# How the requirement of Itô's lemma is satisfied in this theorem about integration by parts?

I'm reading a theorem (about integration by parts) from page 9 of these notes, i.e.,

Theorem Let $$X$$ and $$Y$$ be continuous semi-martingales such that $$\mathbb E \bigg [ \int_0^t X_s^2 \mathrm d \langle Y \rangle_s + \int_0^t Y_s^2 \mathrm d \langle X \rangle_s\bigg ] <\ \infty \quad \forall t \ge 0. \quad (\star)$$ Then $$X_t Y_t - X_0 Y_0 = \int_0^t X_s \mathrm d Y_s + \int_0^t Y_s \mathrm d X_s + \langle X, Y \rangle_s \quad \text{a.s.} \quad \forall t \ge 0.$$

and its proof, i.e.,

Proof. By using Itô's lemma, we have: \begin{aligned} \left(X_t+Y_t\right)^2-\left(X_0+Y_0\right)^2 &= 2 \int_0^t\left(X_s+Y_s\right) \mathrm d\left(X_s+Y_s\right)+\langle X+Y\rangle_t \\ \left(X_t-Y_t\right)^2-\left(X_0-Y_0\right)^2 &= 2 \int_0^t\left(X_s-Y_s\right) \mathrm d\left(X_s-Y_s\right)+\langle X-Y\rangle_t \end{aligned} Subtracting these two formulas gives: $$4 X_t Y_t-4 X_0 Y_0=4 \int_0^t X_s \mathrm d Y_s+4 \int_0^t Y_s \mathrm d X_s+\underbrace{\left(\langle X+Y\rangle_t-\langle X-Y\rangle_t\right)}_{=4\langle X, Y\rangle_t}$$ which completes the proof.

In the proof, the author applies Itô's lemma on the continuous semi-martingale $$X+Y$$ with function $$f(x) = x^2$$. Itô's lemma requires $$\mathbb E \bigg [ \int_0^t (f'(X_s + Y_s))^2 \mathrm d \langle X+Y \rangle_s \bigg ] = 4\mathbb E \bigg [ \int_0^t (X_s + Y_s)^2 \mathrm d \langle X+Y \rangle_s \bigg ] < \infty \quad \forall t\ge 0.$$

Could you explain how this requirement is satisfied from the assumption $$(\star)$$?

Related definition: Let $$(\Omega, \mathcal F, \mathbb P)$$ be a probability space and $$\mathfrak F = (\mathcal{F}_t, t \ge 0)$$ a filtration.

Let $$M$$ be a continuous square-integrable martingale w.r.t. $$\mathfrak F$$. Then $$M^2$$ is a continuous sub-martingale w.r.t. $$\mathfrak F$$. By Doob's decomposition theorem, there exists a unique continuous increasing process $$\langle M \rangle$$ adapted to $$\mathfrak F$$ such that $$\langle M \rangle_0 = 0$$ a.s. and that $$M^2 - \langle M \rangle$$ is a continuous martingale w.r.t. $$\mathfrak F$$. Then $$\langle M \rangle$$ is called the quadratic variation of $$M$$.

A process $$X$$ is called a continuous semi-martingale w.r.t. $$\mathfrak F$$ is that can be written as $$X_t = M_t+V_t$$ where

• $$M$$ is a continuous square-integrable martingale w.r.t. $$\mathfrak F$$.
• $$V$$ is a continuous process that has bounded variation and is adapted to $$\mathfrak F$$ such that $$V_0 = 0$$ a.s.

Then the quadratic variation of $$X$$ is defined as $$\langle X \rangle := \langle M \rangle.$$

Let $$H$$ be a continuous adapted process such that $$\mathbb E \bigg [ \int_0^t H_s^2 \mathrm d \langle X \rangle_s \bigg ] < \infty \quad \forall t\ge 0.$$

Then the stochastic integral of $$H$$ with respect to $$X$$ is defined as $$(H \cdot X)_t := \int_0^t H_s \mathrm d X_s := \underbrace{\int_0^t H_s \mathrm d M_s}_{\text{Itô's integral}} + \underbrace{\int_0^t H_s \mathrm d V_s}_{\text{Riemann-Stieltjes's integral}}.$$

Itô's lemma. Let $$X$$ be a continuous semi-martingale and $$f:\mathbb R \to \mathbb R$$ twice continuously differentiable such that $$\mathbb E \bigg [ \int_0^t (f'(X_s))^2 \mathrm d \langle X \rangle_s \bigg ] < \infty \quad \forall t\ge 0.$$ Then for all $$t \ge 0$$, $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \mathrm d X_s + \frac{1}{2} \int_0^t f''(X_s) \mathrm d \langle X \rangle_s \quad \text{a.s.}$$

I sent an email to the author and have got the reply. The assumption should be $$\mathbb E \bigg [ \int_0^t X_s^2 \mathrm d \langle Y \rangle_s + \int_0^t Y_s^2 \mathrm d \langle X \rangle_s + \int_0^t X_s^2 \mathrm d \langle X \rangle_s + \int_0^t Y_s^2 \mathrm d \langle Y \rangle_s \bigg ] <\ \infty \quad \forall t \ge 0.$$