Proof. In view of the decomposition (2.1), the triangle inequality gives

$$ \begin{align*} \lVert {\mathbb{E}[A_{t_0, t_N} - \sum_{i=1}^N A_{t_{i-1}, t_i} \vert \mathcal{F}_{s}]} \rVert _{L_m(\mathbb{P})} \leq \sum_{n \in \mathbb{N}_0} \sum_{i=0}^{2^n - 1} \lVert {\mathbb{E}[R_i^n \vert \mathcal{F}_{s}]} \rVert _{L_m(\mathbb{P})}. \end{align*} $$

By (1.7) and (2.2),

$$ \begin{align*} \lVert {\mathbb{E}[R_i^n \vert \mathcal{F}_{s}]} \rVert _{L_m(\mathbb{P})} \leq 2 \Gamma_1 (t_0 - s)^{-\alpha} (2^{-n} \lvert {t_N - t_0} \rvert )^{\beta_1} = 2 \Gamma_1 2^{-n \beta_1 } \lvert {t_0 - s} \rvert ^{-\alpha} \lvert {t_N - t_0} \rvert ^{\beta_1}. \end{align*} $$

Therefore, recalling $\beta _1> 1$ from (1.9), the claim follows.

Proof under (1.12).

To simplify the proof, here, we assume (1.12), that is, that the additional technical condition $1+\alpha -\beta _1 < 2\alpha \beta _2 - \alpha $ holds. The proof in the general setting will be given in Appendix A.

We, again, use the representation (2.1). We fix a large $n \in \mathbb {N}$ and set . Fix an integer $L = L_n \in [M + 1, 2^n]$ , which will be chosen later. We have

(2.3)

We estimate the first term of (2.3). By the BDG inequality together with Minkowski’s inequality (see (1.15)), we have

Using (1.8) and (2.2) and noting that we include more terms in the sum by requiring $j \le 2^n/L$ only instead of $Lj+l \le 2^n - 1$ , we get

$$ \begin{align*} \sum_{\substack{j \ge 0:\\ L j + l < 2^n}} \lVert { R^n_{Lj + l} } \rVert _{L_m(\mathbb{P})}^2 \leq 4 \Gamma_2^2 2^{-n (2\beta_2 - 1)} L^{-1} \lvert {t_N - t_0} \rvert ^{2 \beta_2}. \end{align*} $$

Therefore,

We next estimate the second term of (2.3). The triangle inequality yields

$$ \begin{align*} \lVert {\sum_{l = 0}^{L - 1} \sum_{\substack{j \ge 0:\\ L j + l < 2^n}} \mathbb{E}[R^n_{L j + l} \vert \mathcal{F}_{L (j-1) + l + 1}^n]} \rVert _{L_m(\mathbb{P})} \leq \sum_{l = 0}^{L - 1} \sum_{\substack{j \ge 0:\\ L j + l < 2^n}} \lVert {\mathbb{E}[R^n_{L j + l} \vert \mathcal{F}_{L (j-1) + l + 1}^n]} \rVert _{L_m(\mathbb{P})}. \end{align*} $$

By (1.7),

$$ \begin{align*} \lVert {\mathbb{E}[R^n_{L j + l} \vert \mathcal{F}_{L (j-1) + l + 1}^n]} \rVert _{L_m(\mathbb{P})} \leq \Gamma_1 (L-1)^{-\alpha} 2^{-(\beta_1 - \alpha) n} \lvert {t_N - t_0} \rvert ^{\beta_1 - \alpha}. \end{align*} $$

Therefore,

$$ \begin{align*} \lVert {\sum_{l = 0}^{L - 1} \sum_{\substack{j \ge 0:\\ L j + l < 2^n}} \mathbb{E}[R^n_{L j + l} \vert \mathcal{F}_{L (j-1) + l + 1}^n]} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha} \Gamma_1 L^{-\alpha} 2^{-(\beta_1 - \alpha - 1) n} \lvert {t_N - t_0} \rvert ^{\beta_1 - \alpha}. \end{align*} $$

In conclusion,

(2.4) $$ \begin{align} \lVert {\sum_{i=0}^{2^n - 1} R^n_i} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha} \Gamma_1 L^{-\alpha} 2^{-(\beta_1 - \alpha - 1) n} \lvert {t_N - t_0} \rvert ^{\beta_1 - \alpha} + \kappa_{m, d} \Gamma_2 L^{\frac{1}{2}} 2^{-n(\beta_2 - \frac{1}{2})} \lvert {t_N - t_0} \rvert ^{\beta_2}. \end{align} $$

We wish to choose $L = L_n$ so that (2.4) is summable with respect to n. We therefore set , where

(2.5) $$ \begin{align} \alpha \delta + \beta_1 - \alpha - 1> 0, \quad 0 < \delta < \min\{2 \beta_2 - 1, 1 \}. \end{align} $$

Such a $\delta $ exists exactly under the additional technical assumption (1.12), namely, if $1 + \alpha - \beta _1 < 2 \alpha \beta _2 - \alpha $ . Then, (2.4) yields

$$ \begin{align*} \lVert { \sum_{n: 2^{n \delta} \geq M + 2}\sum_{i=0}^{2^n - 1} R^n_i} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha, \beta_1, \beta_2} \Gamma_1 \lvert {t_N - t_0} \rvert ^{\beta_1 - \alpha} + \kappa_{m, d} \Gamma_2 \lvert {t_N - t_0} \rvert ^{\beta_2}. \end{align*} $$

To estimate the contribution coming from the small n with $2^{n\delta } < M+2$ , we apply (1.8) which yields

$$ \begin{align*} \lVert {\sum_{i=0}^{2^n - 1} R^n_i} \rVert _{L_m(\mathbb{P})} \leq 2 \Gamma_2 \sum_{i=0}^{2^n - 1} 2^{-n \beta_2} \lvert {t_N - t_0} \rvert ^{\beta_2} = \Gamma_2 2^{1 + n (1 - \beta_2)} \lvert {t_N - t_0} \rvert ^{\beta_2}. \end{align*} $$

Thus, we conclude

$$ \begin{align*} \lVert { \sum_{n \in \mathbb{N}_0} \sum_{i=0}^{2^n - 1} R^n_i} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha, \beta_1, \beta_2, M} \Gamma_1 \lvert {t_N - t_0} \rvert ^{\beta_1 - \alpha} + \kappa_{m, d} \Gamma_2 \lvert {t_N - t_0} \rvert ^{\beta_2}, \end{align*} $$

where the fact $\kappa _{m, d} \geq 1$ is used.

Sketch of the proof.

Here, we give a sketch of the proof under (1.12). The complete proof is given in Appendix A. The argument is similar to Lemma 2.3.

Write

$$ \begin{align*} \pi' =: \{0 = t_0 < t_1 < \cdots < t_{N-1} < t_N = T\} \end{align*} $$

and

$$ \begin{align*} \{[s, t] \in \pi \mid t_j \leq s < t \leq t_{j+1}\} =: \{t_j = t^j_0 < t^j_1 < \cdots < t^j_{N_j - 1} < t^j_{N_j} = t_{j+1}\}. \end{align*} $$

We set , where $\delta $ satisfies (2.5). We set

As in Lemma 2.3, we consider the decomposition $A^{\pi '}_T - A^{\pi }_T = A + B$ , where

We estimate A by using the BDG inequality, Lemma 2.3, and (2.6), to obtain

$$ \begin{align*} \lVert {A} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha, \beta_1, \beta_2, M, m, d, T} L^{\frac{1}{2}} (\Gamma_1 \lvert {\pi'} \rvert ^{\beta_1 - \alpha - \frac{1}{2}} + \Gamma_2 \lvert {\pi'} \rvert ^{\beta_2 - \frac{1}{2}}). \end{align*} $$

We estimate B by using the triangle inequality, Lemma 2.2, and (2.6), to obtain

$$ \begin{align*} \lVert {B} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha, \beta_1, M, m, d, T} \Gamma_1 L^{-\alpha k} \lvert {\pi'} \rvert ^{\beta_1 - \alpha - 1}. \end{align*} $$

As in Lemma 2.3, we choose with $\delta $ satisfying (2.5). We then obtain the claimed estimate.

Proof. We write $\pi = \{0=t_0 < t_1 < \cdots <t_{N-1} < t_N = T\}$ . We set , and for $l\in \mathbb {N}$ , we inductively set

Set . Then, we define

By construction, $\pi '= \{s_j\}_{j=1}^L$ satisfies the claimed properties: $s_{j+1} - s_j \le 2|\pi |$ if $j<L-2$ , and $s_L - s_{L-1} \le 3 |\pi |$ , so $|\pi '| \le 3 |\pi |$ ; moreover, $\min _{[s,t] \in \pi '}|t-s| \ge |\pi | \ge 3^{-1} |\pi '|$ .

Proof of Theorem 1.1.

We will not write down dependence on $\alpha , \beta _1, \beta _2, M, m, d, T$ . We first prove the convergence of $(A^{\pi }_{\tau })_{\pi }$ . Without loss of generality, we assume $\tau = T$ . Let $\pi _1, \pi _2$ be partitions of $[0, T]$ . By Lemma 2.6, there exist partitions $\pi _1'$ , $\pi _2'$ , such that for $j \in \{1, 2\}$ , the partition $\pi _j$ refines $\pi _j'$ , $ \lvert {\pi ^{\prime }_j} \rvert \leq 3 \lvert {\pi _j} \rvert $ and

$$ \begin{align*} \min_{[s, t] \in \pi^{\prime}_j} \lvert {t- s} \rvert \geq 3^{-1} \lvert {\pi_j'} \rvert. \end{align*} $$

Lemma 2.4 shows that for some $\epsilon> 0$ , we have

$$ \begin{align*} \lVert {A^{\pi_j}_T - A^{\pi_j'}_T} \rVert _{L_m(\mathbb{P})} \lesssim (\Gamma_1 + \Gamma_2) \lvert {\pi_j} \rvert ^{\epsilon}. \end{align*} $$

Therefore, by the triangle inequality,

(2.7) $$ \begin{align} \lVert {A^{\pi_1}_T - A^{\pi_2}_T} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {A^{\pi_1'}_T - A^{\pi_2'}_T} \rVert _{L_m(\mathbb{P})} + (\Gamma_1 + \Gamma_2)( \lvert {\pi_1} \rvert ^{\epsilon} + \lvert {\pi_2} \rvert ^{\epsilon}). \end{align} $$

Let $\pi $ refine both $\pi _1'$ and $\pi _2'$ . Lemma 2.4 implies that

(2.8) $$ \begin{align} \lVert {A^{\pi_1'}_T - A^{\pi_2'}_T} \rVert _{L_m(\mathbb{P})} \leq \lVert {A^{\pi_1'}_T - A^{\pi}_T} \rVert _{L_m(\mathbb{P})} + \lVert {A^{\pi_{{2}}'}_T - A^{\pi}_T} \rVert _{L_m(\mathbb{P})} \lesssim (\Gamma_1 + \Gamma_2) ( \lvert {\pi_1} \rvert ^{\epsilon} + \lvert {\pi_2} \rvert ^{\epsilon}). \end{align} $$

The estimates (2.7) and (2.8) show

$$ \begin{align*} \lVert {A^{\pi_1}_T - A^{\pi_2}_T} \rVert _{L_m(\mathbb{P})} \lesssim (\Gamma_1 + \Gamma_2)( \lvert {\pi_1} \rvert ^{\epsilon} + \lvert {\pi_2} \rvert ^{\epsilon}). \end{align*} $$

Thus, $\{A^{\pi }_T\}_{\pi }$ forms a Cauchy net in $L_m(\mathbb {P})$ . We denote the limit by $\mathscr {S}_T$ . We next prove that $(\mathscr {S}_t)_{t \in [0, T]}$ satisfies (1.10) and (1.11). Let $t_0 < t_1 < t_2$ be such that $M(t_2 - t_1) \leq t_1 - t_0$ . Let $\pi _n = \{ t_1 + k 2^{-n}(t_2-t_1): k=0,\dots , 2^n\}$ be the nth dyadic partition of $[t_1, t_2]$ , and we write

We have

(2.9) $$ \begin{align} \mathbb{E}[\mathscr{S}_{t_1, t_2} - A_{t_1, t_2} \vert \mathcal{F}_{t_0}] = \lim_{n \to \infty} \mathbb{E}[A^n_{t_1, t_2} - A_{t_1, t_2} \vert \mathcal{F}_{t_0}] \quad \text{in } L_m(\mathbb{P}). \end{align} $$

By Lemma 2.2,

$$ \begin{align*} \lVert {\mathbb{E}[A_{t_1, t_2} - A^n_{t_1, t_2} \vert \mathcal{F}_{t_0}]} \rVert _{L_m(\mathbb{P})} \lesssim_{\beta_1} \Gamma_1 \lvert {t_1 - t_0} \rvert ^{-\alpha} \lvert {t_2 - t_1} \rvert ^{\beta_1}. \end{align*} $$

In this estimate, we can replace $A^n_{t_1, t_2}$ by $\mathscr {S}_{t_1, t_2}$ in view of (2.9). Similarly, by Lemma 2.3, we obtain

$$ \begin{align*} \lVert {\mathscr{S}_{t_1, t_2} - A_{t_1, t_2}} \rVert _{L_m(\mathbb{P})} \lesssim_{\alpha, \beta_1, \beta_2, M} \kappa_{m, d} \Gamma_1 \lvert {t_2 - t_1} \rvert ^{\beta_1 - \alpha} + \kappa_{m, d} \Gamma_2 \lvert {t_2 - t_1} \rvert ^{\beta_2}. \end{align*} $$

Under (1.12), we can replace $ \kappa _{m, d} \Gamma _1$ by $\Gamma _1$ .

Finally, let us prove the uniqueness of $\mathcal {A}$ . Let $(\tilde {\mathcal {A}}_t)_{t \in [0, T]}$ be another adapted process satisfying $\tilde {\mathcal {A}}_0 = 0$ , (1.10) and (1.11). It suffices to show $\mathcal {A}_T = \tilde {\mathcal {A}}_T$ almost surely. Let $\pi _n$ be the nth dyadic partition of $[0, T]$ . By Remark 2.5, we have

$$\begin{align*}\lVert {\mathcal{A}_T - \tilde {\mathcal{A}}_T} \rVert _{L_m(\mathbb{P})} \le \lVert {\mathcal{A}_T - A^{\pi^n}_T} \rVert _{L_m(\mathbb{P})} + \lVert {A^{\pi^n}_T - \tilde {\mathcal{A}}_T} \rVert _{L_m(\mathbb{P})} \lesssim 2^{-n\varepsilon}T^\varepsilon. \end{align*}$$

Since $n \in \mathbb {N}$ is arbitrary, we must have $\mathcal {A}_T = \tilde {\mathcal {A}}_T$ almost surely.

Proof. See Appendix A.

Proof. We will not write down dependence on d, H, and m. The filtration $(\mathcal {F}_t)_{t \in \mathbb {R}}$ is generated by the Brownian motion W appearing in the Mandelbrot–Van Ness representation (3.1). We will apply Theorem 1.1 with . Let $m \geq 2$ . We have

$$ \begin{align*} \lVert {A_{s, t}} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{L_{\infty}} \lvert {t - s} \rvert ^H. \end{align*} $$

To estimate conditional expectations, let $0 \leq v < s < t$ be such that $t-s \leq s-v$ and set

We write , if conditioned under $\mathcal {F}_v$ . Namely, we write, for instance

We are going to compute $\mathbb {E}[A_{s, t} \vert \mathcal {F}_v]$ . Conditionally on $\mathcal {F}_v$ , we have the Wiener chaos expansion [Reference Nualart34, Theorem 1.1.1]

$$ \begin{align*} f(B_s) = f(y_s + \tilde{B}_s) = a_0(s) + \sum_{i=1}^d a_i(s) \tilde{B}^i_s + \tilde{B}^{\perp}_s, \end{align*} $$

where $\tilde {B}^{\perp }_s$ is orthogonal in $L_2(\mathbb {P})$ to the subspace spanned by the constant $1$ and

$$ \begin{align*} (\tilde{B}^i_r)_{i=1, \ldots, d; r \geq v}. \end{align*} $$

Note that

$$ \begin{align*} a_0(s) &= \mathbb{E}[f(y_s + \tilde{B}_s)], \\ a_i(s) &= \mathbb{E}[(\tilde{B}_s^i)^2]^{-1} \mathbb{E}[f(y_s + \tilde{B}_s) \tilde{B}_s^i] \overset{\text{Lem. } {{3.2}}}{=} 2H (s-v)^{-2H} \mathbb{E}[f(y_s + \tilde{B}_s) \tilde{B}_s^i]. \end{align*} $$

Then, by the orthogonality of the Wiener chaos decomposition,

$$ \begin{align*} \mathbb{E}[A_{s, t} \vert \mathcal{F}_v] = a_0(s) \cdot Y_{s,t} + \sum_{i=1}^d a_i(s) \cdot \mathbb{E}[\tilde{B_s^i} \tilde{B}_{s, t}]. \end{align*} $$

Hence, for $u \in (s, t)$ ,

$$ \begin{align*} \mathbb{E}[\delta A_{s, u, t} \vert \mathcal{F}_v] = A^0_{s, u, t} + \sum_{i=1}^d A^i_{s, u, t}, \end{align*} $$

where

Here, $\boldsymbol {e}_i$ is the ith unit vector of $\mathbb {R}^d$ . We first estimate $A^0_{s, u, t}$ , for which we begin with estimating $a_0(s) - a_0(u)$ . We set

where X has the standard normal distribution in $\mathbb {R}^d$ . Note that

$$ \begin{align*} a_0(s) = F(Y_s, (2H)^{-\frac{1}{2}}(s-v)^H), \end{align*} $$

and similarly for $a_0(u)$ , we have

$$ \begin{align*} \partial_{m^i} F(m, \sigma) &= \frac{1}{(2 \pi)^{\frac{d}{2}} \sigma^{d+2}} \int_{\mathbb{R}^d} x^i e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} f(x + m) \, \mathrm{d} x, \\ \partial_{\sigma} F(m, \sigma) &= \frac{-d}{(2 \pi)^{\frac{d}{2}} \sigma^{d+1}} \int_{\mathbb{R}^d} f(m + x) e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} x + \frac{1}{(2 \pi)^{\frac{d}{2}} \sigma^{d+3}} \int_{\mathbb{R}^d} \lvert {x} \rvert ^2 f(m + x) e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} x. \end{align*} $$

Therefore,

$$ \begin{align*} \lvert {\partial_m F(m, \sigma)} \rvert + \lvert {\partial_{\sigma} F(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} \sigma^{-1}. \end{align*} $$

This yields

$$ \begin{align*} \lvert {a_0(s) - a_0(u)} \rvert &\leq \lvert {F(Y_s, (2H)^{-\frac{1}{2}}(s - v)^H) - F(Y_u, (2H)^{-\frac{1}{2}}(s - v)^H)} \rvert \\ &\phantom{\leq}+ \lvert {F(Y_u, (2H)^{-\frac{1}{2}}(s - v)^H) - F(Y_u, (2H)^{-\frac{1}{2}}(u - v)^H)} \rvert \\ &\lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)}(s-v)^{-H} \lvert {Y_{s, u}} \rvert + \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} (s-v)^{-H} ( \lvert {u-v} \rvert ^H - \lvert {s - v} \rvert ^H) \\ &\lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)}(s-v)^{-H} \lvert {Y_{s, u}} \rvert + \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)}(s-v)^{-1}(t - s). \end{align*} $$

Therefore,

(3.3) $$ \begin{align} \lvert {A^0_{s, u, t}} \rvert \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)}(s-v)^{-H} \lvert {Y_{s, u}} \rvert \lvert {Y_{u, t}} \rvert + \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} (s-v)^{-1}(t - s) \lvert {Y_{u, t}} \rvert. \end{align} $$

The random variable $Y_{s, u}$ is Gaussian and

(3.4) $$ \begin{align} \mathbb{E}[ \lvert {Y_{s, u}} \rvert ^2] &= d\int_{-\infty}^v (K(s, r) - K(u, r))^2 \, \mathrm{d} r = d \int_{s - v}^{\infty} ((u - s + r)^{H- \frac{1}{2}} - r^{H-\frac{1}{2}})^2 \, \mathrm{d} r \notag \\ &\lesssim (u - s)^2 \int_{s - v}^{\infty} r^{2H - 3} \, \mathrm{d} r \lesssim (s - v)^{2H - 2} (u - s)^2. \end{align} $$

We have a similar estimate for $Y_{u, t}$ . Therefore,

$$ \begin{align*} \lVert {A^0_{s, u, t}} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)}(s - v)^{H - 2} (t - s)^2 \quad \text{if } t-s \leq v - s. \end{align*} $$

Now we move to estimate $A^i_{s, u, t}$ . By Lemma 3.2, we have

$$ \begin{align*} \mathbb{E}[\tilde{B}_s^i \tilde{B}_{s, t}^i] =\int_v^s K(s, r) K(t, r) \, \mathrm{d} r - \int_v^s K(s, r) K(s, r) \, \mathrm{d} r = \frac{1}{2} (s - v)^{2H - 1}(t-s) + O((t-s)^{2H}). \end{align*} $$

Therefore, if we write ,

$$ \begin{align*} A^i_{s, u, t} = \frac{1}{2} \big[a_i^i(s)(s-v)^{2H - 1} - a_i^i(u)(u-v)^{2H - 1} \big](t - u) + O(( \lvert {a_i^i(s)} \rvert + \lvert {a_i^i(u)} \rvert ) \lvert {t-s} \rvert ^{2H}). \end{align*} $$

If we set

then $a_i^i(s) = G_i(Y_s, (2H)^{-\frac {1}{2}} (s - v)^H)$ and similarly for $a_i^i(u)$ . Since

$$ \begin{align*} G_i(m, \sigma) = (2 \pi)^{-\frac{d}{2}} \sigma^{-d-2} \int_{\mathbb{R}^d} f^i(y) (y^i - m^i) e^{-\frac{ \lvert {y-m} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} y, \end{align*} $$

we have

$$ \begin{align*} (2 \pi)^{\frac{d}{2}} \sigma^2 \partial_{m^j} G_i(m, \sigma) &= \int_{\mathbb{R}^d} f^i(m + \sigma x) [- \delta_{ij} + x^i x^j] e^{-\frac{ \lvert {x} \rvert ^2}{2}} \, \mathrm{d} x\\(2 \pi)^{\frac{d}{2}} \sigma^2 \partial_{\sigma} G_i(m, \sigma) &= \int_{\mathbb{R}^d} f^i(m + \sigma x) x^i [- (d+2) + \lvert {x} \rvert ^2] e^{-\frac{ \lvert {x} \rvert ^2}{2}} \, \mathrm{d} x. \end{align*} $$

Therefore,

$$ \begin{align*} \lvert {G_i(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} \sigma^{-1}, \end{align*} $$

$$ \begin{align*} \lvert {\partial_m G_i(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} \sigma^{-2}, \quad \lvert {\partial_{\sigma} G_i(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} \sigma^{-2} \end{align*} $$

and thus

$$ \begin{align*} \lvert {a^i_i(s)} \rvert &\lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} (s-v)^{-H},\\\lvert {a^i_i(s) - a^i_i(u)} \rvert &\lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} (s-v)^{-2H} \big( \lvert {Y_{s,u}} \rvert + (u-v)^H - (s-v)^H \big) \\&\lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} (s-v)^{-2H} \big( \lvert {Y_{s,u}} \rvert + (s-v)^{H-1} (u-s) \big). \end{align*} $$

This yields

$$ \begin{align*} & \lvert {A^i_{s, u, t}} \rvert \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} \big[ (s -v)^{-1} (t-s) \lvert {y_{s, u}} \rvert + (s-v)^{H-2}(t-s)^{2} \\& \quad + (s-v)^{H-2}(t-s)^{2} + (s - v)^{-H}(t - s)^{2H} \big] \end{align*} $$

and

(3.5) $$ \begin{align} \lVert {A^i_{s, u, t}} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} \big[(s-v)^{H-2}(t-s)^{2} + (s - v)^{-H}(t - s)^{2H} \big] \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)}(s - v)^{-H}(t - s)^{2H} \end{align} $$

if $t-s \leq s - v$ .

Therefore, by (3.3) and (3.5),

$$ \begin{align*} \lVert {\mathbb{E}[\delta A_{s, u, t} \vert \mathcal{F}_v]} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{L_{\infty}(\mathbb{R}^d)} (s - v)^{-H}(t - s)^{2H} \end{align*} $$

if $t-s \leq s - v$ . Hence, $(A_{s, t})$ satisfies the assumption of Theorem 1.1 with

$$ \begin{align*} \alpha = H, \quad \beta_1 = 2H, \quad \beta_2 = H, \quad M = 1.\\[-36pt] \end{align*} $$

Proof. We will not write down dependence on d, H, m, and $\gamma $ . The filtration $(\mathcal {F}_t)_{t \in \mathbb {R}}$ is generated by the Brownian motion W appearing in the Mandelbrot–Van Ness representation (3.1). We can assume

We will apply Theorem 1.1 with

We first claim

(3.7) $$ \begin{align} \lVert {\delta A_{s, u, t}} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{C^{\gamma} } \lvert {t-s} \rvert ^{(\gamma+1)H}. \end{align} $$

Observe

$$ \begin{align*} \delta A_{s, u, t} = f(B)_{u, t} \cdot B_{s, u} + f(B)_{u, s} \cdot B_{u, t}. \end{align*} $$

If $H> \frac {1}{4}$ , the claim (3.7) follows from the estimates

$$ \begin{align*} \lvert {f(B)_{u, t}} \rvert \leq \lVert {f} \rVert _{C^{\gamma} } \lvert {B_{u, t}} \rvert ^{\gamma}, \quad \lvert {f(B)_{u, s}} \rvert \leq \lVert {f} \rVert _{C^{\gamma} } \lvert {B_{u, s}} \rvert ^{\gamma}. \end{align*} $$

If $H \leq \frac {1}{4}$ , then $\gamma> 1$ , and we have

$$ \begin{align*} \delta A_{s, u, t} = \Big(f(B)_{u, t} - \sum_{j=1}^d \partial_j f(B_u) B_{u, t}^j\Big) \cdot B_{s, u} + \Big(f(B)_{u, s} - \sum_{j=1}^d \partial_j f(B_u) B_{u, s}^j \Big) \cdot B_{u, t}, \end{align*} $$

where (3.6) is used. Then, the claim (3.7) follows, again, from the Hölder estimate of f. Note that the condition $\gamma> \frac {1}{2H} - 1$ is equivalent to $(\gamma + 1) H> \frac {1}{2}$ .

The rest of the proof consists of estimating the conditional expectation $\mathbb {E}[\delta A_{s, u, t} \vert \mathcal {F}_v]$ . Let $t -s \leq s - v$ . We will use the same notation as in the proof of Proposition 3.3. We have

$$ \begin{align*} \mathbb{E}[\delta A_{s, u, t} \vert \mathcal{F}_v] = D^0_{s, u, t} + \sum_{i=1}^d D^i_{s, u, t}, \end{align*} $$

where

(3.8)

and

We first estimate $D^0_{s, u, t}$ . Suppose that $H> \frac {1}{4}$ . Recall

$$ \begin{align*} \partial_{m^i} F(m, \sigma) &= \frac{1}{(2 \pi)^{\frac{d}{2}} \sigma^{d+2}} \int_{\mathbb{R}^d} x^i e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} [f(x + m) - f(m)] \, \mathrm{d} x, \\ \partial_{\sigma} F(m, \sigma) &= \frac{-d}{(2 \pi)^{\frac{d}{2}} \sigma^{d+1}} \int_{\mathbb{R}^d} [f(m + x) - f(m)] e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} x \\ &\hspace{2cm}+ \frac{1}{(2 \pi)^{\frac{d}{2}} \sigma^{d+3}} \int_{\mathbb{R}^d} \lvert {x} \rvert ^2 [f(m + x) - f(m)] e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} x. \end{align*} $$

Therefore,

$$ \begin{align*} \lvert {\partial_{m^i} F(m, \sigma)} \rvert + \lvert {\partial_{\sigma} F(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } \sigma^{\gamma-1}. \end{align*} $$

This yields

(3.9) $$ \begin{align} \lvert {D^0_{s, u, t}} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } \big[ (s-v)^{(\gamma-1)H} \lvert {Y_{s,u}} \rvert \lvert {Y_{u,t}} \rvert + (s-v)^{\gamma H - 1} (t-s) ( \lvert {Y_{s, u}} \rvert + \lvert {Y_{u, t}} \rvert ) \big]. \end{align} $$

Therefore, by (3.4),

(3.10) $$ \begin{align} \lVert {D^0_{s, u, t}} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{C^{\gamma} } (s - v)^{(\gamma + 1)H - 2} (t - s)^2. \end{align} $$

Now suppose that $H \leq \frac {1}{4}$ . To simplify notation, we write . Since (3.6) gives $\partial _{m^i} I^j = \partial _{m^j} I^i$ for every $i, j$ , we have

$$ \begin{align*} D^0_{s, u, t} &= [I(Y_s, (u - v)^H) - I(Y_u, (u - v)^H) - \sum_{i=1}^d \partial_{m^i} I(Y_u, (u-v)^H) Y_{u, s}^i] \cdot Y_{u, t} \\ &\phantom{=}+ [I(Y_t, (u - v)^H) - I(Y_u, (u - v)^H) - \sum_{i=1}^d \partial_{m^i} I(Y_u, (u-v)^H) Y_{u, t}^i] \cdot Y_{s, u} \\ &\phantom{=}+ [I(Y_s, (s-v)^H) - I(Y_s, (u - v)^H)] \cdot Y_{u, t} + [I(Y_t, (t-v)^H) - I(Y_t, (u - v)^H)] \cdot Y_{s, u}. \end{align*} $$

Since

$$ \begin{align*} \partial_{m^i} \partial_{m^j} F(m, \sigma) = \frac{1}{(2 \pi)^{\frac{d}{2}} \sigma^{d+2}} \int_{\mathbb{R}^d} x^i e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} [\partial_jf(x + m) - \partial_jf(m)] \, \mathrm{d} x, \end{align*} $$

we have

$$ \begin{align*} \lvert {I(Y_s, (u - v)^H) - I(Y_u, (u - v)^H) - \sum_{i=1}^d \partial_{m^i} I(Y_u, (u-v)^H) Y_{u, s}^i} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } (s - v)^{(\gamma - 2)H} \lvert {Y_{s, u}} \rvert ^2. \end{align*} $$

Notice

$$ \begin{align*} \partial_{\sigma} F(m, \sigma) &= \frac{-d}{(2 \pi)^{\frac{d}{2}} \sigma^{d+1}} \int_{\mathbb{R}^d} [f(m + x) - f(m) - \sum_{i=1}^d \partial_i f(m) x^i] e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} x \\ &+ \frac{1}{(2 \pi)^{\frac{d}{2}} \sigma^{d+3}} \int_{\mathbb{R}^d} \lvert {x} \rvert ^2 [f(m + x) - f(m) - \sum_{i=1}^d \partial_i f(m) x^i] e^{-\frac{ \lvert {x} \rvert ^2}{2 \sigma^2}} \, \mathrm{d} x. \end{align*} $$

Therefore,

$$ \begin{align*} \lvert {\partial_{\sigma} F(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } \sigma^{\gamma - 1}. \end{align*} $$

This yields

$$ \begin{align*} \lvert {I(Y_s, (s-v)^H) - I(Y_s, (u - v)^H)} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } (s - v)^{\gamma H - 1} (t - s). \end{align*} $$

Hence, we obtain the estimate (3.10) when $H \leq \frac {1}{4}$ .

We move to estimate $D^i_{s, u, t}$ . By using the identity,

$$ \begin{align*} \mathbb{E}[(\tilde{B}_a^i + \tilde{B}_b^i) \tilde{B}_{a, b}^i] = \mathbb{E}[(\tilde{B}_b^i)^2] - \mathbb{E}[(\tilde{B}_a^i)^2], \end{align*} $$

we obtain

(3.11) $$ \begin{align} &D^i_{s, u, t} = (a_i^i(t) - a_i^i(u)) \mathbb{E}[\tilde{B}_t^i \tilde{B}_{s, t}^i] + (a_i^i(s) - a_i^i(u)) \mathbb{E}[\tilde{B}_s^i \tilde{B}_{s, t}^i] \nonumber\\ &-(a_i^i(s) - a_i^i(u)) \mathbb{E}[\tilde{B}_s^i \tilde{B}_{s, u}^i] -(a_i^i(t) - a_i^i(u)) \mathbb{E}[\tilde{B}_t^i \tilde{B}_{u, t}^i]. \end{align} $$

Since the other terms can be estimated similarly, we only estimate $(a_i^i(t) - a_i^i(u)) \mathbb {E}[\tilde {B}_t^i \tilde {B}_{s, t}^i]$ . By Lemma 3.2,

$$ \begin{align*} \lvert {\mathbb{E}[\tilde{B}_t \tilde{B}_{s, t}]} \rvert \lesssim \lvert {t-s} \rvert ^{2H}. \end{align*} $$

Now we estimate $ \lvert {a_i^i(t) - a_i^i(u)} \rvert $ . Recall $a_i^i(s) = G_i(Y_s, (2H)^{-\frac {1}{2}} (s - v)^H)$ ,

$$ \begin{align*} (2 \pi)^{\frac{d}{2}} \sigma^2 \partial_{m^j} G_i(m, \sigma) &= - \delta_{ij} \int_{\mathbb{R}^d} [f^i(m + \sigma x) - f^i(m)] e^{-\frac{ \lvert {x} \rvert ^2}{2}} \, \mathrm{d} x \\& \quad + \int_{\mathbb{R}^d} [f^i(m + \sigma x) - f^i(m)] x^i x^j e^{-\frac{ \lvert {x} \rvert ^2}{2}} \, \mathrm{d} x, \\(2 \pi)^{\frac{d}{2}} \sigma^2 \partial_{\sigma} G(m, \sigma) &= -(d+2) \int_{\mathbb{R}^d} [f^i(m + \sigma x) - f^i(m)] x^i e^{-\frac{ \lvert {x} \rvert ^2}{2}} \, \mathrm{d} x \\& \quad + \int_{\mathbb{R}^d} [f^i(m + \sigma x) - f^i(m)] x^i \lvert {x} \rvert ^2 e^{-\frac{ \lvert {x} \rvert ^2}{2}} \, \mathrm{d} x. \end{align*} $$

If $H \leq \frac {1}{4}$ , we can replace $f^i(m + \sigma x) - f^i(m)$ by

$$ \begin{align*} f^i(m + \sigma x) - f^i(m) - \sum_{k=1}^d \partial_k f^i(m) \sigma x^k. \end{align*} $$

Therefore,

$$ \begin{align*} \lvert {\partial_{m^j} G_i(m, \sigma)} \rvert + \lvert {\partial_{\sigma} G_i(m, \sigma)} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } \sigma^{\gamma - 2}. \end{align*} $$

This yields

$$ \begin{align*} \lvert {a_i^i(t) - a_i^i(u)} \rvert \lesssim \lVert {f} \rVert _{C^{\gamma} } (s - v)^{(\gamma - 2)H} ( \lvert {Y_{u, t}} \rvert + (s- v)^{H-1} (t-s)) \end{align*} $$

and hence

$$ \begin{align*} \lVert {a_i^i(t) - a_i^i(u)} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{C^{\gamma}} (s-v)^{(\gamma - 1) H - 1} (t - s). \end{align*} $$

Therefore, we obtain

(3.12) $$ \begin{align} \lVert {D^i_{s, u, t}} \rVert _{L_m(\mathbb{P})} \lesssim \lVert {f} \rVert _{C^{\gamma}} (s - v)^{(\gamma - 1) H - 1} (t - s)^{1 + 2H}. \end{align} $$

By (3.10) and (3.12), we conclude

$$ \begin{align*} \lVert {\mathbb{E}[\delta A_{s, u, t} \vert \mathcal{F}_v]} \rVert _{L_m(\mathbb{P})} &\lesssim \lVert {f} \rVert _{C^{\gamma} } [(s - v)^{(1 + \gamma) H - 2} (t - s)^2 + (s - v)^{(\gamma - 1) H - 1} (t - s)^{1 + 2H}] \\ &\lesssim \lVert {f} \rVert _{C^{\gamma} } (s - v)^{(\gamma - 1) H - 1} (t - s)^{1 + 2H} \end{align*} $$

if $t-s \leq s- v$ . Therefore, we can apply Theorem 1.1 with

$$ \begin{align*} \alpha = 1 - (\gamma - 1)H, \quad \beta_1 = 1 + 2H, \quad \beta_2 = (\gamma + 1) H, \quad M = 1.\\[-36pt] \end{align*} $$

Proof. We will not write down dependence on H, $\gamma $ , and m. Without loss of generality, we can assume $\mathbb {E}[B(0)] = 0$ . To apply Theorem 1.1 (for $H < \frac {1}{2}$ ) or Corollary 2.7 (for $H> \frac {1}{2}$ ), respectively, we set

If we set , it suffices to show that the estimates (1.10) and (1.11) are satisfied for $H < \frac {1}{2}$ , and that the estimates (2.14), (2.15), and (2.16) are satisfied for $H> \frac {1}{2}$ . Since the proof is rather long, we split the main arguments into three lemmas.

Lemma 4.4. We have

$$ \begin{align*} \lVert {A_{s, t}} \rVert _{L_m(\mathbb{P})} \lesssim_T \begin{cases} \lvert {t-s} \rvert ^{1 - H} f_H(a), & \text{for all } H \in (0, 1), \\ (\mathbb{E}[B(0)^2] + s^{2H})^{-\frac{1}{2m}} \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}} f_H(a), & \text{if } H> \frac{1}{2}, \end{cases} \end{align*} $$

where in either case there exists a constant $c = c(H)> 0$ , such that

(4.3) $$ \begin{align} \lvert {f_H(a)} \rvert \leq \exp\Big(-\frac{c a^2}{m (\mathbb{E}[B(0)^2] + T^{2H})} \Big), \quad \forall a \in \mathbb{R}. \end{align} $$

Proof. We have

$$ \begin{align*} \lVert {A_{s,t}} \rVert _{L_m(\mathbb{P})} &\leq (t-s)^{1 - (1 + \gamma)H} \lVert { \lvert {B_t - B_s} \rvert ^{2\gamma}} \rVert _{L_m(\mathbb{P})}^{\frac{1}{2}} \mathbb{P}(B_s < a < B_t)^{\frac{1}{2m}} \\ &\lesssim \lvert {t-s} \rvert ^{1-H} e^{- \frac{a^2}{4 m c_H (\mathbb{E}[B(0)^2] + T^{2H})}}. \end{align*} $$

Now we consider the case $H> \frac {1}{2}$ .

Set

Since $H> \frac {1}{2}$ , $\chi _1 \geq 0$ , and

$$ \begin{align*} \lvert {\chi_1} \rvert \lesssim \chi_0^{-1} T^{2H - 1} \lvert {t-s} \rvert , \quad 0 \leq \chi_2 \leq \lvert {t-s} \rvert ^H. \end{align*} $$

Then, if X and Y are two independent standard normal distributions on $\mathbb {R}$ , we have

$$ \begin{align*} (B_s, B_{s, t}) = \sqrt{c_H} ( \chi_0 Y, \chi_1 Y + \chi_2 X) \quad \text{in law.} \end{align*} $$

Therefore, if we set

(4.4)

We first estimate . Using the estimate

$$ \begin{align*} \mathbb{P}(X> x) \lesssim e^{-\frac{x^2}{2}} \quad \text{for } x > 0, \end{align*} $$

we have

(4.5)

Then,

where in the third line we applied

$$ \begin{align*} \sup_{z \in \mathbb{R}} \lvert {z} \rvert ^{m \gamma} e^{- \frac{ \lvert {z} \rvert ^2}{4}} < \infty. \end{align*} $$

And,

Therefore,

We now estimate . Similarly to (4.5), we have

and similarly

Therefore, we conclude

$$ \begin{align*} \lVert {\hat{A}_{s, t}} \rVert _{L_m(\mathbb{P})} \lesssim \Big(\frac{\chi_1 + \chi_2}{\chi_0} \Big)^{\frac{1}{m}} (\chi_1^{\gamma} + \chi_2^{\gamma}) e^{-\frac{a^2}{4 m (\chi_0^2 + \chi_2^2) }} \lesssim_T \chi_0^{-\frac{1}{m}} \lvert {t-s} \rvert ^{(\gamma + \frac{1}{m})H} e^{-\frac{a^2}{4 m (\mathbb{E}[B(0)^2] + T^{2H}) }}, \end{align*} $$

which completes the proof of the lemma.

Recall the Mandelbrot–Van Ness representation (3.1), and recall that W is an $(\mathcal {F}_t)_{t \in \mathbb {R}}$ -Brownian motion.

Lemma 4.6. Let $v < s < t$ , and set

If $\frac {t-s}{s-v}$ is sufficiently small, then

$$ \begin{align*} \mathbb{E}[A_{s, t} \vert \mathcal{F}_v] = \mathfrak{c}_{H, \gamma} \frac{e^{-\frac{(Y_s - a)^2}{2 \sigma_s^2}}}{ \sqrt{2 \pi} \sigma_s}(t-s) + R, \end{align*} $$

where for some $c = c(H, m)> 0$ ,

$$ \begin{align*} \lVert {R} \rVert _{L_m(\mathbb{P})} \lesssim (\mathbb{E}[B(0)^2] + s^{2H})^{-\frac{1}{2m}} e^{- c (\mathbb{E}[B(0)]^2 + T^{2H})^{-1} a^2} \Big(\frac{t-s}{s-v} \Big)^{\min\{1, 2H\} - \frac{H}{m} } \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}}. \end{align*} $$

Proof. As in the proof of Proposition 3.3, for $s> v$ , we set

and we write under the conditioning of $\mathcal {F}_v$ . Then, recalling $\hat {A}_{s,t}$ from (4.4), we have

(4.6)

To compute, we set

By Lemma 3.2,

$$ \begin{align*} \sigma_{s, t}^2 = c_H \lvert {t-s} \rvert ^{2H} + O( \lvert {s-v} \rvert ^{2H - 2} \lvert {t-s} \rvert ^2), \quad \sigma_{s}^2 = \frac{1}{2H} \lvert {s-v} \rvert ^{2H} \end{align*} $$

and

$$ \begin{align*} \rho_{s, t} = \frac{1}{2} \lvert {s-v} \rvert ^{2H - 1} \lvert {t-s} \rvert - \frac{c_H}{2} \lvert {t-s} \rvert ^{2H} + O( \lvert {s-v} \rvert ^{2H - 2} \lvert {t-s} \rvert ^2). \end{align*} $$

We have the decomposition

$$ \begin{align*} \tilde{B}_{s, t} = \sigma^{-2}_s \rho_{s, t} \tilde{B}_s + (\tilde{B}_{s, t} - \sigma_s^{-2} \rho_{s, t} \tilde{B}_s ), \end{align*} $$

where the second term is independent of $\tilde {B}_{s}$ . If we set

(4.7)

and if we write X and Y for two independent standard normal distributions, then the quantity (4.6) equals to

(4.8)

where

For a while, assume $\gamma> 0$ . Using the estimate

$$ \begin{align*} \lvert {(1 + \epsilon)^{\gamma} - 1} \rvert \lesssim \epsilon, \quad \text{if } \lvert {\epsilon} \rvert \leq 1, \end{align*} $$

we have

We set

We have for $y> 0$

$$ \begin{align*} I_{\gamma}'(y) = - \lvert {y} \rvert ^{\gamma} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi}}, \end{align*} $$

and if $q \geq 2 \lvert {p} \rvert $ and $y \in [q - \lvert {p} \rvert , q + \lvert {p} \rvert ]$ , then

$$ \begin{align*} \lvert {I_{\gamma}'(y)} \rvert \lesssim e^{-\frac{1}{4}(q- \lvert {p} \rvert )^2} \leq e^{-\frac{1}{16}q^2}. \end{align*} $$

Therefore, if $q \geq 2 \lvert {p} \rvert $ , we have

$$ \begin{align*} \lvert {I_{\gamma}(q) - I_{\gamma} (p + q)} \rvert \lesssim e^{-\frac{q^2}{16}} \lvert {p} \rvert. \end{align*} $$

Therefore,

(4.9)

When $\gamma = 0$ , we have

and thus (4.9) holds for $\gamma = 0$ . We estimate the expectation (with respect to Y) of each term.

We have

By using the estimate

(4.10)

we obtain

Next, we estimate the second term of (4.9). We have for $n \in \{0, 1\}$ ,

$$ \begin{align*} \mathbb{E}[ \lvert {Y} \rvert ^n e^{-\frac{q(Y)^2}{16}}] &= \frac{\kappa_{s, t}}{\sigma_s + \sigma_s^{-1} \rho_{s,t}} \int_{\mathbb{R}} \left\lvert {\frac{\kappa_{s,t} z + a - y_{s,t} - y_s}{\sigma_s + \sigma_s^{-1} \rho_{s,t}} } \right\rvert ^n e^{-\frac{z^2}{16}} \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \Big(\frac{\kappa_{s,t} z + a - y_{s,t} - y_s}{\sigma_s + \sigma_s^{-1} \rho_{s,t}} \Big)^2} \, \mathrm{d} z \\&\lesssim \frac{\kappa_{s, t}}{\sigma_s + \sigma_s^{-1} \rho_{s, t}} \Big(\frac{\kappa_{s,t} + \lvert {a - y_t} \rvert }{\sigma_s + \sigma_s^{-1} \rho_{s,t}} \Big)^n \\& \quad \times \Big[ e^{-\frac{(y_t - a)^2}{2 (\sigma_s + \sigma_s^{-1} \rho_{s,t})^2}} + e^{-\frac{(y_t - a)^2}{16 (\sigma_s + \sigma_s^{-1} \rho_{s,t})^2}} \frac{\kappa_{s,t}}{\sigma_s + \sigma_s^{-1} \rho_{s,t}} + e^{-\frac{(y_t - a)^2}{32 \kappa_{s,t}^2}} \Big], \end{align*} $$

where we applied (4.10) to get the last inequality. Therefore, we obtain

$$ \begin{align*} \mathbb{E}[ \lvert {p(Y)} \rvert e^{-\frac{q(Y)^2}{16}}] \lesssim \Big(\frac{ \lvert {y_{s, t}} \rvert }{\sigma_s + \sigma_s^{-1} \rho_{s, t}} + \sigma_s^{-1} \lvert {\rho_{s, t}} \rvert \frac{\kappa_{s,t} + \lvert {y_t - a} \rvert }{(\sigma_s + \sigma_s^{-1} \rho_{s,t})^2} \Big) e^{-\frac{(y_t - a)^2}{32 (\sigma_s + \sigma_s^{-1} \rho_{s,t})^2}}. \end{align*} $$

Finally, we estimate the third term of (4.9). Suppose that $\frac {t-s}{s-v}$ is so small that $ \lvert {\sigma _s^{-2} \rho _{s, t}} \rvert \leq \frac {1}{24}$ , and then we have

Hence,

and

This gives the estimate of the third term.

In summary, recalling that $\mathbb {E}[\hat {A}_{s,t} \vert \mathcal {F}_v]$ equals to (4.8), we obtain

$$ \begin{align*} \mathbb{E}[\hat{A}_{s, t} \vert \mathcal{F}_v] = \frac{\kappa_{s,t}^{\gamma+1} e^{-\frac{(Y_s - a)^2}{2 \sigma_s^2}}}{ \sqrt{2 \pi} \sigma_s} \int_0^{\infty} \lvert {x} \rvert ^{\gamma+1} \frac{e^{-\frac{x^2}{2}}}{\sqrt{2 \pi}} \, \mathrm{d} x + R_1, \end{align*} $$

where

Let us estimate $ \lVert {R_1} \rVert _{L_m(\mathbb {P})}$ . Recall that

$$ \begin{align*} \kappa_{s, t} \lesssim \lvert {t-s} \rvert ^{H}, \quad \sigma_s \lesssim \lvert {s - v} \rvert ^{H}, \end{align*} $$

and

$$ \begin{align*} \lvert {\rho_{s, t}} \rvert \lesssim \begin{cases} \lvert {s - v} \rvert ^{2H -1} \lvert {t-s} \rvert , \quad &H> \frac{1}{2}, \\ \lvert {t-s} \rvert ^{2H}, \quad &H < \frac{1}{2}. \end{cases} \end{align*} $$

We have the estimate (3.4) of $Y_{s, t}$ . Since

$$ \begin{align*} \mathbb{E}[Y_s^2] = \mathbb{E}[B(0)^2] + c_H s^{2H} - \frac{1}{2H} \lvert {s-v} \rvert ^{2H} \gtrsim \mathbb{E}[B(0)^2] + s^{2H} =: \chi_s^2, \end{align*} $$

there is a constant $c = c(H)> 0$ , such that

(4.11) $$ \begin{align} \mathbb{E}[ \lvert {Y_s - a} \rvert ^n e^{-\frac{(Y_s - a)^2}{\sigma^2}}] \lesssim_{n} \chi_s^{-1} \sigma^{n+1} e^{-\frac{c a^2}{\chi_s^2 + \sigma^2}}, \end{align} $$

(4.12) $$ \begin{align} \mathbb{E}[ \lvert {Y_{s,t}} \rvert ^n e^{-\frac{(Y_t - a)^2}{\sigma^2}}] \lesssim_{n} \chi_s^{-1} \sigma \lvert {v-s} \rvert ^{n(H - 1)} \lvert {t-s} \rvert ^n e^{-\frac{c a^2}{\chi_s^2 + \sigma^2}}. \end{align} $$

Therefore, for some constant $c_1 = c(H, m)>0$ ,

$$ \begin{align*} \lVert {\sigma_s^{-2} \kappa_{s, t}^{\gamma + 2} e^{-\frac{(Y_s - a)^2}{2 \sigma_s^2}}} \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} \sigma_s^{-2 + \frac{1}{m}} \kappa_{s,t}^{\gamma+2} e^{-\frac{c_1 a^2}{\chi_s^2}} \lesssim \chi_s^{-\frac{1}{m}} \lvert {s-v} \rvert ^{-(2-\frac{1}{m})H} \lvert {t-s} \rvert ^{(\gamma+2)H} e^{-\frac{c_1 a^2}{\chi_s^2}}, \end{align*} $$

$$ \begin{align*} &\lVert {\kappa_{s,t}^{\gamma} \frac{ \lvert {Y_{s, t}} \rvert }{\sigma_s + \sigma_s^{-1} \rho_{s, t}} e^{-\frac{(Y_t - a)^2}{32 (\sigma_s + \sigma_s^{-1} \rho_{s,t})^2}}} \rVert _{L_m(\mathbb{P})} \\& \quad \lesssim \chi_s^{-\frac{1}{m}} \kappa_{s,t}^{\gamma} \sigma_s^{-1 + \frac{1}{m}} \lvert {s-v} \rvert ^{H-1} \lvert {t-s} \rvert e^{-\frac{c_1 a^2}{\chi_s^2}} \lesssim \chi_s^{-\frac{1}{m}} \lvert {s-v} \rvert ^{\frac{H}{m} - 1} \lvert {t-s} \rvert ^{\gamma H + 1}e^{-\frac{c_1 a^2}{\chi_s^2}}, \end{align*} $$

$$ \begin{align*} & \lVert { \kappa_{s,t}^{\gamma} \sigma_s^{-1} \lvert {\rho_{s, t}} \rvert \frac{\kappa_{s,t} + \lvert {Y_t - a} \rvert }{(\sigma_s + \sigma_s^{-1} \rho_{s,t})^2} e^{-\frac{(Y_t - a)^2}{32 (\sigma_s + \sigma_s^{-1} \rho_{s,t})^2}}} \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} \kappa_{s, t}^{\gamma} \sigma_s^{-2 + \frac{1}{m}} \lvert {\rho_{s,t}} \rvert e^{-\frac{c_1 a^2}{\chi_s^2}} \\& \quad \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \begin{cases} \lvert {s-v} \rvert ^{-1 + \frac{H}{m}} \lvert {t-s} \rvert ^{\gamma H + 1}, \quad &H> \frac{1}{2}, \\\lvert {s-v} \rvert ^{-2H + \frac{H}{m} } \lvert {t-s} \rvert ^{(\gamma+2)H}, & H < \frac{1}{2}, \end{cases} \end{align*} $$

$$ \begin{align*} &\lVert { \kappa_{s,t}^{\gamma} e^{-\frac{(Y_s - a)^2}{5 \sigma_s^2}} (\sigma_s^{-1} \lvert {Y_{s,t}} \rvert + \sigma_s^{-3} \lvert {\rho_{s,t}} \rvert \lvert {Y_s - a} \rvert ) } \rVert _{L_m(\mathbb{P})} \\& \quad \lesssim \chi_s^{-1} \kappa_{s,t}^{\gamma}(\sigma_s^{\frac{1}{m} - 1} \lvert {s-v} \rvert ^{H-1} \lvert {t-s} \rvert + \sigma_s^{-2+\frac{1}{m}} \lvert {\rho_{s,t}} \rvert ) e^{-\frac{c_1 a^2}{\chi_s^2}}\\& \quad \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \begin{cases} \lvert {s-v} \rvert ^{\frac{H}{m} - 1} \lvert {t-s} \rvert ^{2-H}, &H> \frac{1}{2}, \\ ( \lvert {s-v} \rvert ^{\frac{H}{m} -1} \lvert {t-s} \rvert ^{2-H} + \lvert {s-v} \rvert ^{(-2+\frac{1}{m})H} \lvert {t-s} \rvert ^{1+H}), &H < \frac{1}{2}, \end{cases} \end{align*} $$

$$ \begin{align*} &\lVert { \lvert {Y_{s,t}} \rvert ^{\gamma} e^{-\frac{(Y_s - a)^2}{5 \sigma_s^2}} (\sigma_s^{-1} \lvert {Y_{s,t}} \rvert + \sigma_s^{-3} \lvert {\rho_{s,t}} \rvert \lvert {Y_s - a} \rvert ) } \rVert _{L_m(\mathbb{P})} \\& \quad \lesssim \sigma_s^{-1} \lVert { \lvert {Y_{s,t}} \rvert ^{\gamma + 1} e^{-\frac{(Y_s - a)^2}{5 \sigma_s^2}}} \rVert _{L_m(\mathbb{P})} + \sigma_s^{-2} \rho_{s, t} \lVert { \lvert {Y_{s,t}} \rvert ^{\gamma} e^{-\frac{(Y_s - a)^2}{6 \sigma_s^2}}} \rVert _{L_m(\mathbb{P})} \\& \quad \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \big[ \sigma_s^{\frac{1}{m} - 1} \lvert {s-v} \rvert ^{(\gamma +1)(H - 1)} \lvert {t-s} \rvert ^{(\gamma + 1) } \\& \qquad\quad\qquad\qquad + \sigma_s^{-2+\frac{1}{m}} \lvert {\rho_{s,t}} \rvert \lvert {v-s} \rvert ^{\gamma(H-1)} \lvert {t-s} \rvert ^{\gamma} \big]\\& \quad \lesssim_{H, m} \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \\& \qquad \times \begin{cases} \lvert {s-v} \rvert ^{(\gamma+\frac{1}{m})H - (\gamma+1)} \lvert {t-s} \rvert ^{\gamma+1}, &H> \frac{1}{2}, \\ ( \lvert {s-v} \rvert ^{(\gamma+\frac{1}{m})H - (\gamma+1)} \lvert {t-s} \rvert ^{\gamma+1} + \lvert {s-v} \rvert ^{(-2 + \frac{1}{m} + \gamma)H - \gamma} \lvert {t-s} \rvert ^{\gamma+ 2H}), &H< \frac{1}{2}, \end{cases} \end{align*} $$

$$ \begin{align*} &\lVert { (\sigma_s^{-1} \rho_{s,t})^{\gamma} e^{-\frac{Y_s^2}{5 \sigma_s^2}} (\sigma_s^{-1} \lvert {Y_{s,t}} \rvert + \sigma_s^{-3} \lvert {\rho_{s,t}} \rvert \lvert {Y_s} \rvert ) } \rVert _{L_m(\mathbb{P})} \\&\quad\lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \big[ \sigma_s^{\frac{1}{m} - 1 - \gamma} \lvert {\rho_{s,t}} \rvert ^{\gamma} \lvert {s-v} \rvert ^{H-1} \lvert {t-s} \rvert + \sigma_s^{-2 + \frac{1}{m} - \gamma} \lvert {\rho_{s,t}} \rvert ^{\gamma+ 1} \big] \\&\quad\lesssim \chi_{s}^{-\frac{1}{m}}e^{-\frac{c_1 a^2}{\chi_s^2}} \\&\qquad\times \begin{cases} \lvert {s-v} \rvert ^{(\gamma + \frac{1}{m})H - (\gamma + 1)} \lvert {t-s} \rvert ^{\gamma + 1}, & H> \frac{1}{2}, \\ \lvert {s-v} \rvert ^{(\frac{1}{m} - \gamma)H - 1} \lvert {t-s} \rvert ^{1+ 2 H \gamma} + \lvert {s-v} \rvert ^{(-2 + \frac{1}{m} - \gamma)H} \lvert {t-s} \rvert ^{2H(\gamma+1)}, & H < \frac{1}{2}, \\ \end{cases} \end{align*} $$

and finally

After this long calculation, we conclude

(4.13) $$ \begin{align} \lVert {R_1} \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \Big(\frac{t-s}{s-v} \Big)^{\min\{1, 2H\} - \frac{H}{m}} \lvert {t-s} \rvert ^{(\gamma + \frac{1}{m}) H} \end{align} $$

if $\frac {t-s}{s-v}$ is sufficiently small.

By (4.7), we have

$$ \begin{align*} \frac{\kappa_{s,t}^{\gamma+1} e^{-\frac{(y_s - a)^2}{2 \sigma_s^2}}}{ \sqrt{2 \pi} \sigma_s} \int_0^{\infty} \lvert {x} \rvert ^{\gamma+1} \frac{e^{-\frac{x^2}{2}}}{\sqrt{2 \pi}} \, \mathrm{d} x = \mathfrak{c}_{H, \gamma} \frac{e^{-\frac{(y_s - a)^2}{2 \sigma_s^2}} \lvert {t-s} \rvert ^{(\gamma+1)H}}{ \sqrt{2 \pi} \sigma_s} + R_2, \end{align*} $$

where

(4.14) $$ \begin{align} \lvert {R_2} \rvert \lesssim_H \frac{e^{-\frac{(y_s - a)^2}{2 \sigma_s^2}} }{ \sigma_s} \Big(\frac{t-s}{s-v} \Big)^{2 \min\{H, 1-H\}} \lvert {t-s} \rvert ^{(\gamma+1)H}. \end{align} $$

Therefore,

$$ \begin{align*} \lVert {R_2} \rVert _{L_m(\mathbb{P})} \lesssim_{H} \chi_s^{-\frac{1}{m}} e^{-\frac{c_1 a^2}{\chi_s^2}} \begin{cases} \lvert {s-v} \rvert ^{(\frac{1}{m} + 1)H - 2} \lvert {t-s} \rvert ^{2 + (\gamma - 1) H}, &H> \frac{1}{2},\\ \lvert {s-v} \rvert ^{(\frac{1}{m} - 3)H} \lvert {t-s} \rvert ^{(\gamma + 3) H}, &H < \frac{1}{2}. \end{cases} \end{align*} $$

This completes the proof of the lemma.

Lemma 4.7. We have

(4.15) $$ \begin{align} \lVert {L_t(a) - L_s(a)} \rVert _{L_m(\mathbb{P})} \lesssim_T \begin{cases} \lvert {t-s} \rvert ^{1 - H} f_H(a), & \hspace{-2cm}\text{for all } H \in (0, 1), \\ (\mathbb{E}[B(0)^2] + s^{2H})^{-\frac{1}{2m}} \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}} f_H(a), & \text{if } H> \frac{1}{2}, \end{cases} \end{align} $$

where $f_H(a)$ satisfies the estimate (4.3). Moreover, if $\frac {t-s}{s-v}$ is sufficiently small, then

$$ \begin{align*} \mathbb{E}[L_t(a) - L_s(a) \vert \mathcal{F}_v] = \frac{e^{-\frac{ \lvert {Y_s - a} \rvert ^2}{2 \sigma_s^2}}}{\sqrt{2 \pi} \sigma_s} (t-s) + \tilde{R}, \end{align*} $$

where for some $c = c(H, m)> 0$ ,

(4.16) $$ \begin{align} \lVert {\tilde{R}} \rVert _{L_m(\mathbb{P})} \lesssim (\mathbb{E}[B(0)^2] + s^{2H})^{-\frac{1}{2m}} e^{-c (\mathbb{E}[B(0)^2] + T^{2H})^{-1} a^2} \Big(\frac{t-s}{s-v} \Big)^{\frac{1}{m} + (1 - \frac{1}{m})H} \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}}. \end{align} $$

Proof. The estimate in (4.15) follows from [Reference Ayache, Wu and Xiao3, (3.38)]. However, since this is not entirely obvious, we sketch here an alternative derivation, which is motivated by [Reference Butkovsky, Lê and Mytnik7]. In view of the formal expression $L_t(a) = \int _0^t \delta _a(B_r) \, \mathrm {d} r$ , we set

We note $\mathbb {E}[\delta \bar {A}_{s,u,t}(a) \vert \mathcal {F}_s] =0$ . By Lê’s stochastic sewing lemma [Reference Lê24], to prove (4.15), it suffices to show

• • the estimate

  $$ \begin{align*} \lVert {e^{-\frac{ H(Y_s - a)^2}{c_H (r-s)^{2H}}}} \rVert _{L_m(\mathbb{P})} \lesssim_{T} \begin{cases} f_H(a), & H < \frac{1}{2}, \\ (\mathbb{E}[B(0)^2] + s^{2H})^{-\frac{1}{2m}} \lvert {r-s} \rvert ^{\frac{H}{m}} f_H(a), & H> \frac{1}{2}, \end{cases} \end{align*} $$

• • and the identity $L_t(a) = \lim _{ \lvert {\pi } \rvert \to 0} \sum _{[u,v] \in \pi } \bar {A}_{u, v}(a)$ , where $\pi $ is a partition of $[0, t]$ .

The first point is essentially given in Lemma 4.4. The second point follows from the identity

$$ \begin{align*} \int_{\mathbb{R}} \Big(\sum_{[u,v] \in \pi} \bar{A}_{u,v}(a) \Big) f(a) \, \mathrm{d} a = \sum_{[u,v] \in \pi} \int_u^v \mathbb{E}[f(B_r) \vert \mathcal{F}_u] \, \mathrm{d} r. \end{align*} $$

Thus, we now focus on the estimate (4.16). We have the identity

$$ \begin{align*} \mathbb{E}[L_t - L_s \vert \mathcal{F}_v] = \int_s^t \frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_r^2}}}{\sqrt{2 \pi} \sigma_r} \, \mathrm{d} r. \end{align*} $$

Indeed, we can convince ourselves of the validity of the identity from the formal expression

$$ \begin{align*} L_t - L_s = \int_s^t \delta(B_r - a) \, \mathrm{d} r. \end{align*} $$

We have the decomposition

$$ \begin{align*} &\frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_r^2}}}{\sigma_r} - \frac{e^{-\frac{ \lvert {Y_s - a} \rvert ^2}{2 \sigma_s^2}}}{\sigma_s} = \Big[\frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_r^2}}}{\sigma_r} - \frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_r^2}}}{\sigma_s} \Big] \\& \quad + \Big[ \frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_r^2}}}{\sigma_s} - \frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_s^2}}}{\sigma_s} \Big] + \Big[\frac{e^{-\frac{ \lvert {Y_r - a} \rvert ^2}{2 \sigma_s^2}}}{\sigma_s} - \frac{e^{-\frac{ \lvert {Y_s - a} \rvert ^2}{2 \sigma_s^2}}}{\sigma_s} \Big] =: R_3 + R_4 + R_5. \end{align*} $$

By (4.11), we obtain

$$ \begin{align*} \lVert {R_3} \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}} \sigma_r^{\frac{1}{m}} \lvert {\sigma_r^{-1} - \sigma_s^{-1}} \rvert \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}} \lvert {s-v} \rvert ^{-1-H} \lvert {r-s} \rvert ^{1+\frac{H}{m}}. \end{align*} $$

We have

$$ \begin{align*} \sigma_s \lvert {R_4} \rvert = e^{-\frac{ \lvert {Y_r} \rvert ^2}{2 \sigma_r^2}} \lvert {1 - e^{-\frac{Y_r^2}{2} (\frac{1}{\sigma_s} - \frac{1}{\sigma_r})}} \rvert \lesssim \lvert {\sigma_s^{-2} - \sigma_r^{-2}} \rvert Y_r^2 e^{-\frac{ \lvert {Y_r} \rvert ^2}{2 \sigma_r}}. \end{align*} $$

Hence, by (4.11),

$$ \begin{align*} \lVert {R_4} \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}}\sigma_s^{-1 + \frac{1}{m}} (1 - \sigma_s^2 \sigma_r^{-2}) \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}} \lvert {s-v} \rvert ^{-1 + (-1+\frac{1}{m})H} \lvert {t-s} \rvert. \end{align*} $$

To estimate $R_5$ , observe

By (4.12),

$$ \begin{align*} \lVert {e^{-\frac{ \lvert {Y_s - a} \rvert ^2}{8 \sigma_s^2}} \sigma_s^{-2} \lvert {Y_{s,r}} \rvert } \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}}\sigma_s^{-2 + \frac{1}{m}} \lvert {s-v} \rvert ^{H-1} \lvert {t-s} \rvert \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}} \lvert {s-v} \rvert ^{(-1 + \frac{1}{m})H - 1} \lvert {t-s} \rvert. \end{align*} $$

To estimate $\mathbb {P}( \lvert {Y_s - a} \rvert \leq 2 \lvert {Y_{s,r}} \rvert )$ , consider the decomposition

$$ \begin{align*} Y_{s,r} = \mathbb{E}[Y_s^2]^{-1} \mathbb{E}[Y_s Y_{s,r}] Y_s + (Y_{s,r} - \mathbb{E}[Y_s^2]^{-1} \mathbb{E}[Y_s Y_{s,r}] Y_s). \end{align*} $$

If $\frac {t-s}{s-v}$ is sufficiently small, then $ \lvert {\mathbb {E}[Y_s^2]^{-1} \mathbb {E}[Y_s Y_{s,r}]} \rvert \leq \frac {1}{4}$ . Therefore,

$$ \begin{align*} \mathbb{P}( \lvert {Y_s - a} \rvert \leq 2 \lvert {Y_{s,r}} \rvert ) \leq \mathbb{P}( \chi_s \lvert {Y - \chi_s^{-1} a} \rvert \lesssim_H \lvert {s-v} \rvert ^{H-1} \lvert {t-s} \rvert \lvert {X} \rvert ) \lesssim \chi_s^{-1} e^{-\frac{c a^2}{2 \chi_s^2}} \lvert {s-v} \rvert ^{H-1} \lvert {t-s} \rvert. \end{align*} $$

This gives an estimate of $R_5$ . Thus, we conclude

$$ \begin{align*} \mathbb{E}[L_t - L_s \vert \mathcal{F}_v] = \frac{e^{-\frac{ \lvert {Y_s - a} \rvert ^2}{2 \sigma_s^2}}}{\sqrt{2 \pi} \sigma_s} \lvert {t-s} \rvert + R_6, \end{align*} $$

where

$$ \begin{align*} \lVert {R_6} \rVert _{L_m(\mathbb{P})} \lesssim \chi_s^{-\frac{1}{m}} e^{-\frac{c a^2}{2 \chi_s^2}} \Big(\frac{t-s}{s-v} \Big)^{\frac{1}{m} + (1 - \frac{1}{m})H} \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}}. \end{align*} $$

This completes the proof of the lemma.

Now we can complete the proof of Theorem 4.1. The above lemmas show

$$ \begin{align*} \lVert {\mathfrak{c}_{H, \gamma}(L_t(a) - L_t(a)) - A_{s, t}(a)} \rVert _{L_m(\mathbb{P})} \begin{cases} \lesssim_T \lvert {t-s} \rvert ^{1 - H}, & \text{for all } H \in (0, 1), \\ \lesssim_{T} s^{-\frac{H}{m}} \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}}, & \text{if } H> \frac{1}{2}, \end{cases} \end{align*} $$

and if $\frac {t-s}{s-v}$ is sufficiently small, then

$$ \begin{align*} & \lVert {\mathbb{E}[\mathfrak{c}_{H, \gamma}(L_t(a) - L_s(a)) - A_{s, t}(a) \vert \mathcal{F}_v]} \rVert _{L_m(\mathbb{P})} \\ & \quad \begin{cases} \lesssim_T e^{- c (\mathbb{E}[B(0)^2] + T^{2H})^{-1} a^2} \Big(\frac{t-s}{s-v} \Big)^{\min\{2H, \frac{1}{m} + H\} - \frac{H}{m}} \lvert {t-s} \rvert ^{1 - H}, & H < \frac{1}{2}, \\ \lesssim_{T} s^{-\frac{H}{m}} e^{- c (\mathbb{E}[B(0)^2] + T^{2H})^{-1} a^2} \Big(\frac{t-s}{s-v} \Big)^{\min\{1, \frac{1}{m} + H\} - \frac{H}{m}} \lvert {t-s} \rvert ^{1 - H + \frac{H}{m}}, & H> \frac{1}{2}. \end{cases} \end{align*} $$

Noting that the exponents satisfy the assumption of Theorem 1.1 or Corollary 2.7, Remark 1.4 implies

(4.17) $$ \begin{align} \lVert {\mathfrak{c}_{H, \gamma} L_T(a) - \sum_{[s,t] \in \pi} A_{s, t}(a)} \rVert _{L_m(\mathbb{P})} \lesssim_{T} e^{- c (\mathbb{E}[B(0)^2] + T^{2H})^{-1} a^2} \lvert {\pi} \rvert ^{\epsilon}. \end{align} $$

Hence, we complete the proof of Theorem 4.1.

Proof. The claim is a special case of Theorem 4.1 with $\gamma = 0$ . When $m = 2$ , it is proved in [Reference Azaïs4, Theorem 5].

Proof. Set

It suffices to show that $\sum _{[s,t] \in \pi _n} \tilde {A}_{s, t}(\cdot )$ converge weakly to $0$ in $L_m(\mathbb {R})$ . Since

and since is bounded in $L_m(\mathbb {P})$ by assumption, it suffices to show that

$$ \begin{align*} \lim_{n \to \infty} \int_{\mathbb{R}} \Big( \sum_{[s,t] \in \pi_n} \tilde{A}_{s, t}(a) \Big) g(a) \, \mathrm{d} a = 0 \end{align*} $$

for every compactly supported continuous function g. Since for $B_s < B_t$ we have

$$ \begin{align*} B_{s,t}^{-1}\int_{B_s}^{B_t} \lvert {B_t - a} \rvert ^{\frac{1}{H} - 1} \, \mathrm{d} a = H \lvert {B_{s, t}} \rvert ^{\frac{1}{H} - 1}, \end{align*} $$

we obtain

$$ \begin{align*} \int_{B_s}^{B_t} \tilde{A}_{s, t}(a) g(a) \, \mathrm{d} a = \int_{B_s}^{B_t} \lvert {B_t - a} \rvert ^{\frac{1}{H} - 1} (g(a) - B_{s,t}^{-1} \int_{B_s}^{B_t} g(x) \, \mathrm{d} x) \, \mathrm{d} a. \end{align*} $$

Therefore,

$$ \begin{align*} \int_{\mathbb{R}} \Big( \sum_{[s,t] \in \pi_n} \tilde{A}_{s, t}(a) \Big) g(a) \, \mathrm{d} a &= \sum_{[s, t] \in \pi_n} \int_{B_s}^{B_t} \lvert {B_t - a} \rvert ^{\frac{1}{H} - 1} (g(a) - B_{s,t}^{-1} \int_{B_s}^{B_t} g(x) \, \mathrm{d} x) \, \mathrm{d} a \\ &\leq \int_{\mathbb{R}} \tilde{L}^{\pi_n}_T(a) \, \mathrm{d} a \times \sup_{ \lvert {x-y} \rvert \leq \sup_{[s,t] \in \pi} \lvert {B_t - B_s} \rvert } \lvert {g(x) - g(y)} \rvert \end{align*} $$

which converges to $0$ .

Proof. The argument is very similar to that of [Reference Lê24, Theorem 6.2]. Let $x^{(i)}$ ( $i = 1, 2$ ) be two solutions to (5.1). Then,

$$ \begin{align*} x^{(1)}_t - x^{(2)}_t = \int_0^t \{ b (x^{(1)}_s) - b (x^{(2)}_s) \} \, \mathrm{d} s + \int_0^t \{ \sigma (x^{(1)}_s) - \sigma (x^{(2)}_s) \} \, \mathrm{d} y_s = \int_0^t \{ x^{(1)}_s - x^{(2)}_s \} \, \mathrm{d} v_s, \end{align*} $$

where

Note that the second term is well-defined as a Young integral since

$$\begin{align*}s \mapsto \nabla \sigma (\theta x^{(1)}_s + (1 - \theta) x^{(2)}_s) \end{align*}$$

is $\delta \alpha $ -Hölder continuous and $\delta \alpha + \alpha> 1$ by our assumption of $\delta $ . Therefore, $x^{(1)} - x^{(2)}$ is a solution of the Young differential equation

$$\begin{align*}\, \mathrm{d} z_t = z_t \, \mathrm{d} v_t, \quad z_0 = 0. \end{align*}$$

The uniqueness of this linear Young differential equation is known. Hence, $x^{(1)} - x^{(2)} = 0.$

Proof. Here, we consider dyadic partitions. Fix $s < t$ , and set

Since $\mathcal {A}_{s,t} = \lim _{n \to \infty } A^n_{s, t}$ , it suffices to show

$$ \begin{align*} \lVert {A^{n}_{s,t} - A^{n+1}_{s, t}} \rVert _{L_m(\mathbb{P})} \lesssim 2^{-n \delta} (\Gamma_1 | t - s |^{\beta_1 - \alpha} + \Gamma_2 | t - s |^{\gamma + \beta_2} + \kappa_{m, d} \Gamma_3 | t - s |^{\beta_3} ) \end{align*} $$

for some $\delta> 0$ and all sufficiently large n. As in the proof of Theorem 1.1, we decompose

$$ \begin{align*} A^n_{s, t} - A^{n + 1}_{s, t} = \!\sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1}^{}\! \{ \delta A^n_{l + j L} -\mathbb{E} [\delta A^n_{l + j L} | \mathcal{F}_{l + (j - 1) L + 1}^n] \} + \!\sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1} \!\mathbb{E} [\delta A^n_{l + j L} | \mathcal{F}_{l + (j - 1) L + 1}^n]. \end{align*} $$

By the BDG inequality,

$$ \begin{align*} &\Big\| \sum_{j : l + j L \leq 2^n - 1}^{} \{ \delta A^n_{l + j L} -\mathbb{E} [\delta A^n_{l + j L} | \mathcal{F}_{l + (j - 1) L + 1}^n] \} \Big\|_{L_m (\mathbb{P})} \\& \quad \lesssim \kappa_{m, d} \Big( \sum_{j : l + j L \leq 2^n - 1} \| \delta A^n_{l + j L} \|^2_{L_m (\mathbb{P})} \Big)^{\frac{1}{2}} \leq \kappa_{m, d} \Gamma_3 \Big( \sum_{j : l + j L \leq 2^n - 1} (2^{- n} | t - s |)^{2 \beta_3} \Big)^{\frac{1}{2}}. \end{align*} $$

Thus,

$$\begin{align*}\Big\| \sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1}^{} \{ \delta A^n_{l + j L} -\mathbb{E} [\delta A^n_{l + j L} | \mathcal{F}_{l + (j - 1) L + 1}^n] \} \Big\|_{L_m(\mathbb{P})} \lesssim \kappa_{m, d} \Gamma_3 L^{\frac{1}{2}} 2^{- n ( \beta_3 - \frac{1}{2} )} | t - s |^{\beta_3}. \end{align*}$$

Furthermore,

$$ \begin{align*} & \Big\| \sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1} \mathbb{E} [\delta A^n_{l + j L} | \mathcal{F}_{l + (j - 1) L + 1}^n] \Big\|_{L_m (\mathbb{P})} \\& \quad \leq \sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1} \| \mathbb{E} [\delta A^n_{l + j L} | \mathcal{F}_{l + (j - 1) L + 1}^n] \|_{L_m (\mathbb{P})}\\& \quad \leq \Gamma_1 \sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1} \Big( \frac{L - 1}{2^n} | t - s | \Big)^{- \alpha} (2^{- n} | t - s |)^{\beta_1} \\& \qquad \phantom{\leq} + \Gamma_2 \sum_{l = 0}^L \sum_{j : l + j L \leq 2^n - 1} \Big( \frac{L }{2^n} | t - s | \Big)^{\gamma} (2^{- n} | t - s |)^{\beta_2}\\& \quad \lesssim \Gamma_1 2^{- n (\beta_1 - \alpha - 1)} L^{- \alpha} | t - s |^{\beta_1 - \alpha} + \Gamma_2 2^{- n (\gamma + \beta_2 - 1)} L^{\gamma} | t - s |^{\gamma + \beta_2}. \end{align*} $$

Therefore,

$$ \begin{align*} &\| A^n_{s, t} - A^{n + 1}_{s, t} \|_{L_m (\mathbb{P})} \lesssim \Gamma_1 2^{- n (\beta_1 - \alpha - 1)} L^{- \alpha} | t - s |^{\beta_1 - \alpha} \\& \quad + \Gamma_2 2^{- n (\gamma + \beta_2 - 1)} L^{\gamma} | t - s |^{\gamma + \beta_2} + \kappa_{m, d} \Gamma_3 L^{\frac{1}{2}} 2^{- n ( \beta_3 - \frac{1}{2} )} | t - s |^{\beta_3}. \end{align*} $$

We choose $L = \lfloor 2^{n \varepsilon } \rfloor $ with $\varepsilon \in (0, 1)$ so that

$$\begin{align*}\min \Big\{ \beta_1 - \alpha - 1 + \alpha \varepsilon, \gamma + \beta_2 - 1 - \gamma \varepsilon, \beta_3 - \frac{1 + \varepsilon}{2} \Big\}> 0. \end{align*}$$

Namely,

$$\begin{align*}\frac{1 + \alpha - \beta_1}{\alpha} < \varepsilon < \min \Big\{ \frac{\gamma + \beta_2 - 1}{\gamma}, 2 \beta_3 - 1 \Big\}. \end{align*}$$

Such an $\epsilon $ exists exactly under our assumption (5.6).

Proof. Our tool is Lemma 5.5. Since arguments are similar, we only prove the claim for $Z \in \mathcal {D}_1(\alpha , \beta )$ . We set

We have

$$\begin{align*}\delta A_{s, u, t} = - \frac{f (Z_u) - f (Z_s)}{2} B_{u, t} + \frac{f (Z_t) - f (Z_u)}{2} B_{s, u}. \end{align*}$$

Hence

(5.11) $$ \begin{align} \| \delta A_{s, u, t} \|_{L_m (\mathbb{P})} \lesssim \| f \|_{C^{\delta}} \lVert {\lVert{Z}\rVert_{C^{\alpha}}^{\delta}} \rVert _{L_{2m}(\mathbb{P})} | t - s |^{H + \alpha \delta}. \end{align} $$

Let $v < s$ with $t-s \leq s- v$ . As $Z \in \mathcal {D}_1 (\alpha , \beta )$ , we write

$$\begin{align*}Z_r = z_{}^{(1)} (v) + z^{(2)}_{} (v) B_r + z^{(3)} (v) (r - v) + R_{v, r}, \quad r \in [s, t]. \end{align*}$$

Then, if we write ,

$$\begin{align*}| f (Z_r) - f (\hat{Z}_r) | \leq \| f \|_{C^{\delta}} | R_{v, r} |^{\delta} \leq \| f \|_{C^{\delta}} \lVert {Z} \rVert _{\mathcal{D}(\alpha,\beta)}^{\delta} | v - r |^{\delta \beta}. \end{align*}$$

Hence, if we write , then

(5.12) $$ \begin{align} \| \delta A_{s, u, t} - \delta \hat{A}_{s, u, t} \|_{L_m (\mathbb{P})} \lesssim_{\alpha, \delta, T} \| f \|_{C^{\delta}} \lVert {\lVert{Z}\rVert_{\mathcal{D}(\alpha,\beta)}^{\delta}} \rVert _{L_{2m}(\mathbb{P})} | t - v |^{\delta \beta} | t - s |^H. \end{align} $$

Next, we will estimate $ \lVert {\mathbb {E} [\delta \hat {A}_{s, u, t} \vert \mathcal {F}_v]} \rVert _{L_m(\mathbb {P})}$ . The rest of the calculation resembles the proof of Proposition 3.5. We write and as before. We can decompose

where $\hat {Y}_r$ is $\mathcal {F}_v$ -measurable and $\tilde {B}_r$ is independent of $\mathcal {F}_v$ . We set

Then, as in the proof of Proposition 3.5, we have the decomposition

$$\begin{align*}\mathbb{E} [\delta \hat{A}_{s, u, t} | \mathcal{F}_v] = \hat{D}^0_{s, u, t} + \sum_{i = 1}^d \hat{D}_{s, u, t}^i, \end{align*}$$

where, as in (3.8) and (3.11),

The map $\mathbb {R}^d \ni x \mapsto f (z^{(2)} (v)^{} x) \in \mathbb {R}^d$ belongs to $C^{\delta } (\mathbb {R}^d)$ with its norm bounded by

$$\begin{align*}| z^{(2)} (v) |^{\delta} \| f \|_{C^{\delta} (\mathbb{R}^d)}. \end{align*}$$

Therefore, by repeating the argument used to obtain (3.9), we obtain

$$ \begin{align*} &| \hat{D}^0_{s, u, t} | \leq | z^{(2)} (v) |^{\delta} \| f \|_{C^{\delta} (\mathbb{R}^d)} \Big\{ (s - v)^{\delta H - 1} (t - s) (| Y_{s, u} | + | Y_{u, t} |) \\ & \quad + (s - v)^{(\delta - 1) H} \Big( | (z^{(2)} (v))^{- 1} \hat{Y}_{s, u} | | Y_{u, t} | + | z^{(2)} (v)^{- 1} \hat{Y}_{u, t} | | Y_{s, u} | \Big) \Big\}. \end{align*} $$

Referring to (3.4), we have

$$\begin{align*}\| (z^{(2)} (v))^{- 1} \hat{Y}_{s, u} \|_{L_m (\mathbb{P})} \lesssim (s - v)^{H - 1} (t - s) + \| z^{(2)}(v)^{-1} z^{(3)} (v) \|_{L_m (\mathbb{P})} (t - s). \end{align*}$$

Therefore,

$$ \begin{align*} &\| \hat{D}^0_{s, u, t} \|_{L_m (\mathbb{P})} \lesssim \| f \|_{C^{\delta}} \| | z^{(2)} (v) |^{\delta} \|_{L_{3 m} (\mathbb{P})} \\ & \quad \times \{ (s - v)^{(\delta + 1) H - 2} (t - s)^2 + \| z^{(2)}(v)^{-1} z^{(3)} (v) \|_{L_{3 m} (\mathbb{P})} (s - v)^{\delta H - 1} (t - s)^2 \}. \end{align*} $$

Similarly as before, we have

$$\begin{align*}| a^i_i (t) - a^i_i (s) | \lesssim | z^{(2)} (v) |^{\delta} \| f \|_{C^{\delta} (\mathbb{R}^d)} (s - v)^{(\delta - 2) H} (| (z^{(2)} (v))^{- 1} \hat{Y}_{s, u} | + (s - v)^{H - 1} (t - s)). \end{align*}$$

As $H> \frac {1}{2}$ , by Lemma 3.2, we have

$$\begin{align*}| \mathbb{E} [\tilde{B}_s \tilde{B}_{s, t}] | \lesssim (s - v)^{2 H - 1} (t - s). \end{align*}$$

Therefore,

$$ \begin{align*} &\| \hat{D}^i_{s, u, t} \|_{L_m (\mathbb{P})} \lesssim \| f \|_{C^{\delta}} \| | z^{(2)} (v) |^{\delta} \|_{L_{3 m} (\mathbb{P})} \\ &\quad \times \{ (s - v)^{(\delta + 1) H - 2} (t - s)^2 + \| z^{(2)}(v)^{-1} z^{(3)} (v) \|_{L_{3 m} (\mathbb{P})} (s - v)^{\delta H - 1} (t - s)^2 \}. \end{align*} $$

Combining our estimates, we have

(5.13) $$ \begin{align} &\| \mathbb{E} [\delta \hat{A}_{s, u, t} | \mathcal{F}_v] \|_{L_m (\mathbb{P})} \lesssim_T \| f \|_{C^{\delta}} \| | z^{(2)} (v) |^{\delta} \|_{L_{3 m} (\mathbb{P})} \nonumber\\ & \quad \times (1 + \| z^{(2)}(v)^{-1} z^{(3)} (v) \|_{L_{3 m} (\mathbb{P})}) (s - v)^{(\delta + 1) H - 2} (t - s)^2. \end{align} $$

Hence, combining (5.11), (5.12), and (5.13)

$$\begin{align*}\| \delta A_{s, u, t} \|_{L_m (\mathbb{P})} \lesssim \| f \|_{C^{\delta}} \lVert {\lVert{Z}\rVert_{C^{\alpha}}^{\delta}} \rVert _{L_{2m}(\mathbb{P})} | t - s |^{H + \alpha \delta}, \end{align*}$$

and

$$ \begin{align*} &\| \mathbb{E} [\delta A_{s, u, t} | \mathcal{F}_v] \|_{L_m (\mathbb{P})} \lesssim_{\alpha, \delta, T} \| f \|_{C^{\delta}} \| | z^{(2)} (v) |^{\delta} \|_{L_{3 m} (\mathbb{P})} \\& \quad \times (1 + \| z^{(2)}(v)^{-1} z^{(3)} (v) \|_{L_{3 m} (\mathbb{P})}) (s - v)^{(\delta + 1) H - 2} (t - s)^2 \\& \quad +\| f \|_{C^{\delta}} \lVert {\lVert{Z}\rVert_{\mathcal{D}(\alpha,\beta)}^{\delta}} \rVert _{L_{2m}(\mathbb{P})} | t - v |^{\delta \beta} | t - s |^H. \end{align*} $$

To apply Lemma 5.5, we need

$$\begin{align*}\delta \beta + H> 1, \quad \frac{1 - (1 + \delta) H}{2 - (1 + \delta) H} < \min \Big\{ \frac{\delta \beta + H - 1}{\delta \beta}, 2 (H + \alpha \delta) - 1 \Big\} , \end{align*}$$

which, if $\alpha $ is sufficiently close to H, are fulfilled under (5.10).

Proof of Theorem 5.2.

We first prove the pathwise uniqueness. We suppose that the assumption in the case 2 mentioned in Theorem 5.2 holds, and the other cases will be discussed later. Let $X^{(1)}$ and $X^{(2)}$ be $(\mathcal {F}_t)$ -adapted solutions. Our strategy is similar to Proposition 5.1, but, here, we must construct the integral (5.5) stochastically. For each $k \in \mathbb {N}$ , we set

we let $\sigma ^{k} \in C^{1 + \delta }(\mathbb {R}^d; \mathcal {M}_d)$ be such that $\sigma ^k = \sigma $ in $\{x \mid \lvert {x} \rvert \leq k\}$ and

$$ \begin{align*} \inf_{x \in \mathbb{R}^d} \inf_{y: \lvert {y} \rvert = 1} y \cdot \sigma^k(x) y \geq \frac{\lambda_k}{2}, \end{align*} $$

and we set

If we write

then in the event $\Omega _k$ , we have $X^{(i)}_t = X^{(i), k}_t$ , $t \in [0, T]$ .

Let $\{\sigma ^{k, n}\}_{n=1}^{\infty }$ be a smooth approximation of $\sigma ^k$ . In general, we can only guarantee the convergence in $C^{1+\delta '}(\mathbb {R}^d, \mathcal {M}_d)$ for any $\delta '<\delta $ , which is still sufficient to make the following argument work. To simplify the notation, we assume that we can take $\delta '=\delta $ .

We have

$$ \begin{align*} \int_{0}^t \sigma^k(X^{(i)}_r) \, \mathrm{d} B_r = \lim_{n \to \infty} \int_{0}^t \sigma^{k, n}(X^{(i)}_r) \, \mathrm{d} B_r \end{align*} $$

and in $\Omega _k$

$$ \begin{align*} \int_{0}^t \{ \sigma^{k, n}(X^{(1)}_r) - \sigma^{k, n}(X^{(2)}_r)\} \, \mathrm{d} B_r = \int_0^t \{X^{(1), k}_r - X^{(2), k}_r \} \, \mathrm{d} V^{k, n}_r, \end{align*} $$

where

For a fixed $\theta \in (0, 1)$ , we set

By the a priori estimate (5.9), for $\alpha \in (\frac {1}{2}, H)$ , we have

$$ \begin{align*} Z_t^{\theta, k} - Z_s^{\theta, k} &= \{\theta \sigma^k(X_s^{(1), k}) + (1 - \theta) \sigma^k(X_s^{(1), k})\} B_{s,t} \\ & \quad + \{\theta b(X_s^{(1), k}) + (1 - \theta) b (X_s^{(1), k})\}(t-s) + R_{s, t} \end{align*} $$

with

$$ \begin{align*} \lvert {R_{s, t}} \rvert \lesssim_{T, \alpha} (1 + \lvert {x} \rvert + \lVert {b} \rVert _{C^1_b} + \lVert {\sigma} \rVert _{C^1_b} \lVert {B} \rVert _{C^{\alpha}})^3 \lvert {t-s} \rvert ^{2 \alpha}. \end{align*} $$

Note that we have

$$ \begin{align*} \inf_{y: \lvert {y} \rvert = 1} y \cdot \{\theta \sigma^k(X_s^{(1), k}) + (1 - \theta) \sigma^k(X_s^{(1), k})\} y \geq \frac{\lambda_k}{2}, \end{align*} $$

and hence

$$ \begin{align*} \lvert {\{\theta \sigma^k(X_s^{(1), k}) + (1 - \theta) \sigma^k(X_s^{(1), k})\}^{-1} } \rvert \lesssim \lambda_k^{-1}. \end{align*} $$

Therefore, we have $Z^{\theta , k} \in \mathcal {D}_1(\alpha , 2 \alpha )$ with

$$ \begin{align*} \lVert {Z^{\theta, k}} \rVert _{\mathcal{D}_1(\alpha, 2 \alpha)} \lesssim_{T, \alpha} (1 + \lvert {x} \rvert + \lVert {b} \rVert _{C^1_b} + \lVert {\sigma} \rVert _{C^1_b} \lVert {B} \rVert _{C^{\alpha}})^3 + (1 + \lambda_k^{-1}) ( \lVert {b} \rVert _{L_{\infty}} + \lVert {\sigma} \rVert _{L_{\infty}}). \end{align*} $$

Since

$$ \begin{align*} \max \Big\{ \frac{1 - H}{2 \alpha}, \frac{2 - 3 H + H^2}{2 \alpha + H - H^2}, \frac{3 - \sqrt{3}}{2 H} - 1 \Big\} = \frac{2 - 3 H + H^2}{2 \alpha + H - H^2}< \delta \end{align*} $$

if $\alpha $ is sufficiently close to H, by Proposition 5.8,

$$ \begin{align*} \lVert { V^{k, n_1}_{s, t} - V^{k, n_2}_{s, t}} \rVert _{L_m(\mathbb{P})} \lesssim_{T, \alpha, \delta, b, \sigma, k, m} \lVert {\sigma^{k, n_1} - \sigma^{k, n_2}} \rVert _{C^{1 + \delta}(\mathbb{R}^d, \mathcal{M}_d)} \lvert {t-s} \rvert ^{H}. \end{align*} $$

By Kolmogorov’s continuity theorem, the sequence $(V^{k, n})_{n \in \mathbb {N}}$ converges to some $V^k$ in $C^{\alpha }([0, T], \mathbb {R}^d)$ .

Therefore, we conclude that almost surely in $\Omega _k$ , the path $z = X^{(1)} - X^{(2)}$ solves the linear Young equation

and hence $X^{(1)} = X^{(2)}$ . Since $\mathbb {P}(\Omega _k) \to 1$ , we conclude $X^{(1)} = X^{(2)}$ almost surely. Thus, we completed the proof of the uniqueness under the case 2. The other cases can be handled similarly. Indeed, under the case 1, we have $X^{(i)} \in \mathcal {D}_0(\alpha , 2 \alpha )$ , and under the case 3, we have $X^{(i)} \in \mathcal {D}_0(\alpha , 1)$ .

Now, it remains to prove the existence of a strong solution. However, in view of the Yamada-Watanabe theorem (Proposition B.2), it suffices to show the existence of a weak solution, which will be proved in Lemma B.3 based on a standard compactness argument.