\) is also a centered Gaussian process. From the bound 35 , we also have \[H_F(n) = \sum_{m=n}^{\infty}2^{-\frac{m}{2}}\sup_{k\in I_m}|\theta_{m,k}^F| \le C\sum_{m=n}^{\infty}\sqrt m\,2^{-\frac{m}{p}}(m+1)^{-\frac{2}{p}} \le C_p\sqrt n\,2^{-\frac{n}{p}}(n+1)^{-\frac{2}{p}}\] for all sufficiently large \(n\), and therefore \(H_F\in\ell^p(\mathbb{N}_0)\) almost surely. Hence, \(F\in\mathcal{E}^p\) almost surely. The proof for \(G\in\mathcal{E}^q\) is identical, and the same argument shows that \(G\) is a continuous centered Gaussian process. Corollary 3 implies that the Faber–Schauder integrals \(I_{F,G}\) and \(I_{G,F}\) exist on \([0,1]\) almost surely.

Moreover, this random pair is still outside the finite-variation Young regime. For every \(1<r<p\), \[2^{-\frac{nr}{2}}\sum_{k\in I_n} \big|\theta^F_{n,k}\big|^r = a_n^r\sum_{k\in I_n}|\xi_{n,k}|^r.\] For each \(n\), the variables \((|\xi_{n,k}|^r)_{k\in I_n}\) are independent and identically distributed with finite second moment. Hence, \[\mathbb{E}\Big[ 2^{-n}\sum_{k\in I_n} |\xi_{n,k}|^r \Big] = \mathbb{E}|\xi_{0,0}|^r, \qquad \text{Var}\Big( 2^{-n}\sum_{k\in I_n} |\xi_{n,k}|^r \Big) = 2^{-n}\text{Var}( |\xi_{0,0}|^r).\] By Chebyshev’s inequality, \[\sum_{n=0}^{\infty} \mathbb{P}\bigg( \Big| 2^{-n}\sum_{k\in I_n}|\xi_{n,k}|^r- \mathbb{E}|\xi_{0,0}|^r \Big|>\varepsilon \bigg) \le \varepsilon^{-2} \text{Var}(|\xi_{0,0}|^r) \sum_{n=0}^{\infty}2^{-n} < \infty.\] The Borel–Cantelli lemma yields \[2^{-n}\sum_{k\in I_n}|\xi_{n,k}|^r \longrightarrow \mathbb{E}|\xi_{0,0}|^r \quad\text{almost surely},\] hence, almost surely, for all sufficiently large \(n\) \[2^{-\frac{nr}{2}}\sum_{k\in I_n}\big|\theta^F_{n,k}\big|^r \ge c_r 2^n a_n^r = c_r2^{n(1-\frac{r}{p})}(n+1)^{-\frac{2r}{p}} \xrightarrow{n\to\infty}\infty\] for some constant \(c_r>0\). Applying 16 with this fixed exponent \(r\) gives \(\|F\|_{r\text{-var}}=\infty\) almost surely. Taking the intersection over all rational \(r\in(1,p)\) and using the monotonicity of variation exponents, we obtain that \(F\) has infinite \(r\)-variation for every \(1\le r<p\) almost surely. Similarly, \(G\) has infinite \(s\)-variation for every \(1\le s<q\). Therefore, there are no exponents \(r,s\ge1\) such that \(F\) has finite \(r\)-variation, \(G\) has finite \(s\)-variation, and \[\frac{1}{r}+\frac{1}{s}>1.\] Thus, the finite-variation Young theorem cannot be applied to this pair. ◻

Proof. By stationary increments and self-similarity of fBM, \[2B^H_{t_{n+1,2k+1}}-B^H_{t_{n,k}}-B^H_{t_{n,k+1}} \stackrel{d}{=} 2^{-nH}\big(2B^H_{1/2}-B^H_0-B^H_1\big).\] Multiplying by \(2^{n/2}\) gives \[\label{eq:fBM-increments-dist} \theta^{B^H}_{n,k}\stackrel{d}{=}2^{n(\frac{1}{2}-H)}\big(2B^H_{1/2}-B^H_0-B^H_1\big).\tag{37}\] The random variable on the right-hand side is centered Gaussian and nondegenerate for \(H\in(0,1)\). Taking absolute moments and setting \(C_{H,r} := \mathbb{E} |2B^H_{1/2}-B^H_0-B^H_1|^r\) proves the claim. ◻

Proof. We work with the continuous modification of \(B^H\), which exists for every \(H\in(0,1)\).

We first assume \(H > 1/p\). The affine coefficient \(\theta^{B^H}_{-1,0}=B^H_1-B^H_0\) is finite almost surely. For \(n \ge 0\), Lemma 9 gives \[\mathbb{E}\Big[\big(\zeta_{B^H}^{(p)}(n)\big)^p\Big] = 2^{-\frac{np}{2}}\sum_{k\in I_n}\mathbb{E} \big|\theta^{B^H}_{n,k}\big|^p = c_{H,p}2^{n(1-pH)}.\] Since \(pH>1\), we have \[\sum_{n=0}^{\infty}\mathbb{E}\Big[\big(\zeta_{B^H}^{(p)}(n)\big)^p\Big] < \infty.\] Hence, by Fubini’s theorem, \(\zeta_{B^H}^{(p)}\in\ell^p(\mathbb{N}_{-1})\) almost surely.

For the Schauder tail condition, define \[\label{def:Znk} Z_{n,k}:=2^{-n(\frac{1}{2}-H)}\theta^{B^H}_{n,k}, \qquad n\ge0, \quad k\in I_n.\tag{38}\] By 37 , the random variables \(Z_{n,k}\) are centered Gaussian with variance independent of \(n\) and \(k\). Hence, by the Gaussian tail bound and the union bound, one can choose \(C>0\) such that \[\sum_{n=1}^{\infty} \mathbb{P} \Big(\max_{k\in I_n}|Z_{n,k}|>C\sqrt n\Big) < \infty.\] By the Borel–Cantelli lemma, almost surely, for all sufficiently large \(n\), \[\max_{k\in I_n}|Z_{n,k}|\le C\sqrt n, \qquad \text{equivalently}, \qquad 2^{-\frac{n}{2}}\sup_{k\in I_n}\big|\theta^{B^H}_{n,k}\big| \le C\sqrt n\,2^{-Hn}.\] Hence, almost surely, for all sufficiently large \(n\), \[H_{B^H}(n) = \sum_{m=n}^{\infty}2^{-\frac{m}{2}}\sup_{k\in I_m}\big|\theta^{B^H}_{m,k}\big| \le C\sum_{m=n}^{\infty}\sqrt m\,2^{-Hm} \le C_H\sqrt n\,2^{-Hn},\] where \(C_H>0\) is a constant depending only on \(H\) and on \(C\). Since \((\sqrt n\,2^{-Hn})_{n\ge0}\in\ell^p(\mathbb{N}_0)\) for every \(p>1\), the tail of \(H_{B^H}\) belongs to \(\ell^p(\mathbb{N}_0)\) almost surely. The remaining finitely many values are finite almost surely, hence \(H_{B^H}\in\ell^p(\mathbb{N}_0)\) almost surely. Therefore, \(B^H\in\mathcal{E}^p\) almost surely.

Conversely, suppose \(H\le 1/p\). We will show that \(\zeta^{(p)}_{B^H}\notin \ell^p(\mathbb{N}_{-1})\) almost surely, as this implies \(B^H\notin \mathcal{E}^p\) almost surely. For \(n\ge0\) and \(k\in I_n\), we recall \[Z_{n,k}:=2^{-n(\frac{1}{2}-H)}\theta^{B^H}_{n,k} = 2^{nH}\big(2B^H_{t_{n+1,2k+1}}-B^H_{t_{n,k}}-B^H_{t_{n,k+1}}\big).\] By stationary increments and self-similarity of fBM, for each fixed \(n\), \[\label{eq:row-wise-dist} (Z_{n,0},\ldots,Z_{n,2^n-1}) \overset{d}{=} (Y_0,\ldots,Y_{2^n-1}),\tag{39}\] where \[Y_k:=2\widetilde{B}^H_{k+1/2}-\widetilde{B}^H_k-\widetilde{B}^H_{k+1},\] and \(\widetilde{B}^H\) is an fBM with the same Hurst index, defined on \([0,\infty)\). Thus \((Y_k)_{k\ge0}\) is a centered stationary Gaussian sequence.

We first estimate its covariance. Let \(\rho(m):=\mathbb{E}[Y_0Y_m]\) for \(m\ge0\). Define \[X_j:=\widetilde{B}^H_{(j+1)/2}-\widetilde{B}^H_{j/2}, \qquad j\ge0.\] Then \(Y_k=X_{2k}-X_{2k+1}\). If \(\gamma(r):=\mathbb{E}[X_0X_r]\), then from 36 \[\gamma(r)=2^{-2H-1}\left(|r+1|^{2H}+|r-1|^{2H}-2|r|^{2H}\right), \qquad r\ge0.\] Therefore, for \(m\ge1\), \[\rho(m) = \mathbb{E}\big[ (X_0-X_1)(X_{2m}-X_{2m+1}) \big] = 2\gamma(2m)-\gamma(2m-1)-\gamma(2m+1).\] For \(m\ge2\), substituting the explicit formula for \(\gamma\) gives \[\rho(m)=2^{-2H-1}\left(-(2m+2)^{2H}+4(2m+1)^{2H}-6(2m)^{2H}+4(2m-1)^{2H}-(2m-2)^{2H}\right).\] Writing \(u(x)=x^{2H}\) for \(x>0\), the expression in parentheses is \(-\Delta^4u(2m-2)\), where \[\Delta^4u(x):=u(x+4)-4u(x+3)+6u(x+2)-4u(x+1)+u(x).\] Let \(u^{(4)}\) denote the fourth derivative of \(u\). Since \[u^{(4)}(x)=2H(2H-1)(2H-2)(2H-3)x^{2H-4}, \qquad x>0,\] we have \[|u^{(4)}(x)|\le C_Hx^{2H-4}, \qquad x>0,\] where, here and below, \(C_H>0\) denotes a constant depending only on \(H\), whose value may change from line to line. Moreover, by repeated use of the fundamental theorem of calculus, \[\Delta^4u(x)=\int_{[0,1]^4}u^{(4)}(x+s_1+s_2+s_3+s_4)\,ds_1ds_2ds_3ds_4.\] Therefore, for \(m\ge2\), \[|\rho(m)|\le C_H\sup_{y\in[2m-2,2m+2]} y^{2H-4}\le C_Hm^{2H-4}.\] Since \(H<1\), we have \(2H-4<-2\), and \[\label{ineq:rho-summable} \sum_{m=0}^{\infty}|\rho(m)|<\infty,\tag{40}\] where the finitely many terms \(m=0,1\) are irrelevant for summability.

We next show that the empirical averages of \(|Z_{n,k}|^p\) converge almost surely. Set \[\label{def:Qn} Q_n:=2^{-n}\sum_{k=0}^{2^n-1}|Z_{n,k}|^p.\tag{41}\] By the row-wise distributional identity 39 , we have \(\mathbb{E}Q_n=\mathbb{E}|Y_0|^p\). Moreover, the covariance sequence of \((|Y_k|^p)_{k\ge0}\) is summable. Indeed, since \(Y_0\) is nondegenerate, \(\rho(0)>0\). For \(m\ge1\), the random variables \[\frac{Y_0}{\sqrt{\rho(0)}} \quad \text{and} \quad \frac{Y_m}{\sqrt{\rho(0)}}\] form a standard bivariate normal vector with correlation \(\rho(m)/\rho(0)\). By Gebelein’s inequality for functions of a bivariate Gaussian vector [13] (see also [14]), applied to the centered square-integrable function \[x\mapsto |\rho(0)^{1/2}x|^p-\mathbb{E}|Y_0|^p,\] we obtain \[\big|\text{Cov}(|Y_0|^p,|Y_m|^p)\big| \le \frac{|\rho(m)|}{\rho(0)}\,\text{Var}(|Y_0|^p) \le C_{H,p}|\rho(m)|,\] for some constant \(C_{H,p} > 0\). Here \(\text{Var}(|Y_0|^p)<\infty\) because \(Y_0\) is Gaussian. From 40 , we obtain \[\sum_{m=0}^{\infty}\big|\text{Cov}(|Y_0|^p,|Y_m|^p)\big|<\infty.\] By the distributional identity 39 , the variance of \(Q_n\) is equal to the variance of \(2^{-n}\sum_{k=0}^{2^n-1}|Y_k|^p\). Hence \[\text{Var}(Q_n)=2^{-2n}\sum_{k,\ell=0}^{2^n-1}\text{Cov}(|Y_k|^p,|Y_\ell|^p).\] By stationarity, \(\text{Cov}(|Y_k|^p,|Y_\ell|^p)\) depends only on \(|\ell-k|\). More precisely, for \(m\ge1\), there are \(2^n-m\) pairs \((k,\ell)\) with \(\ell=k+m\) and \(2^n-m\) pairs with \(k=\ell+m\). Hence \[\begin{align} \text{Var}(Q_n) &= 2^{-2n}\bigg(2^n\text{Var}(|Y_0|^p)+2\sum_{m=1}^{2^n-1}(2^n-m)\text{Cov}(|Y_0|^p,|Y_m|^p)\bigg) \\ & \le 2^{-2n}\bigg(2^n\text{Var}(|Y_0|^p)+2\sum_{m=1}^{2^n-1}(2^n-m)\big|\text{Cov}(|Y_0|^p,|Y_m|^p)\big|\bigg) \\ & \le 2^{-n}\bigg(\text{Var}(|Y_0|^p)+2\sum_{m=1}^{\infty}\big|\text{Cov}(|Y_0|^p,|Y_m|^p)\big|\bigg) \le C_{H,p}\,2^{-n}. \end{align}\] Here the constant \(C_{H,p}\) is finite because \(Y_0\) is Gaussian, so \(\text{Var}(|Y_0|^p)<\infty\). Hence, for every \(\varepsilon>0\), Chebyshev’s inequality yields \[\sum_{n=0}^{\infty}\mathbb{P}\left(\left|Q_n-\mathbb{E}|Y_0|^p\right|>\varepsilon\right)\le \varepsilon^{-2}\sum_{n=0}^{\infty}\text{Var}(Q_n)<\infty.\] By the Borel–Cantelli lemma, \[Q_n\xrightarrow{n \to \infty} \mathbb{E}|Y_0|^p \quad \text{a.s.}\] Since \(Y_0\) is nondegenerate Gaussian, \(\mathbb{E}|Y_0|^p>0\).

Finally, by the definition of the level energy in Definition 2, 38 , and 41 \[\Big(\zeta^{(p)}_{B^H}(n)\Big)^p = 2^{-\frac{np}{2}}\sum_{k=0}^{2^n-1}|\theta^{B^H}_{n,k}|^p=2^{n(1-pH)}Q_n.\] If \(H<1/p\), then \((\zeta^{(p)}_{B^H}(n))^p\to\infty\) as \(n\to\infty\) almost surely. If \(H=1/p\), then \[\Big(\zeta^{(p)}_{B^H}(n)\Big)^p = Q_n\longrightarrow \mathbb{E}|Y_0|^p>0 \quad \text{a.s.}\] In both cases, the sequence \(((\zeta^{(p)}_{B^H}(n))^p)_{n\ge0}\) is not summable. Therefore \(\zeta^{(p)}_{B^H} \notin \ell^p(\mathbb{N}_{-1})\) and hence \(B^H\notin \mathcal{E}^p\) almost surely. ◻

Proof. Suppose first that \(H_1+H_2>1\). Then we can choose \(p>1\) such that \[1-H_2<\frac{1}{p}<H_1.\] Let \(q\) be the conjugate exponent of \(p\). Then \[\frac{1}{q}=1-\frac{1}{p}<H_2.\] Hence \(H_1>1/p\) and \(H_2>1/q\). Theorem 20 gives \(B^{H_1}\in\mathcal{E}^p\) and \(\widetilde{B}^{H_2}\in\mathcal{E}^q\) almost surely.

Conversely, suppose that there exist conjugates \(p,q>1\) such that \(B^{H_1}\in\mathcal{E}^p\) and \(\widetilde{B}^{H_2}\in\mathcal{E}^q\) almost surely. By the necessity part of Theorem 20, we must have \[H_1>\frac{1}{p} \quad \text{and} \quad H_2>\frac{1}{q}, \qquad \text{hence} \qquad H_1+H_2>\frac{1}{p}+\frac{1}{q}=1.\] When these equivalent conditions hold, the existence of the integrals follows from Corollary 3. ◻

Proof. We first compute the level energies. For \(n \ge 0\) \[\theta^F_{n,k}= \begin{cases} 2^{n(\frac{1}{2}-\alpha)}\xi_{n,k}, & k\in\Lambda_n,\\ ~~~0, & k\notin\Lambda_n. \end{cases}\] Hence \[\mathbb{E}\big[\big(\zeta_{F}^{(p)}(n)\big)^p\big] = 2^{-\frac{np}{2}} \sum_{k\in\Lambda_n}2^{np(\frac{1}{2}-\alpha)}\mathbb{E}|\xi_{n,k}|^p \le C2^{(\eta-p\alpha)n}.\] Since \(p\alpha>\eta\), Fubini’s theorem gives \(\zeta_{F}^{(p)}\in\ell^p(\mathbb{N}_{-1})\) almost surely.

We next prove the Schauder tail condition. By the Gaussian tail bound and the union bound, choosing \(C_0>0\) sufficiently large gives \[\sum_{n=1}^{\infty} \mathbb{P}\bigg( \max_{k\in\Lambda_n}|\xi_{n,k}|>C_0\sqrt n \bigg) < \infty.\] By the Borel–Cantelli lemma, almost surely, for all sufficiently large \(n\), \[\max_{k\in\Lambda_n}|\xi_{n,k}|\le C_0\sqrt n, \qquad \text{hence} \qquad 2^{-\frac{n}{2}}\sup_{k\in I_n}\big|\theta^F_{n,k}\big| \le C_0\sqrt n\,2^{-\alpha n}.\] Consequently, \[H_F(n) = \sum_{m=n}^{\infty}2^{-\frac{m}{2}}\sup_{k\in I_m}|\theta_{m,k}^F| \le C_0\sum_{m=n}^{\infty}\sqrt m\,2^{-\alpha m} \le C_\alpha\sqrt n\,2^{-\alpha n}\] for a constant \(C_{\alpha} > 0\) and for all sufficiently large \(n\). Since \((\sqrt n\,2^{-\alpha n})_{n\ge0}\in\ell^p(\mathbb{N}_0)\) and the remaining finitely many values are finite almost surely, we obtain \(H_F\in\ell^p(\mathbb{N}_0)\) almost surely.

Thus, \(F\in\mathcal{E}^p\) almost surely. The same estimates also imply uniform convergence of the defining Faber–Schauder series, so \(F\) has continuous sample paths. Since each finite partial sum is a centered Gaussian process and the partial sums converge in \(L^2\) at every fixed time, the limit \(F\) is a centered Gaussian process.

It remains to prove the sharpness of the Hölder exponent. For each \(n\), choose one index \(k_n\in\Lambda_n\). Since the random variables \((\xi_{n,k_n})_{n\ge0}\) are independent standard Gaussians, the second Borel–Cantelli lemma gives \[\mathbb{P}\big(|\xi_{n,k_n}|\ge1\;\text{infinitely often}\big)=1.\] Hence, almost surely, along infinitely many levels \(n\), \[\sup_{k\in I_n}\big|\theta^F_{n,k}\big| \ge 2^{n(\frac{1}{2}-\alpha)}.\] Fix a sample path \(\omega\) in this probability-one event. If \(F(\omega) \in C^\gamma([0,1])\) for some \(\gamma>\alpha\), then the estimate 34 gives \[\sup_{k\in I_n}\big|\theta^{F(\omega)}_{n,k}\big| \le C_\gamma(\omega)2^{n(\frac{1}{2}-\gamma)}, \qquad n\ge0,\] for some finite constant \(C_\gamma(\omega)>0\). This contradicts the preceding lower bound along infinitely many levels, since it would imply \(2^{n(\gamma-\alpha)}\le C_\gamma(\omega)\) along infinitely many \(n\). Thus, almost surely, \(F\) is not \(\gamma\)-Hölder continuous for any \(\gamma\in(\alpha,1]\). ◻

Proof. Since \(\eta(1-H)<\alpha\), we can choose \(p>1\) such that \[1-H<\frac{1}{p}<\frac{\alpha}{\eta}\] when \(\eta>0\). If \(\eta=0\), choose any \(p>1\) such that \[1-H<\frac{1}{p}<1.\] Let \(q\) be the conjugate exponent of \(p\). Then \[\frac{1}{q}=1-\frac{1}{p}<H.\] Also, by the choice of \(p\), we have \(p\alpha>\eta\).

Hence, Proposition 21 gives \(F\in\mathcal{E}^p\) almost surely and Theorem 20 implies \(B^H\in\mathcal{E}^q\) almost surely. Therefore, the Faber–Schauder integral conclusions follow from Corollary 3.

Finally, we show that the classical Hölder–Young condition cannot be verified. Proposition 21 implies, almost surely, that \(F\) is not \(\gamma\)-Hölder continuous for any \(\gamma \in (\alpha, 1]\). Also, almost surely, fBM \(B^H\) is not \(\delta\)-Hölder continuous for any \(\delta \in (H,1]\) (see, e.g., [15]). If the Hölder–Young condition were verifiable, then there would exist exponents \(\gamma,\delta>0\) such that \[F\in C^\gamma([0,1]), \qquad B^H\in C^\delta([0,1]), \qquad \gamma+\delta>1.\] Necessarily \(\gamma\le\alpha\) and \(\delta\le H\), and therefore \(\gamma+\delta\le\alpha+H\le1\), a contradiction. ◻

Proof. Writing \[\Delta_{n,k}f:=f(t_{n,k+1})-f(t_{n,k})=f_{n+1,L}(k)+f_{n+1,R}(k), \qquad n \ge 0, \quad k\in I_n,\] Lemma 3 and Minkowski’s inequality give \[\bigg( \sum_{k\in I_n}\big|\Delta_{n,k}f\big|^p \bigg)^{\frac{1}{p}} \le \mathcal{A}_f^{(p)}(n).\] Since \(f\in\mathcal{E}^p\), we have \(\mathcal{A}_f^{(p)}\in\ell^p(\mathbb{N}_0)\), and hence \(\mathcal{A}_f^{(p)}(n)\to0\).

If \(r\ge p\), then \[\bigg(\sum_{k\in I_n}\big|\Delta_{n,k}f\big|^r\bigg)^{\frac{1}{r}} \le \bigg(\sum_{k\in I_n}\big|\Delta_{n,k}f\big|^p\bigg)^{\frac{1}{p}} \le \mathcal{A}_f^{(p)}(n).\] Therefore \[[f]^{(r)}_{\mathbb{T}_n}(1) = \sum_{k\in I_n}\big|\Delta_{n,k}f\big|^r \le \big(\mathcal{A}_f^{(p)}(n)\big)^r \xrightarrow{n\to\infty}0.\] For general \(t\in[0,1]\), the quantity \([f]^{(r)}_{\mathbb{T}_n}(t)\) contains all full level-\(n\) increments before \(t\) and at most one additional partial increment of length at most \(2^{-n}\). Hence \[[f]^{(r)}_{\mathbb{T}_n}(t) \le [f]^{(r)}_{\mathbb{T}_n}(1)+\omega_f(2^{-n})^r.\] Taking the supremum over \(t\in[0,1]\) and using the uniform continuity of \(f\) proves the first assertion.

For the covariation estimate, Hölder’s inequality gives, for every \(j=0,\dots,2^n\), \[\bigg| \sum_{k=0}^{j-1}\Delta_{n,k}f\,\Delta_{n,k}g \bigg| \le \bigg( \sum_{k\in I_n}\big|\Delta_{n,k}f\big|^p \bigg)^{\frac{1}{p}} \bigg( \sum_{k\in I_n}\big|\Delta_{n,k}g\big|^q \bigg)^{\frac{1}{q}} \le \mathcal{A}_f^{(p)}(n) \, \mathcal{A}_g^{(q)}(n).\] Since \(\mathcal{A}_f^{(p)}(n)\to0\) and \(\mathcal{A}_g^{(q)}(n)\to0\) as \(n \to \infty\), the right-hand side tends to zero. ◻