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# A matrix Burkholder-Davis-Gundy inequality

Tom Maître

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Tom Maître. A matrix Burkholder-Davis-Gundy inequality. 2025. hal-05063068

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A matrix Burkholder-Davis-Gundy inequality

Tom Maitre

May 11, 2025

Abstract

We prove an inequality for the spectral norm of matrix valued stochastic integrals. This in-
equality can be seen either as a non-commutative version of the
Burkholder-Davis-Gundy inequality or as an extension of the non-commutative Khintchine
inequality of Lust-Piquard to stochastic integrals. The proof relies on a version of Freedman's
inequality for matrix valued martingales.

# 1 Introduction and preliminaries

The non-commutative Khintchine inequality, discovered by Lust-Piquard, asserts that there exists two universal constant c and C such that for a random variable X of the form $\sum_{i=1}^{N} H_i \gamma_i$ where $\gamma_i$ are independent standard Gaussian variables and $H_i$ are deterministic symmetric matrices of size $n \times n$,

$$
c \left\| \sum_{i=1}^{N} H_i^2 \right\|^{1/2} \le \mathbb{E} \|X\| \le C \sqrt{\log n} \left\| \sum_{i=1}^{N} H_i^2 \right\|^{1/2} \quad (1)
$$

where $\|A\|$ denotes the spectral norm of the symmetric matrix A; For further details on inequality,
we refer to [Lus91] or [Pis03].

The Burkholder-Davis-Gundy's (BDG) inequality asserts that

$$
c_p \left( \mathbb{E} \langle X \rangle_t^{p/2} \right)^{1/p} \le \mathbb{E} \sup_{0 \le s \le t} |X_s|^p \le C_p \left( \mathbb{E} \langle X \rangle_t^{p/2} \right)^{1/p} \quad (2)
$$

where X is a continuous martingale, $\langle X \rangle$ is its quadratic variation, and $c_p, C_p$ are constants depending only on $p$. It is an important result in stochastic analysis that has numerous applications. Our main result of this paper generalizes both (1) and (2).

Throughout this paper, we consider a local martingale X that can be written, for all $t \ge 0$, as

$$
X_t = \int_0^t \sum_{i=1}^N H_{i,s} dB_s^i \quad (3)
$$

where $(B_t = (B_t^1, \dots, B_t^N))$ is a standard $N$-dimensional Brownian motion, and $(H_{i,t})_t$ are pro-
gressively measurable processes starting from 0 and taking values in the space $n \times n$ symmetric
matrices. For every matrix $A$, we denote $\text{Tr}(A)$ its trace. We assume that $(H_{i,s})_s$ satisfies

$$
\mathrm{Tr} \left( \int_0^t \sum_{i=1}^N H_{i,s}^2 ds \right) < \infty
$$

almost surely, which ensures that the stochastic integral (3) is well-defined. We define the quadratic variation of X, denoted $\langle X \rangle$, as follows:

$$
\langle X \rangle_t = \int_0^t \sum_{i=1}^N H_{i,s}^2 ds, \quad \forall t \ge 0.
$$

This process plays a significant role in our main theorem, which we now state.
**Theorem 1.1.** Let $(X_t)$ be a stochastic process of the form (3). Then, there exists a universal constant $C$ such that, for all $p \in \mathbb{N}^*$ and all $t \in \mathbb{R}_+^*$,

$$ \left( \mathbb{E} \sup_{0 \le s \le t} \|X_t\|^p \right)^{\frac{1}{p}} \le C \sqrt{p} \left( \mathbb{E} \|\langle X \rangle_t\|^p \right)^{\frac{1}{2p}} + \sqrt{\log n} \left( \mathbb{E} \|\langle X \rangle_t\|^\frac{p}{2} \right)^{\frac{1}{p}}. \quad (4) $$

Our proof gives $C = 2\sqrt{2}$.

The $\log^{1/2} n$ correction is necessary in general, as we shall see below. We can compare Theorem 1.1 with the Burlholder-Davis-Gundy's inequality. In fact, this theorem allows us to control the moments of the supremum of the spectral norm of our process by the spectral norm of its quadratic variation. Thus this extends the BDG inequality, in the case of symmetric matrices. However, let us remark that the power $1/2$ in the first term of the right-hand side of inequality (4) is outside the expectation, rather than inside. Because of this gap Theorem 1.1 does not fully recover the classical BDG inequality in dimension 1. It is likely possible to pull this power $1/2$ inside, but we did not achieve that.

Furthermore, if we take $p = 1$ and apply Jensen's inequality, we get

$$ \mathbb{E} \sup_{0 \le s \le t} \|X_t\| \lesssim \sqrt{\log n} \left( \mathbb{E} \left\| \int_0^t \sum_{i=1}^N H_{i,s}^2 ds \right\| \right)^{1/2}. \quad (5) $$

where $a \lesssim b$ means there exists an universal constant $k$ such that $a \le kb$. If we consider the special case where the matrices $(H_{i,t})$ are deterministic and constant over time, applying (5) at $t = 1$ yields in particular

$$ \mathbb{E} \left\| \sum_{i=1} \gamma_i H_i \right\| \lesssim \sqrt{\log n} \left\| \sum_{i=1}^N H_i^2 \right\|^{1/2}, $$

where the variables $\gamma_i$ are independent standard Gaussian variables. This is the non-commutative Khintchine inequality of Lust-Piquard, in the form put forward by Tropp [Tro16]. Note that the $\sqrt{\log n}$ factor is in general necessary (see e.g. [Tro16] for the details), which shows that also in (4) the $\sqrt{\log n}$ is needed.

Noncommutative versions of the Burkholder-Davis-Gundy inequalities have been established in the context of free probability, notably in [Bia98] and [Pis97]. However, these results do not imply our main theorem, which relies on features that are specific to the matrix-valued setting.

The proof of Theorem 1.1 relies on the following inequality, due to Freedman. If $(M_t)_{t \ge 0}$ is a continuous local martingale starting from 0, then, for all $u > 0$ and $\sigma \in \mathbb{R}$,

$$ \mathbb{P} (\exists t > 0, M_t \ge u \text{ and } \langle M \rangle_t \le \sigma^2) \le \exp \left( - \frac{u^2}{2\sigma^2} \right). $$

(See [Fre75] for more details). Tropp extended this result to the matrix-valued case, but only in discrete time (see [Tro11]). Our version of this inequality reads as follows

**Theorem 1.2** (Freedmann's matrix inequality). Let $(X_t)_{t \ge 0}$ be a stochastic process of the form (3). Let $\sigma, u \in \mathbb{R}$. Then,

$$ \mathbb{P} (\exists t > 0, \lambda_{max}(X_t) \ge u \text{ and } \|\langle X \rangle_t\| \le \sigma^2) \le ne^{-\frac{u^2}{2\sigma^2}}. $$

Using different methods, we also prove an inequality for Schatten norm of matrix of the form (3). Recall that for all $p \ge 1$, the Schatten $p$-norm of a symmetric matrix $A$ is given by $\|A\|_p^p = \sum_{i=1}^n |\lambda_i|^p$ where $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $A$. This indeed defines a norm on the set of symmetric matrices.

**Theorem 1.3.** Let $(X_t)_{t \ge 0}$ be a stochastic process of the form given in (3), and $p \in \mathbb{N}^*$. Then,

$$ \mathbb{E} \|X_t\|_{2p}^2 \le (2p - 1) \left( \mathbb{E} \int_0^t \left\| \sum_{i=1}^N H_{i,s}^2 \right\|_p ds \right). $$
A classical approach to studying the spectral norm of a matrix is to use the well-known inequalities, valid for all $p \ge 1$:

$$ \|A\| \le \|A\|_p \le n^{1/p} \|A\| \quad (6) $$

where the upper bound is derived by bounding the trace as $n$ times the maximal eigenvalue. Using Theorem 1.3 with $p \approx \log n$, Doob's maximal inequality and (6), we see that Theorem 1.3 yields the following

$$ \mathbb{E} \sup_{0 \le s \le t} \|X_s\| \le C \sqrt{\log n} \left( \int_0^t \mathbb{E} \left\| \sum_{i=1}^N H_{i,s}^2 \right\| ds \right)^{\frac{1}{2}}. $$

However, this approach yields a less precise inequality than Theorem 1.1. Indeed, the square root is outside the expectation and integral again, but the spectral norm is now inside the integral in contrast with (5). To improve this result, we can hope for the existence of an inequality of this type, with the square root inside the expectation.

This paper is organized as follows. First, we prove Theorem 1.3. The proof of this theorem uses Itô's lemma. In the second section we present the proof of Theorem 1.2. Finally, we use this inequality to establish our main theorem.

## 2 A bound for the Schatten norms

In this section, we prove Theorem 1.3. Our proof is inspired by Tropp's proof of the matrix Khintchine inequality, which relies on the Gaussian integration by parts formula. In our context, integration by parts is replaced by the use of Itô's formula. For this, we require the following lemma.

**Lemma 2.1.** Suppose that $H$ and $A$ are Hermitian matrices of the same size. Let $q$ and $r$ be integers that satisfy $0 \le q \le r$. Then,

$$ \mathrm{Tr}(HA^qHA^{r-q}) \le \mathrm{Tr}(H^2|A|^r). $$

See e.g. [Kla24, section 7] for proof of this lemma. We are in a position to prove the main result of this section.

*Proof of Theorem 2.1.* By Itô's lemma, for all sufficiently regular $f$,

$$ df(X_t) = \sum_{i=1}^{N} \langle \nabla f(X_t); H_{i,t} \rangle dB_t^i + \frac{1}{2} \sum_{i=1}^{N} \langle \nabla^2 f(X_t) H_{i,t}; H_{i,t} \rangle dt, \quad (7) $$

where $\nabla^2 f$ is the Hessian of $f$, $\langle A; B \rangle = \mathrm{Tr}(AB)$ denotes the usual scalar product on the space of symmetric matrices, and $d$ represents the Itô derivative.

Now, consider $f(X) = \mathrm{Tr}(X^{2p})^{1/p}$, which is twice differentiable. For all Hermitian matrices $X, H$, we have

$$ \langle \nabla f(X); H \rangle = 2\mathrm{Tr}(X^{2p})^{\frac{1}{p}-1}\mathrm{Tr}(X^{2p-1}H), $$

and

$$ \begin{aligned} \langle \nabla^2 f(X)H; H \rangle &= 2\mathrm{Tr}(X^{2p})^{\frac{1}{p}-1} \sum_{k=0}^{2p-2} \mathrm{Tr}(X^k H X^{2p-2-k} H) + 4p \left(\frac{1}{p}-1\right) \mathrm{Tr}(X^{2p-1} H)^2 \mathrm{Tr}(X^{2p})^{\frac{1}{p}-2} \\ &\le 2(2p-1)\mathrm{Tr}(X^{2p})^{\frac{1}{p}-1} \mathrm{Tr}(X^{2p-2} H^2) \end{aligned} $$

by Lemma 2.1, and since the last term on the right-hand side of the equality is non-positive. Plugging this back into (7) yields

$$ df(X_t) \le 2\mathrm{Tr}(X_t^{2p})^{\frac{1}{p}-1} \sum_{i=1}^{N} \mathrm{Tr}(X_t^{2p-1} H_{i,t}) dB_t^i + (2p-1) \|X_t\|_{2p}^{2(1-p)} \left\langle \sum_{i=1}^{N} H_{i,t}^2; X_t^{2p-2} \right\rangle dt. \quad (8) $$
Applying Hölder's inequality for the Schatten norms (see e.g. [Ba05r]), we get

$$ \left\langle \sum_{i=1}^{N} H_{i,t}^{2}; X_{t}^{2p-2} \right\rangle \leq \|X_{t}\|_{2p}^{2(p-1)} \left\| \sum_{i=1}^{N} H_{i,t}^{2} \right\|_{p}. $$

Plugging this back in (8), we obtain

$$ df(X_t) \leq 2\mathrm{Tr}(X_t^{2p})^{\frac{1}{p}-1} \sum_{i=1}^{N} \mathrm{Tr}(X_t^{2p-1} H_{i,t}) dB_t^i + (2p-1) \left\| \sum_{i=1}^{N} H_{i,t}^2 \right\|_p dt. $$

Since the first process on the right-hand side of the inequality is a local martingale, there exists a sequence of stopping times $(T_m)_{m \geq 1}$ such that, for all $m$, this process stopped at $T_m$ is a martingale with zero expectation, and $T_m \uparrow \infty$. Applying expectation at time $t \wedge T_m$, we get

$$ \mathbb{E}\|X_{t \wedge T_m}\|_{2p}^2 \leq (2p-1)\mathbb{E}\int_0^{t \wedge T_m} \left\|\sum_{i=1}^N H_{i,s}^2\right\|_p ds \leq (2p-1)\mathbb{E}\int_0^t \left\|\sum_{i=1}^N H_{i,s}^2\right\|_p ds. $$

Letting $m$ tend to infinity and using Fatou's lemma yields the result. □

**Remark 1.** If we consider $f = \|\cdot\|_{2p}$, the same proof yields

$$ \left( \mathbb{E} \|X_t\|_{2p}^{2p} \right)^{1/p} \leq (2p-1) \int_0^t \left( \mathbb{E} \left\| \sum_{i=1}^N H_{i,s}^2 \right\|_p^p \right)^{1/p} ds. $$

Nonetheless, the issue here is the placement of the power $p$.

In fact, the symmetry assumption is not essential, it can be removed by a standard trick, see e.g. [Tro15]. We denote $M_{n_1 \times n_2}$ the space of rectangular matrices of $n_1$ rows and $n_2$ columns. For $A \in M_{n_1 \times n_2}$, we apply the symmetric case to the Hermitian dilatation, namely the matrix $\mathcal{H}(A)$ given by

$$ \mathcal{H}(A) = \begin{pmatrix} 0 & A \\ A^* & 0 \end{pmatrix}. \qquad (9) $$

Recall that for $A \in M_{n_1 \times n_2}$, the Schatten $p$-norm of $A$ is defined as $\|A\|_p^p = \sum_{i=1}^{\min\{n_1, n_2\}} |\sigma_i(A)|^p$ where $\sigma_i(A), \dots, \sigma_{\min\{n_1, n_2\}}(A)$ are the singular values of $A$. Moreover, $A^*$ design the conjugate transpose of $A$.

**Theorem 2.2.** Let $p \geq 1$, and $(X_t)$ to a process of the form (3), where $(H_{i,s})$ are not necessary symmetric and possibly rectangular matrices. Then

$$ \mathbb{E}\|X_t\|_{2p}^2 \leq 2^{-1/p} \sqrt{2p-1} \int_0^t \mathbb{E} \left( \left\| \sum_{i=1}^N H_{i,s} H_{i,s}^* \right\|_p^p + \left\| \sum_{i=1}^N H_{i,s}^* H_{i,s} \right\|_p^p \right)^{1/p} ds $$

*Proof.* We observe that $\mathcal{H}(X_t) = \int_0^t \sum_{i=1}^N \mathcal{H}(H_{i,s}) dB_s^i$. Applying Theorem 1.3 to the self-adjoint matrix $\mathcal{H}(X_t)$, we obtain

$$ \mathbb{E}\|\mathcal{H}(X_t)\|_{2p}^2 \leq (2p-1) \int_0^t \mathbb{E} \left\| \sum_{i=1}^N \mathcal{H}(H_{i,s})^2 \right\|_p ds. \qquad (10) $$

But, for every matrix $A$,

$$ \mathcal{H}(A)^2 = \begin{pmatrix} AA^* & 0 \\ 0 & A^*A \end{pmatrix}. $$
Thus,

$$ \left\| \sum_{i=1}^{N} \mathcal{H}(H_{i,s})^2 \right\|_{p} = \left( \left\| \sum_{i=1}^{N} H_{i,s} H_{i,s}^* \right\|_{p}^{p} + \left\| \sum_{i=1}^{N} H_{i,s}^* H_{i,s} \right\|_{p}^{p} \right)^{1/p} . \quad (11) $$

On the other hand, note that

$$ \|\mathcal{H}(X_t)\|_{2p}^p = (\|X_t^* X_t\|_p^p + \|X_t X_t^*\|_p^p)^{1/p} = 2^{1/p} \|X_t^* X_t\|_p = 2^{1/p} \|X_t\|_{2p}^2. \quad (12) $$

By combining (10), (11) and (12), we obtain the desired result. □

This result is consistent with the non-symmetric version of matrix Khintchine inequality, see [Lus91].

# 3 The matrix Freedman inequality

In this section, we prove Theorem 1.2. To proceed, we first introduce the following lemma:

**Lemma 3.1.** Let $f: M \to \text{Tre}^M$ where $M$ is taking values in the space of symmetric matrices. Then, for every symmetric matrices $M, H$;

$$ \langle \nabla^2 f(M) H; H \rangle \le \langle \nabla f(M); H^2 \rangle = \text{Tr}(e^M H^2) $$

where $\nabla^2 f(M)$ stands for the Hessian matrix of $f$ at $M$.

The proof of Lemma 3.1 can be found in Section 7 of [Kla24]. The key lemma is as follows :

**Lemma 3.2.** Let $(X_t)_{t \ge 0}$ be a local martingale of the form (3). Then, the process

$$ \left( \text{Tr}(e^{X_t - \frac{1}{2}\langle X \rangle_t}) \right)_{t \ge 0} $$

is a supermartingale.

*Proof.* Consider the process $Y_t = X_t - \frac{1}{2}\langle X \rangle_t$ and let $f: A \to \text{Tr}(e^A)$ as defined in the previous lemma. By Itô's lemma, and Lemma 3.1, we have

$$ \begin{aligned} df(Y_t) &= \sum_{i=1}^{N} \left[ \langle \nabla f(Y_t); H_{i,t} \rangle dB_t^i - \frac{1}{2} \langle \nabla f(Y_t); H_{i,t}^2 \rangle dt \right] + \frac{1}{2} \sum_{i=1}^{N} \langle \nabla^2 f(Y_t) H_{i,t}; H_{i,t} \rangle dt \\ &\le \text{Tr} \left( e^{Y_s} \left( \sum_{i=1}^{N} H_{i,s} dB_s^i - \frac{1}{2} \sum_{i=1}^{N} H_{i,s}^2 ds \right) \right) + \frac{1}{2} \text{Tr} \left( e^{Y_s} \left( \sum_{i=1}^{N} H_{i,s} dB_s^i \right)^2 \right) \\ &= \sum_{i=1}^{N} \text{Tr} (e^{Y_s} H_{i,s}) dB_s^i. \end{aligned} $$

Therefore the nonnegative process $(\text{Tr}(e^{Y_t}))_{t \ge 0}$ is the sum of a local martingale, and a decreasing adapted process. Fatou's lemma ensures that it is a supermartingale. □

**Corollary 3.3.** Let $(X_t)$ be a stochastic process as defined in (3). Let $\sigma, u \in \mathbb{R}$ and $t > 0$. If

$$ \mathbb{P}(\|\langle X \rangle_t\| \le \sigma^2) = 1. $$

Then,

$$ \mathbb{P}(\lambda_{\max}(X_t) \ge u) \le n e^{-\frac{u^2}{2\sigma^2}}. $$
*Proof.* We almost surely have $\langle X \rangle_t^2 \asymp \sigma^2 I_n$, where $A \asymp B$ means that the matrix $B - A$ is a positive semi-definite symmetric matrix. Fix $\beta \ge 0$. Then

$$ \mathrm{Tr} \left( e^{\beta X_t - \frac{\beta^2}{2} \langle X \rangle_t} \right) \ge \mathrm{Tr} \left( e^{\beta X_t - \beta^2 \frac{\sigma^2}{2} I_n} \right) = \mathrm{Tr} \left( e^{\beta X_t} \right) e^{-\frac{\beta^2 \sigma^2}{2}} \ge e^{\beta \lambda_{\max}(X_t) - \frac{\beta^2 \sigma^2}{2}}. $$

Combining with Lemma 3.2, we obtain

$$ \mathbb{E}e^{\beta\lambda_{\max}(X_t)} \le e^{-\frac{\beta^2\sigma^2}{2}} \mathrm{Tr} \left( e^{\beta X_0 - \frac{\beta^2\langle X \rangle_0}{2}} \right) = ne^{-\frac{\beta^2\sigma^2}{2}}. $$

We conclude by Chernoff's inequality. □

*Proof of Theorem 1.2.* Let $\epsilon > 0$ and define the stopping time

$$ \tau = \inf\{t > 0, \lambda_{\max}(X_t) \ge u \text{ or } \|\langle X \rangle_t\| \ge \sigma^2 + \epsilon\}. $$

Let $X^\tau$ be the process $X$ stopped at time $\tau$, namely $X^\tau = X_{t\wedge\tau}$. We observe that $\langle X^\tau \rangle_t = \langle X \rangle_{t\wedge\tau}$. By definition of $\tau$ and by continuity of $t \to \langle X \rangle_t$, we have

$$ \|\langle X_t^\tau \rangle\| \le \sigma^2 + \epsilon $$

almost surely. Then by Corollary 3.3,

$$ \mathbb{P}(\lambda_{\max}(X_t^\tau) \ge u) \le ne^{-\frac{u^2}{2(\sigma^2+\epsilon)}}. \qquad (13) $$

Now, if there exists $s \le t$ such that $\lambda_{\max}(X_s) \ge u$ and $\|\langle X \rangle_s\| \le \sigma^2$, then $\tau \le s \le t$ and

$$ \|\langle X \rangle_{\tau}\| \le \|\langle X \rangle_s\| \le \sigma^2 < \sigma^2 + \epsilon. $$

Then, by definition of $\tau$, we must have

$$ \lambda_{\max}(X_{\tau}) = \lambda_{\max}(X_{t\wedge\tau}) \ge u. $$

Combining with (13), we get

$$ \mathbb{P}(\exists s \le t, \lambda_{\max}(X_s) \ge u \text{ and } \lambda_{\max}(\langle X \rangle_s) \le \sigma^2) \le \mathbb{P}(\lambda_{\max}(X_t^\tau) \ge u) \\ \le ne^{-\frac{u^2}{2(\sigma^2+\epsilon)^2}}. $$

We obtain Theorem 1.2 by letting $\epsilon$ tend to 0 and $t$ to $\infty$ and using monotone convergence. □

This result can also be extended to non-symmetric and rectangular matrices.

**Theorem 3.4.** Consider $(X_t)$ of the form (3) but taking values in $M_{n_1 \times n_2}$ the space of rectangular matrices of $n_1$ rows and $n_2$ columns. Then,

$$ \mathbb{P}(\exists t > 0, \|X_t\| \ge u \text{ and } \Lambda_t \le \sigma^2) \le (n_1 + n_2)e^{-u^2/(2\sigma^2)} $$

where

$$ \Lambda_t = \max \left\{ \left\| \int_0^t \sum_{i=1}^N H_{i,s} H_{i,s}^* ds \right\| ; \left\| \int_0^t \sum_{i=1}^N H_{i,s}^* H_{i,s} ds \right\| \right\}. \qquad (14) $$

As for Theorem 2.2, the proof consists in applying the symmetric case to $\mathcal{H}(X_t)$ where $\mathcal{H}$ is defined by (9), and using the relation

$$ \lambda_{\max}(\mathcal{H}(A)) = \|\mathcal{H}(A)\| = \|A\|. \qquad (15) $$

The details are left to the reader.
## 4 A Burkholder-Davis-Gundy type inequality for the spectral norm

*Proof of Theorem 1.1.* Let $p \in \mathbb{N}^*$ and $t > 0$. We can write

$$ \mathbb{P}(\exists s \in [0, t], \|X_s\| \ge u) \le \mathbb{P}(\exists s \in [0, t], \|X_s\| \ge u \text{ and } \|\langle X \rangle_s\| \le \sigma^2) + \mathbb{P}(\|\langle X \rangle_t\| > \sigma^2) $$

because $\langle X \rangle_s \preceq \langle X \rangle_t$. Thus, by definition of the spectral norm,

$$ \mathbb{P}(\exists s \in [0, t], \|X_s\| \ge u \text{ and } \|\langle X \rangle_s\| \le \sigma^2) \le \mathbb{P}(\exists s \in [0, t], \lambda_{\max}(X_s) \ge u \text{ and } \|\langle X \rangle_s\| \le \sigma^2) \\ + \mathbb{P}(\exists s \in [0, t], \lambda_{\max}(-X_s) \ge u \text{ and } \|\langle -X \rangle_s\| \le \sigma^2). $$

As $X$ has same quadratic variation of $-X$, applying Theorem 1.2 twice yields

$$ \mathbb{P}(\exists s \in [0, t], \|X_s\| \ge u) \le 2ne^{-\frac{u^2}{2\sigma^2}} + \mathbb{P}(\|\langle X \rangle_t\| > \sigma^2). \quad (16) $$

Now, if we set

$$ \sigma^2 = \frac{u^2}{2(\log n + \lambda u)} $$

with $\lambda > 0$ to be fixed later, we obtain, using (16),

$$ \mathbb{P}\left(\sup_{0 \le s \le t} \|X_s\| \ge u\right) \le e^{-\lambda u} + \mathbb{P}\left(\|\langle X \rangle_t\| > \frac{u^2}{2(\log n + \lambda u)}\right) $$

Observe that if $u \ge 0$,

$$ \frac{u^2}{2} - \lambda \|\langle X \rangle_t\| u - \log n \|\langle X \rangle_t\| \le 0 \iff u \le \lambda \|\langle X \rangle_t\| + \sqrt{\lambda^2 \|\langle X \rangle_t\|^2 + \log n \|\langle X \rangle_t\|} \\ \iff u \le 2\lambda \|\langle X \rangle_t\| + \sqrt{\log n \|\langle X \rangle_t\|}. $$

Hence,

$$ \mathbb{P}\left(\sup_{0 \le s \le t} \|X_s\| \ge u\right) \le e^{-\lambda u} + \mathbb{P}\left(2\lambda \|\langle X \rangle_t\| + \sqrt{\log n \|\langle X \rangle_t\|} > u\right). $$

Multiplying by $pu^{p-1}$ and integrating on $[0, \infty[$

$$ \mathbb{E} \sup_{0 \le s \le t} \|X_s\|^p \le \frac{p!}{\lambda^p} + \mathbb{E} \left( 2\lambda \|\langle X \rangle_t\| + \sqrt{\log n \|\langle X \rangle_t\|} \right)^p. $$

Since $p! \le p^p$ and using Minkowski's inequality

$$ \left( \mathbb{E} \sup_{0 \le s \le t} \|X_s\|^p \right)^{\frac{1}{p}} \le \frac{p}{\lambda} + 2\lambda \left( \mathbb{E} \|\langle X \rangle_t\|^p \right)^{\frac{1}{p}} + \sqrt{\log n} \left( \mathbb{E} \|\langle X \rangle_t\|^2 \right)^{\frac{1}{p}}. $$

Choosing $\lambda = \left( \frac{p}{2\mathbb{E}[\|\langle X \rangle_t\|^p]^{1/p}} \right)^{1/2}$ allows us to obtain

$$ \left( \mathbb{E} \sup_{0 \le s \le t} \|X_s\|^p \right)^{\frac{1}{p}} \le 2\sqrt{2p} \left( \mathbb{E} \|\langle X \rangle_t\|^p \right)^{\frac{1}{2p}} + \sqrt{\log n} \left( \mathbb{E} \|\langle X \rangle_t\|^2 \right)^{\frac{1}{p}}, $$

which is the result.

□

As in Theorem 1.3, the symmetric assumption is not essential. Indeed, using the same notation from section 1, and applying Theorem 1.1 to $\mathcal{H}(X_t)$. We obtain the following theorem.

**Theorem 4.1.** Assuming that for all integers $i$, $(H_{i,s})_s$ are progressively measurable processes taking values in the space of rectangular matrices with entries consisting of $n_1$ rows and $n_2$ columns. Then,

$$ \left( \mathbb{E} \sup_{0 \le s \le t} \|X_s\|^p \right)^{1/p} \le C \sqrt{p} \left( \mathbb{E} \Lambda_t^p \right)^{1/2p} + \sqrt{2 \log(n_1 + n_2)} \left( \mathbb{E} \Lambda_t^{p/2} \right)^{1/p} $$

where $\Lambda_t$ is defined on (14).

The proof is left to the reader, as it follows similar arguments to those of Theorem 2.2.
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